1 integer is 7 times another. If the product of the 2 integers is 448, then find the integers. Let the first integer be x and the second integer be y. We have the following two equations: [LIST=1] [*]x = 7y [*]xy = 448 [/LIST] Substitute (1) into (2), we have: (7y)y = 448 7y^2 = 448 Divide each side by 7 y^2 = 64 y = -8, 8 We use 8, since 8*7 = 56, and 56*8 =448. So the answer is [B](x, y) = (8, 56)[/B]

10 times the first of 2 consecutive even integers is 8 times the second. Find the integers. Let the first integer be x. Let the second integer be y. We're given: [LIST=1] [*]10x = 8y [*]We also know a consecutive even integer means we add 2 to x to get y. y = x + 2 [/LIST] Substitute (1) into (2): 10x = 8(x + 2) Multiply through: 10x = 8x + 16 To solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=10x%3D8x%2B16&pl=Solve']we type this equation into our search engine[/URL] and we get: [B]x = 8[/B] Since y = x + 2, we plug in x = 8 to get: y = 8 + 2 [B]y = 10 [/B] Now, let's check our work. Does x = 8 and y = 10 make equation 1 hold? 10(8) ? 8(10) 80 = 80 <-- Yes!

2 baseball players hit 60 home runs combined last season. The first player hit 3 more home runs than twice a number of home runs the second player hit. how many home runs did each player hit? Declare variables: Let the first players home runs be a Let the second players home runs be b We're given two equations: [LIST=1] [*]a = 2b + 3 [*]a + b = 60 [/LIST] To solve this system of equations, we substitute equation (1) into equation (2) for a: 2b + 3 + b = 60 Using our math engine, we [URL='https://www.mathcelebrity.com/1unk.php?num=2b%2B3%2Bb%3D60&pl=Solve']type this equation[/URL] in and get: b = [B]19 [/B] To solve for a, we substitute b = 19 into equation (1): a = 2(19) + 3 a = 38 + 3 a = [B]41[/B]

2 numbers that add up makes 5 but multiplied makes -36 Let the first number be x and the second number be y. We're given two equations: [LIST=1] [*]x + y = 5 [*]xy = -36 [/LIST] Rearrange equation (1) by subtracting y from each side: [LIST=1] [*]x = 5 - y [*]xy = -36 [/LIST] Substitute equation (1) for x into equation (2): (5 - y)y = -36 5y - y^2 = -36 Add 36 to each side: -y^2 + 5y + 36 = 0 We have a quadratic equation. To solve this, we [URL='https://www.mathcelebrity.com/quadratic.php?num=-y%5E2%2B5y%2B36%3D0&pl=Solve+Quadratic+Equation&hintnum=0']type it in our search engine and solve[/URL] to get: y = ([B]-4, 9[/B]) We check our work for each equation: [LIST=1] [*]-4 + 9 = -5 [*]-4(9) = -36 [/LIST] They both check out

2 numbers that are equal have a sum of 60 Let's choose 2 arbitrary variables for the 2 numbers x, y Were given 2 equations: [LIST=1] [*]x = y <-- Because we have the phrase [I]that are equal[/I] [*]x + y = 60 [/LIST] Because x = y in equation (1), we can substitute equation (1) into equation (2) for x: y + y = 60 Add like terms to get: 2y = 60 Divide each side by 2: 2y/2 = 60/2 Cancel the 2's and we get: y = [B]30 [/B] Since x = y, x = y = 30 x = [B]30[/B]

2 times as many dimes as quarters and they have a combined value of 180 cents, how many of each coin does he have? Let d be the number of dimes. Let q be the number of quarters. We're given two equations: [LIST=1] [*]d = 2q [*]0.1d + 0.25q = 180 [/LIST] Substitute (1) into (2): 0.1(2q) + 0.25q = 180 0.2q + 0.25q = 180 [URL='https://www.mathcelebrity.com/1unk.php?num=0.2q%2B0.25q%3D180&pl=Solve']Typing this equation into the search engine[/URL], we get: [B]q = 400[/B] Now substitute q = 400 into equation 1: d = 2(400) [B]d = 800[/B]

2 consecutive even integers such that the smaller added to 5 times the larger gives a sum of 70. Let the first, smaller integer be x. And the second larger integer be y. Since they are both even, we have: [LIST=1] [*]x = y - 2 <-- Since they're consecutive even integers [*]x + 5y = 70 <-- Smaller added to 5 times the larger gives a sum of 70 [/LIST] Substitute (1) into (2): (y - 2) + 5y = 70 Group like terms: (1 + 5)y - 2 = 70 6y - 2 = 70 [URL='https://www.mathcelebrity.com/1unk.php?num=6y-2%3D70&pl=Solve']Typing 6y - 2 = 70 into our search engine[/URL], we get: [B]y = 12 <-- Larger integer[/B] Plugging this into Equation (1) we get: x = 12 - 2 [B]x = 10 <-- Smaller Integer[/B] So (x, y) = (10, 12)

46 people showed up to the party. There were 8 less men than women present. How many men were there? Let the number of men be m. Let the number of women be w. We're given two equations: [LIST=1] [*]m = w - 8 [I](8 less men than women)[/I] [*]m + w = 46 [I](46 showed up to the party)[/I] [/LIST] Substitute equation (1) into equation (2) for m: w - 8 + w = 46 To solve for w in this equation, we [URL='https://www.mathcelebrity.com/1unk.php?num=w-8%2Bw%3D46&pl=Solve']type in the equation into our search engine [/URL]and we get: w = 27 To solve for men (m), we substitute w = 27 into equation (1): m = 27 - 8 m = [B]19[/B]

5 books and 5 bags cost $175. What is the cost of 2 books and 2 bags Let the cost of each book be b and the cost of each bag be c. We're given 5b + 5c = 175 We can factor this as: 5(b + c) = 175 Divide each side of the equation by 5, we get: (b + c) = 35 The problem asks for 2b + 2c Factor out 2: 2(b + c) we know from above that (b + c) = 35, so we substitute: 2(35) [B]70[/B]

508 people are there, the daily price is $1.25 for kids and $2.00 for adults. The receipts totaled $885.50. How many kids and how many adults were there? Assumptions: [LIST] [*]Let the number of adults be a [*]Let the number of kids be k [/LIST] Given with assumptions: [LIST=1] [*]a + k = 508 [*]2a + 1.25k = 885.50 (since cost = price * quantity) [/LIST] Rearrange equation (1) by subtracting c from each side to isolate a: [LIST=1] [*]a = 508 - k [*]2a + 1.25k = 885.50 [/LIST] Substitute equation (1) into equation (2): 2(508 - k) + 1.25k = 885.50 Multiply through: 1016 - 2k + 1.25k = 885.50 1016 - 0.75k = 885.50 To solve for k, we [URL='https://www.mathcelebrity.com/1unk.php?num=1016-0.75k%3D885.50&pl=Solve']type this equation into our search engine[/URL] and we get: k = [B]174[/B] Now, to solve for a, we substitute k = 174 into equation 1 above: a = 508 - 174 a = [B]334[/B]

Substitute s for x 7(s) = 7s

A 124-inch length of ribbon is to be cut into three pieces. The longest piece is to be 36 inches longer than the shortest piece, and the third piece is to be half the length of the longest piece. Find the length of each piece of ribbon. [LIST] [*]Let the longest piece be l. [*]The shortest piece is s = l - 36 [*]The third medium piece m = 0.5l [/LIST] We know s + m + l = 124. Now substitute for s and m (l - 36) + 0.5l + l = 124 Combine like terms: 2.5l - 36 = 124 Type [URL='http://www.mathcelebrity.com/1unk.php?num=2.5l-36%3D124&pl=Solve']2.5l - 36 = 124 into our search engine[/URL], we get l = [B]64[/B] Shortest piece s = 64 - 36 = [B]28[/B] Medium piece m = 0.5(64) = [B]32[/B]

A 98-inch piece of wire must be cut into two pieces. One piece must be 10 inches shorter than the other. How long should the pieces be? The key phrase in this problem is [B]two pieces[/B]. Declare Variables: [LIST] [*]Let the short piece length be s [*]Let the long piece length be l [/LIST] We're given the following [LIST=1] [*]s = l - 10 [*]s + l = 98 (Because the two pieces add up to 98) [/LIST] Substitute equation (1) into equation (2) for s: l - 10+ l = 98 Group like terms: 2l - 10 = 98 Solve for [I]l[/I] in the equation 2l - 10 = 98 [SIZE=5][B]Step 1: Group constants:[/B][/SIZE] We need to group our constants -10 and 98. To do that, we add 10 to both sides 2l - 10 + 10 = 98 + 10 [SIZE=5][B]Step 2: Cancel 10 on the left side:[/B][/SIZE] 2l = 108 [SIZE=5][B]Step 3: Divide each side of the equation by 2[/B][/SIZE] 2l/2 = 108/2 l = [B]54[/B] To solve for s, we substitute l = 54 into equation (1): s = 54 - 10 s = [B]44[/B] Check our work: The shorter piece is 10 inches shorter than the longer piece since 54 - 44 = 10 Second check: Do both pieces add up to 98 54 + 44 ? 98 98 = 98

A bag contains 120 marbles. Some are red and the rest are black. There are 19 red marbles for every black marble. How many red marbles are in the bag? Let the red marbles be r Let the black marbles be b. A 19 to 1 red to black is written as: r = 19b We're also given: b + r = 120 Substitute r = 19b into this equation and we get: b + 19b = 120 Combine like terms: 20b = 120 To solve this equation, [URL='https://www.mathcelebrity.com/1unk.php?num=20b%3D120&pl=Solve']we type it in our search engine [/URL]and we get: b = 6 Since r = 19b, we substitute b = 6 into this equation to solve for r: r = 19(6) r = [B]114[/B]

A boy is 10 years older than his brother. In 4 years he will be twice as old as his brother. Find the present age of each? Let the boy's age be b and his brother's age be c. We're given two equations: [LIST=1] [*]b = c + 10 [*]b + 4 = 2(c + 4) [/LIST] Substitute equation (1) into equation (2): (c + 10) + 4 = 2(c + 4) Simplify by multiplying the right side through and grouping like terms: c + 14 = 2c + 8 [URL='https://www.mathcelebrity.com/1unk.php?num=c%2B14%3D2c%2B8&pl=Solve']Type this equation into our search engine[/URL] and we get: c = [B]6[/B] Now plug c = 6 into equation (1): b = 6 + 10 b = [B]16[/B]

a boy purchased a party-length sandwich 57 inches long. he wants to cut it into three pieces so that the middle piece is 6inches longer than the shortest piece and the shortest piece is 9 inches shorter than the longest price. how long should the three pieces be? Let the longest piece be l. The middle piece be m. And the short piece be s. We have 2 equations in terms of the shortest piece: [LIST=1] [*]l = s + 9 (Since the shortest piece is 9 inches shorter, this means the longest piece is 9 inches longer) [*]m = s + 6 [*]s + m + l = 57 [/LIST] We substitute equations (1) and (2) into equation (3): s + (s + 6) + (s + 9) = 57 Group like terms: (1 + 1 + 1)s + (6 + 9) = 57 3s + 15 = 57 To solve for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=3s%2B15%3D57&pl=Solve']type this equation into our search engine[/URL] and we get: s = [B]14 [/B] [U]Plug s = 14 into equation 2 to solve for m:[/U] m = 14 + 6 m = [B]20 [/B] [U]Plug s = 14 into equation 1 to solve for l:[/U] l = 14 + 9 l = [B]23 [/B] Check our work for equation 3: 14 + 20 + 23 ? 57 57 = 57 <-- checks out [B][/B]

A business owner spent $4000 for a computer and software. For bookkeeping purposes, he needs to post the price of the computer and software separately. The computer costs 4 times as much as the software. What is the cost of the software? Let c be the cost of the computer and s be the cost of the software. We have two equations: [LIST=1] [*]c + s = 4000 [*]c = 4s [/LIST] Substitute (2) into (1) (4s) + s = 4000 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=4s%2Bs%3D4000&pl=Solve']equation solver[/URL], we get [B]s = 800[/B]. Substitute this into Equation (2), we get: c = 4(800) [B]c = 3,200[/B]

A cashier has 44 bills, all of which are $10 or $20 bills. The total value of the money is $730. How many of each type of bill does the cashier have? Let a be the amount of $10 bills and b be the amount of $20 bills. We're given two equations: [LIST=1] [*]a + b = 44 [*]10a + 20b = 730 [/LIST] We rearrange equation 1 in terms of a. We subtract b from each side and we get: [LIST=1] [*]a = 44 - b [*]10a + 20b = 730 [/LIST] Now we substitute equation (1) for a into equation (2): 10(44 - b) + 20b = 730 Multiply through to remove the parentheses: 440 - 10b + 20b = 730 Group like terms: 440 + 10b = 730 Now, to solve for b, we [URL='https://www.mathcelebrity.com/1unk.php?num=440%2B10b%3D730&pl=Solve']type this equation into our search engine[/URL] and we get: b = [B]29 [/B] To get a, we take b = 29 and substitute it into equation (1) above: a = 44 - 29 a = [B]15 [/B] So we have [B]15 ten-dollar bills[/B] and [B]29 twenty-dollar bills[/B]

A class has 35 boys and girls. There are 7 more girls than boys. Find the number of girls and boys in the class Let the number of boys be b and the number of girls be g. We're given two equations: [LIST=1] [*]b + g = 35 [*]g = b + 7 (7 more girls means we add 7 to the boys) [/LIST] To solve for b, we substitute equation (2) into equation (1) for g: b + b + 7 = 35 To solve for b, we type this equation into our search engine and we get: b = [B]14[/B] Now, to solve for g, we plug b = 14 into equation (2) above: g = 14 + 7 g = [B]21[/B]

A cookie recipe uses 10 times as much flour as sugar. If the total amount of these ingredients is 8 1/4 cups, how much flour and how much sugar would it be? Let f be the number of cups of sugar. And let f be the number of cups of flour. We're given two equations: [LIST=1] [*]f = 10s [*]s + f = 8 & 1/4 [/LIST] Substitute (1) into (2): s + 10s = 8 & 1/4 11fs= 33/4 <-- 8 & 1/4 = 33/4 Cross multiply: 44s = 33 Divide each side by 44: s= 33/44 Divide top and bottom by 11 and we get s [B]= 3/4 or 0.75[/B] Now substitute this into (1): f = 10(33/44) [B]f = 330/44 or 7 & 22/44 or 7.5[/B]

A dog and a cat together cost $100. If the price of the dogs $90 more than the cat, what is the cost of the cat? Set up givens and equations [LIST] [*]Let the cost of the dog be d [*]Let the cost of the cat be c [/LIST] We're given 2 equations: [LIST=1] [*]c + d = 100 [*]d = c + 90 [/LIST] Substitute equation (2) into equation (1) for d c + c + 90 = 100 Using our [URL='https://www.mathcelebrity.com/1unk.php?num=c%2Bc%2B90%3D100&pl=Solve']math engine[/URL], we see that: c = [B]5 [/B] Substitute c = 5 into equation (2) above: d = 5 + 90 d = [B]95[/B]

A first number plus twice a second number is 10. Twice the first number plus the second totals 35. Find the numbers. [U]The phrase [I]a number[/I] means an arbitrary variable[/U] A first number is written as x A second number is written as y [U]Twice a second number means we multiply y by 2:[/U] 2y [U]A first number plus twice a second number:[/U] x + 2y [U]A first number plus twice a second number is 10 means we set x + 2y equal to 10:[/U] x + 2y = 10 [U]Twice the first number means we multiply x by 2:[/U] 2x [U]Twice the first number plus the second:[/U] 2x + y [U]Twice the first number plus the second totals 35 means we set 2x + y equal to 35:[/U] 2x + y = 35 Therefore, we have a system of two equations: [LIST=1] [*]x + 2y = 10 [*]2x + y = 35 [/LIST] Since we have an easy multiple of 2 for the x variable, we can solve this by multiply the first equation by -2: [LIST=1] [*]-2x - 4y = -20 [*]2x + y = 35 [/LIST] Because the x variables are opposites, we can add both equations together: (-2 + 2)x + (-4 + 1)y = -20 + 35 The x terms cancel, so we have: -3y = 15 To solve this equation for y, we [URL='https://www.mathcelebrity.com/1unk.php?num=-3y%3D15&pl=Solve']type it in our search engine[/URL] and we get: y = [B]-5 [/B] Now we substitute this y = -5 into equation 2: 2x - 5 = 35 To solve this equation for x, we[URL='https://www.mathcelebrity.com/1unk.php?num=2x-5%3D35&pl=Solve'] type it in our search engine[/URL] and we get: x = [B]20[/B]

A first number plus twice a second number is 14. Twice the first number plus the second totals 40. Find the numbers. [B][U]Givens and assumptions:[/U][/B] [LIST] [*]Let the first number be x. [*]Let the second number be y. [*]Twice means multiply by 2 [*]The phrases [I]is[/I] and [I]totals[/I] mean equal to [/LIST] We're given two equations: [LIST=1] [*]x + 2y = 14 [*]2x + y = 40 [/LIST] To solve this system, we can take a shortcut, and multiply the top equation by -2 to get our new system: [LIST=1] [*]-2x - 4y = -28 [*]2x + y = 40 [/LIST] Now add both equations together (-2 _ 2)x (-4 + 1)y = -28 + 40 -3y = 12 To solve this equation, we [URL='https://www.mathcelebrity.com/1unk.php?num=-3y%3D12&pl=Solve']type it in our search engine[/URL] and we get: y = [B]-4 [/B] We substitute this back into equation 1 for y = -4: x + 2(-4) = 14 x - 8 = 14 To solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=x-8%3D14&pl=Solve']type it in our search engine[/URL] and we get: x = [B]22[/B]

A first number plus twice a second number is 7. Twice the first number plus the second totals 23. Find the numbers Let the first number be a and the second number be b. We have: [LIST=1] [*]a + 2b = 7 [*]2a + b = 23 [/LIST] Rearrange (1) into (3) (3) a = 7 - 2b Substitute (3) into (2): 2(7 - 2b) + b = 23 Multiply through: 14 - 4b + b = 23 Combine like terms: 14 - 3b = 23 Subtract 14 from each side: -3b = 9 Divide each side by -3 [B]b = -3[/B] Substitute this into (3) a = 7 - 2b a = 7 - 2(-3) a = 7 + 6 [B]a = 13[/B] [B](a, b) = (13, -3)[/B]

A food truck sells salads for $6.50 each and drinks for $2.00 each. The food trucks revenue from selling a total of 209 salads and drinks in one day was $836.50. How many salads were sold that day? Let the number of drinks be d. Let the number of salads be s. We're given two equations: [LIST=1] [*]2d + 6.50s = 836.50 [*]d + s = 209 [/LIST] We can use substitution to solve this system of equations quickly. The question asks for the number of salads (s). Therefore, we want all expressions in terms of s. Rearrange Equation 2 by subtracting s from both sides: d + s - s = 209 - s Cancel the s's, we get: d = 209 - s So we have the following system of equations: [LIST=1] [*]2d + 6.50s = 836.50 [*]d = 209 - s [/LIST] Substitute equation (2) into equation (1) for d: 2(209 - s) + 6.50s = 836.50 Multiply through to remove the parentheses: 418 - 2s + 6.50s = 836.50 To solve this equation for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=418-2s%2B6.50s%3D836.50&pl=Solve']type it into our search engine and we get[/URL]: s = [B]93[/B]

A fraction has a value of 3/4. If 7 is added to the numerator, the resulting fraction is equal to the reciprocal of the original fraction. Find the original fraction. Let the fraction be x/y. We're given two equations: [LIST=1] [*]x/y = 3/4 [*](x + 7)/y = 4/3. [I](The reciprocal of 3/4 is found by 1/(3/4)[/I] [/LIST] Cross multiply equation 1 and equation 2: [LIST=1] [*]4x = 3y [*]3(x + 7) = 4y [/LIST] Simplifying, we get: [LIST=1] [*]4x = 3y [*]3x + 21 = 4y [/LIST] If we divide equation 1 by 4, we get: [LIST=1] [*]x = 3y/4 [*]3x + 21 = 4y [/LIST] Substitute equation (1) into equation (2) for x: 3(3y/4) + 21 = 4y 9y/4 + 21 = 4y Multiply the equation by 4 on both sides to eliminate the denominator: 9y + 84 = 16y To solve this equation for y, we type it in our math engine and we get: y = [B]12 [/B] We then substitute y = 12 into equation 1 above: x = 3 * 12/4 x = 36/4 x = [B]9 [/B] So our original fraction x/y = [B]9/12[/B]

A garden table and a bench cost $977 combined. The garden table costs $77 more than the bench. What is the cost of the bench? Let the garden table cost be g and the bench cost be b. We're given [LIST=1] [*]b + g = 977 [*]g = b + 77 <-- The phrase [I]more than[/I] means we add [/LIST] Substitute (2) into (1): b + (b + 77) = 977 Combine like terms: 2b + 77 = 977 [URL='https://www.mathcelebrity.com/1unk.php?num=2b%2B77%3D977&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]b = $450[/B]

A high school graduating class is made up of 440 students. There are 168 more girls than boys. How many boys are in the class? Let b be the number of boys and g be the number of girls. We're given 2 equations: [LIST=1] [*]b + g = 440 [*]g = b + 168 [/LIST] Substitute (2) into (1) b + (b + 168) = 440 Combine like terms: 2b + 168 = 440 [URL='https://www.mathcelebrity.com/1unk.php?num=2b%2B168%3D440&pl=Solve']Type this equation into the search engine[/URL], and we get: [B]b = 136[/B]

A home is to be built on a rectangular plot of land with a perimeter of 800 feet. If the length is 20 feet less than 3 times the width, what are the dimensions of the rectangular plot? [U]Set up equations:[/U] (1) 2l + 2w = 800 (2) l = 3w - 20 [U]Substitute (2) into (1)[/U] 2(3w - 20) + 2w = 800 6w - 40 + 2w = 800 [U]Group the w terms[/U] 8w - 40 = 800 [U]Add 40 to each side[/U] 8w = 840 [U]Divide each side by 8[/U] [B]w = 105 [/B] [U]Substitute w = 105 into (2)[/U] l = 3(105) - 20 l = 315 - 20 [B]l = 295[/B]

a horse and a saddle cost $5,000. if the horse cost 4 times as much as the saddle, what was the cost of each? Let the cost of the horse be h, and the cost of the saddle be s. We're given: [LIST=1] [*]h + s = 5000 [*]h = 4s [/LIST] Substitute equation (2) into equation (1): 4s + s = 5000 [URL='https://www.mathcelebrity.com/1unk.php?num=4s%2Bs%3D5000&pl=Solve']Type this equation into the search engine[/URL], we get: [B]s = 1,000[/B] Substitute s = 1000 into equation (2): h = 4(1000) [B]h = 4,000[/B]

A house costs 3.5 times as much as the lot. Together they sold for $135,000. Find the cost of each. Let the house cost be h, and the lot cost be l. We have the following equations: [LIST=1] [*]h = 3.5l [*]h + l = 135,000 [/LIST] Substitute (1) into (2) 3.5l + l = 135,000 Combine like terms: 4.5l = 135,000 Divide each side by 4.5 to isolate l [B]l = 30,000[/B] Substitute this back into equation (1) h = 3.5(30,000) [B]h = 105,000[/B]

a is 2 years older than b who is twice as old as c. if the total ages of a,b and c is 42, then how old is b We're given 3 equations: [LIST=1] [*]a = b + 2 [*]b = 2c [*]a + b + c = 42 [/LIST] Substituting equation (2) into equation (1), we have: a = 2c + 2 Since b = 2c, we substitute both of these into equation (3) to get: 2c + 2 + 2c + c = 42 To solve for c, we [URL='https://www.mathcelebrity.com/1unk.php?num=2c%2B2%2B2c%2Bc%3D42&pl=Solve']type this equation into our math engine[/URL] and we get: c = 8 Now take c = 8 and substitute it into equation (2) above: b = 2(8) b = [B]16[/B]

a large fry has 120 more calories than a small. 5 large fries is the same amount of calories as 7 small. How many calories does each size fry have? Let the number of calories in large fries be l. Let the number of calories in small fries be s. We're given two equations: [LIST=1] [*]l = s + 120 [*]5l = 7s [/LIST] Substitute equation (1) into equation (2): 5(s + 120) = 7s [URL='https://www.mathcelebrity.com/1unk.php?num=5%28s%2B120%29%3D7s&pl=Solve']Type this equation into the search engine[/URL] and we get: s = [B]300[/B] Substitute s = 300 into equation (1): l = 300 + 120 l = [B]420[/B]

A local shop sold 499 hamburgers and cheese burgers. There were 51 fewer cheese burgers sold. How many hamburgers were sold? Let h = number of hamburgers sold and c be the number of cheeseburgers sold. We have two equations: (1) c = h - 51 (2) c + h = 499 Substitute (1) into (2) h - 51 + h = 499 Combine like terms 2h - 51 = 499 Add 51 to both sides 2h = 550 Divide each side by 2 to isolate h [B]h = 275[/B]

A man is 5 years older than his wife, and the daughter age is half of the mother, and if you add their ages is equal 100 Let the man's age be m. Let the wife's age be w. Let the daughter's age be d. We're given: [LIST=1] [*]m = w + 5 [*]d = 0.5m [*]d + m + w = 100 [/LIST] Rearrange equation 1 in terms of w my subtracting 5 from each side: [LIST=1] [*]w = m - 5 [*]d = 0.5m [*]d + m + w = 100 [/LIST] Substitute equation (1) and equation (2) into equation (3) 0.5m + m + m - 5 = 100 We [URL='https://www.mathcelebrity.com/1unk.php?num=0.5m%2Bm%2Bm-5%3D100&pl=Solve']type this equation into our search engine[/URL] to solve for m and we get: m = [B]42 [/B] Now, substitute m = 42 into equation 2 to solve for d: d = 0.5(42) d = [B]21 [/B] Now substitute m = 42 into equation 1 to solve for w: w = 42 - 5 w = [B]37 [/B] To summarize our ages: [LIST] [*]Man (m) = 42 years old [*]Daughter (d) = 21 years old [*]Wife (w) = 37 years old [/LIST]

A man is four times as old as his son. In five years time he will be three times as old. Find their present ages. Let the man's age be m, and the son's age be s. We have: [LIST=1] [*]m = 4s [*]m + 5 = 3(s + 5) [/LIST] Substitute (1) into (2) 4s + 5 = 3s + 15 Use our [URL='http://www.mathcelebrity.com/1unk.php?num=4s%2B5%3D3s%2B15&pl=Solve']equation calculator[/URL], and we get [B]s = 10[/B]. m = 4(10) [B]m = 40[/B]

A man purchased 20 tickets for a total of $225. The tickets cost $15 for adults and $10 for children. What was the cost of each ticket? Declare variables: [LIST] [*]Let a be the number of adult's tickets [*]Let c be the number of children's tickets [/LIST] Cost = Price * Quantity We're given two equations: [LIST=1] [*]a + c = 20 [*]15a + 10c = 225 [/LIST] Rearrange equation (1) in terms of a: [LIST=1] [*]a = 20 - c [*]15a + 10c = 225 [/LIST] Now that I have equation (1) in terms of a, we can substitute into equation (2) for a: 15(20 - c) + 10c = 225 Solve for [I]c[/I] in the equation 15(20 - c) + 10c = 225 We first need to simplify the expression removing parentheses Simplify 15(20 - c): Distribute the 15 to each term in (20-c) 15 * 20 = (15 * 20) = 300 15 * -c = (15 * -1)c = -15c Our Total expanded term is 300-15c Our updated term to work with is 300 - 15c + 10c = 225 We first need to simplify the expression removing parentheses Our updated term to work with is 300 - 15c + 10c = 225 [SIZE=5][B]Step 1: Group the c terms on the left hand side:[/B][/SIZE] (-15 + 10)c = -5c [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] -5c + 300 = + 225 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 300 and 225. To do that, we subtract 300 from both sides -5c + 300 - 300 = 225 - 300 [SIZE=5][B]Step 4: Cancel 300 on the left side:[/B][/SIZE] -5c = -75 [SIZE=5][B]Step 5: Divide each side of the equation by -5[/B][/SIZE] -5c/-5 = -75/-5 c = [B]15[/B] Recall from equation (1) that a = 20 - c. So we substitute c = 15 into this equation to solve for a: a = 20 - 15 a = [B]5[/B]

A math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 points. How many problems of each point value are on the test? Let's call the 5 point questions m for multiple choice. Let's call the 2 point questions t for true-false. We have two equations: [LIST=1] [*]m + t = 38 [*]5m + 2t = 100 [/LIST] Rearrange (1) to solve for m - subtract t from each side: 3. m = 38 - t Now, substitute (3) into (2) 5(38 - t) + 2t = 100 190 - 5t + 2t = 100 Combine like terms: 190 - 3t = 100 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=190-3t%3D100&pl=Solve']equation solver[/URL], we get [B]t = 30[/B]. Plugging t = 30 into (1), we get: 30 + t = 38 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=m%2B30%3D38&pl=Solve']equation solver[/URL] again, we get [B]m = 8[/B]. Check our work for (1) 8 + 30 = 38 <-- Check Check our work for (2) 5(8) + 2(30) ? 100 40 + 60 ? 100 100 = 100 <-- Check You could also use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+38&term2=5m+%2B+2t+%3D+100&pl=Cramers+Method']simultaneous equations calculator[/URL]

A motorboat travels 408 kilometers in 8 hours going upstream and 546 kilometers in 6 hours going downstream. What is the rate of the boat in still water and what is the rate of the current? [U]Assumptions:[/U] [LIST] [*]B = the speed of the boat in still water. [*]S = the speed of the stream [/LIST] Relative to the bank, the speeds are: [LIST] [*]Upstream is B - S. [*]Downstream is B + S. [/LIST] [U]Use the Distance equation: Rate * Time = Distance[/U] [LIST] [*]Upstream: (B-S)6 = 258 [*]Downstream: (B+S)6 = 330 [/LIST] Simplify first by dividing each equation by 6: [LIST] [*]B - S = 43 [*]B + S = 55 [/LIST] Solve this system of equations by elimination. Add the two equations together: (B + B) + (S - S) = 43 + 55 Cancelling the S's, we get: 2B = 98 Divide each side by 2: [B]B = 49 mi/hr[/B] Substitute this into either equation and solve for S. B + S = 55 49 + S = 55 To solve this, we [URL='https://www.mathcelebrity.com/1unk.php?num=49%2Bs%3D55&pl=Solve']type it in our search engine[/URL] and we get: S = [B]6 mi/hr[/B]

A movie theater has a seating capacity of 143. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. There are half as many adults as there are children. If the total ticket sales was $ 1030, How many children, students, and adults attended? Let c be the number of children's tickets, s be the number of student's tickets, and a be the number of adult's tickets. We have 3 equations: [LIST=1] [*]a + c + s = 143 [*]a = 0.5c [*]12a + 5c + 7s =1030 [/LIST] Substitute (2) into (1) 0.5c + c + s = 143 1.5c + s = 143 Subtract 1.5c from each side 4. s = 143 - 1.5c Now, take (4) and (2), and plug it into (3) 12(0.5c) + 5c + 7(143 - 1.5c) = 1030 6c + 5c + 1001 - 10.5c = 1030 Combine like terms: 0.5c + 1001 = 1030 Use our [URL='http://www.mathcelebrity.com/1unk.php?num=0.5c%2B1001%3D1030&pl=Solve']equation calculator[/URL] to get [B]c = 58[/B]. Plug this back into (2) a = 0.5(58) [B]a = 29 [/B] Now take the a and c values, and plug it into (1) 29 + 58 + s = 143 s + 87 = 143 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=s%2B87%3D143&pl=Solve']equation calculator[/URL] again, we get [B]s = 56[/B]. To summarize, we have: [LIST] [*]29 adults [*]58 children [*]56 students [/LIST]

A new car worth $30,000 is depreciating in value by $3,000 per year. After how many years will the cars value be $9,000 Step 1, the question asks for Book Value. Let y be the number of years since purchase. We setup an equation B(y) which is the Book Value at time y. B(y) = Sale Price - Depreciation Amount * y We're given Sale price = $30,000, depreciation amount = 3,000, and B(y) = 9000 30000 - 3000y = 9000 To solve for y, we [URL='https://www.mathcelebrity.com/1unk.php?num=30000-3000y%3D9000&pl=Solve']type this in our math engine[/URL] and we get: y = [B]7 [/B] To check our work, substitute y = 7 into B(y) B(7) = 30000 - 3000(7) B(7) = 30000 - 21000 B(7) = 9000 [MEDIA=youtube]oCpBBS7fRYs[/MEDIA]

A parking meter contains 27.05 in quarters and dimes. All together there are 146 coins. How many of each coin are there? Let d = the number of dimes and q = the number of quarters. We have two equations: (1) d + q = 146 (2) 0.1d + 0.25q = 27.05 Rearrange (1) into (3) solving for d (3) d = 146 - q Substitute (3) into (2) 0.1(146 - q) + 0.25q = 27.05 14.6 - 0.1q + 0.25q = 27.05 Combine q's 0.15q + 14.6 = 27.05 Subtract 14.6 from each side 0.15q = 12.45 Divide each side by 0.15 [B]q = 83[/B] Plugging that into (3), we have: d = 146 - 83 [B]d = 63[/B]

A person will devote 31 years to be sleeping and watching tv. The number of years sleeping will exceed the number of years watching tv by 19. How many years will the person spend on each of these activities Let s be sleeping years and t be tv years, we have two equations: [LIST=1] [*]s + t = 31 [*]s = t + 19 [/LIST] Substitute (2) into (1) (t + 19) + t = 31 Combine like terms: 2t + 19 = 31 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2t%2B19%3D31&pl=Solve']equation solver[/URL], we get [B]t = 6[/B]. Using equation (2), we have s = 6 + 19 s = [B]25[/B]

A pile of coins, consisting of quarters and half dollars, is worth 11.75. If there are 2 more quarters than half dollars, how many of each are there? Let h be the number of half-dollars and q be the number of quarters. Set up two equations: (1) q = h + 2 (2) 0.25q + 0.5h = 11.75 [U]Substitute (1) into (2)[/U] 0.25(h + 2) + 0.5h = 11.75 0.25h + 0.5 + 0.5h = 11.75 [U]Group h terms[/U] 0.75h + 0.5 = 11.75 [U]Subtract 0.5 from each side[/U] 0.75h = 11.25 [U]Divide each side by h[/U] [B]h = 15[/B] [U]Substitute h = 15 into (1)[/U] q = 15 + 2 [B]q = 17[/B]

A private jet flies the same distance in 4 hours that a commercial jet flies in 2 hours. If the speed of the commercial jet was 154 mph less than 3 times the speed of the private jet, find the speed of each jet. Let p = private jet speed and c = commercial jet speed. We have two equations: (1) c = 3p - 154 (2) 4p =2c Plug (1) into (2): 4p = 2(3p - 154) 4p = 6p - 308 Subtract 4p from each side: 2p - 308 = 0 Add 308 to each side: 2p = 308 Divide each side by 2: [B]p = 154[/B] Substitute this into (1) c = 3(154) - 154 c = 462 - 154 [B]c = 308[/B]

a rectangle has an area of 238 cm 2 and a perimeter of 62 cm. What are its dimensions? We know the rectangle has the following formulas: Area = lw Perimeter = 2l + 2w Given an area of 238 and a perimeter of 62, we have: [LIST=1] [*]lw = 238 [*]2(l + w) = 62 [/LIST] Divide each side of (1) by w: l = 238/w Substitute this into (2): 2(238/w + w) = 62 Divide each side by 2: 238/w + w = 31 Multiply each side by w: 238w/w + w^2 = 31w Simplify: 238 + w^2 = 31w Subtract 31w from each side: w^2 - 31w + 238 = 0 We have a quadratic. So we run this through our [URL='https://www.mathcelebrity.com/quadratic.php?num=w%5E2-31w%2B238%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL] and we get: w = (14, 17) We take the lower amount as our width and the higher amount as our length: [B]w = 14 l = 17 [/B] Check our work for Area: 14(17) = 238 <-- Check Check our work for Perimeter: 2(17 + 14) ? 62 2(31) ? 62 62 = 62 <-- Check

A rectangular field is to be enclosed with 1120 feet of fencing. If the length of the field is 40 feet longer than the width, then how wide is the field? We're given: [LIST=1] [*]l = w + 40 [/LIST] And we know the perimeter of a rectangle is: P = 2l + 2w Substitute (1) into this formula as well as the given perimeter of 1120: 2(w + 40) + 2w = 1120 Multiply through and simplify: 2w + 80 + 2w = 1120 Group like terms: 4w + 80 = 1120 [URL='https://www.mathcelebrity.com/1unk.php?num=4w%2B80%3D1120&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 260[/B]

A rectangular football pitch has its length equal to twice its width and a perimeter of 360m. Find its length and width. The area of a rectangle (A) is: A = lw --> where l is the length and w is the width We're given l = 2w, so we substitute this into the Area equation: A = (2w)w A = 2w^2 We're given the area of the pitch is 360, so we set: 2w^2 = 360 We [URL='https://www.mathcelebrity.com/1unk.php?num=2w%5E2%3D360&pl=Solve']type this equation into our search engine[/URL], follow the links, and get: w = [B]6*sqrt(5) [/B] Now we take this, and substitute it into this equation: 6*sqrt(5)l = 360 Dividing each side by 6*sqrt(5), we get: l = [B]60/sqrt(5)[/B]

A rectangular garden is 5 ft longer than it is wide. Its area is 546 ft2. What are its dimensions? [LIST=1] [*]Area of a rectangle is lw. lw = 546ft^2 [*]We know that l = w + 5. [/LIST] Substitute (2) into (1) (w + 5)w = 546 w^2 + 5w = 546 Subtract 546 from each side w^2 + 5w - 546 = 0 Using the positive root in our [URL='http://www.mathcelebrity.com/quadratic.php?num=w%5E2%2B5w-546%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get [B]w = 21[/B]. This means l = 21 + 5. [B]l = 26[/B]

A rectangular parking lot has a perimeter of 152 yards. If the length of the parking lot is 12 yards greater than the width. What is the width of the parking lot? The perimeter of a rectangle is: 2l + 2w = P. We're given 2 equations: [LIST=1] [*]2l + 2w = 152 [*]l = w + 12 [/LIST] Substitute equation (2) into equation (1) for l: 2(w + 12) + 2w = 152 2w + 24 + 2w = 152 Combine like terms: 4w + 24 = 152 To solve for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=4w%2B24%3D152&pl=Solve']type this equation into our search engine[/URL] and we get: w =[B] 32[/B]

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters Given l = length and w = width, The perimeter of a rectangle is 2l + 2w, we have: [LIST=1] [*]l = 3w [*]2l + 2w = 56 [/LIST] Substitute equation (1) into equation (2) for l: 2(3w) + 2w = 56 6w + 2w = 56 To solve this equation for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B2w%3D56&pl=Solve']type it in our math engine[/URL] and we get: w = [B]7 [/B] To solve for l, we substitute w = 7 into equation (1): l = 3(7) l = [B]21[/B]

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. We're given the following: [LIST] [*]l = 3w [/LIST] We know the Perimeter (P) of a rectangle is: P = 2l + 2w Substituting l = 3w and P = 56 into this equation, we get: 2(3w) + 2w = 56 Multiplying through, we get: 6w + 2w = 56 (6 +2)w = 56 8w = 56 [URL='https://www.mathcelebrity.com/1unk.php?num=8w%3D56&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 7[/B] Substitute w = 7 into l = 3w, we get: l = 3(7) [B]l = 21[/B]

A rectangular room is 3 times as long as it is wide, and its perimeter is 56 meters. Find the dimension of the room. We're given: l = 3w The Perimeter (P) of a rectangle is: P = 2l + 2w With P = 56, we have: [LIST=1] [*]l = 3w [*]2l + 2w = 56 [/LIST] Substitute equation (1) into equation (2) for l: 2(3w) + 2w = 56 6w + 2w = 56 To solve this equation for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B2w%3D56&pl=Solve']type it in our search engine[/URL] and we get: w = [B]7 [/B] Now we plug w = 7 into equation (1) above to solve for l: l = 3(7) l = [B]21[/B]

a son is 1/4 of his fathers age. the difference in their ages is 30. what is the fathers age. Declare variables: [LIST] [*]Let f be the father's age [*]Let s be the son's age [/LIST] We're given two equations: [LIST=1] [*]s = f/4 [*]f - s = 30. [I]The reason why we subtract s from f is the father is older[/I] [/LIST] Using substitution, we substitute equaiton (1) into equation (2) for s: f - f/4 = 30 To remove the denominator/fraction, we multiply both sides of the equation by 4: 4f - 4f/4 = 30 *4 4f - f = 120 3f = 120 To solve for f, we divide each side of the equation by 3: 3f/3 = 120/3 Cancel the 3's on the left side and we get: f = [B]40[/B]

A store sells small notebooks for $6 and large notebooks for $12. If a student buys 6 notebooks and spends $60, how many of each did he buy? Let the amount of small notebooks be s. Let the amount of large notebooks be l. We're given two equations: [LIST=1] [*]l + s = 6 [*]12l + 6s = 60 [/LIST] Multiply equation (1) by -6 [LIST=1] [*]-6l - 6s = -36 [*]12l + 6s = 60 [/LIST] Now add equation 1 to equation 2: 12l - 6l + 6s - 6s = 60 - 36 Cancel the 6s terms, and we get: 6l = 24 To solve for l, we [URL='https://www.mathcelebrity.com/1unk.php?num=6l%3D24&pl=Solve']type this equation into our search engine[/URL] and we get: l = [B]4 [/B] Now substitute this into equation 1: 4 + s = 6 To solve for s, [URL='https://www.mathcelebrity.com/1unk.php?num=4%2Bs%3D6&pl=Solve']we type this equation into our search engine[/URL] and we get: s = [B]2[/B]

a textbook store sold a combined total of 296 sociology and history text books in a week. the number of history textbooks sold was 42 less than the number of sociology textbooks sold. how many text books of each type were sold? Let h = history book and s = sociology books. We have 2 equations: (1) h = s - 42 (2) h + s = 296 Substitute (1) to (2) s - 42 + s = 296 Combine variables 2s - 42 = 296 Add 42 to each side 2s = 338 Divide each side by 2 s = 169 So h = 169 - 42 = 127

A textbook store sold a combined total of 307 biology and chemistry textbooks in a week. The number of chemistry textbooks sold was 71 less than the number of biology textbooks sold. How many textbooks of each type were sold? Let b be the number of biology books and c be the number of chemistry books. We have two equations: [LIST=1] [*]b + c = 307 [*]c = b - 71 [/LIST] Substitute (2) into (1) for c b + (b - 71) = 307 Combine like terms: 2b - 71 = 307 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2b-71%3D307&pl=Solve']equation solver[/URL], we get: [B]b = 189[/B] Now substitute that into (2): c = 189 - 71 [B]c = 118[/B]

A washer and a dryer cost 600 combined. The cost of the washer is 3 times the cost of the dryer. What is the cost of the dryer? Let w be the cost of the washer. Let d be the cost of the dryer. We have 2 given equations: [LIST=1] [*]w + d = 600 [*]w = 3d [/LIST] Substitute (2) into (1) (3d) + d = 600 4d = 600 [URL='http://www.mathcelebrity.com/1unk.php?num=4d%3D600&pl=Solve']Run it through our equation calculator[/URL], to get [B]d = 150[/B].

A woman is one-half as old as her mother. The sum of their ages is 75. What are their ages? Let the woman's age be w. Let the mother's age be m. We're given two equations: [LIST=1] [*]w = m/2 [*]m + w = 75 [/LIST] Substitute equation (1) into equation (2) for w: m + m/2 = 75 To solve for m, we [URL='https://www.mathcelebrity.com/1unk.php?num=m%2Bm%2F2%3D75&pl=Solve']type this equation into our search engine [/URL]and we get: m = [B]50 [/B] To solve for w, we plug m = 50 into equation (1): w = 50/2 w = [B]25[/B]

A+B+D=255 B+15=A D+12=B A= [LIST=1] [*]A + B + D = 255 [*]B + 15 = A [*]D + 12 = B [*]A = ? [*]Rearrange (3) to solve for D by subtracting 12 from each side: D = B - 12 [/LIST] Substitute (2) and (5) into 1 (B + 15) + B + (B - 12) = 255 Combine like terms: 3B + 3 = 255 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3b%2B3%3D255&pl=Solve']equation solver[/URL], b = 84 Substitute b = 84 into equation (2): A = 84 + 15 [B]A = 99[/B]

According to the American Bureau of Labor Statistics, you will devote 32 years to sleeping and eating. The number of years sleeping will exceed the number of years eating by 24. Over your lifetime, how many years will you spend on each of these activities? Assumptions: [LIST] [*]Let years eating be e [*]Let years sleeping be s [/LIST] We're given: [LIST=1] [*]s = e + 24 [*]e + s = 32 [/LIST] To solve this system of equations, we substitute equation (1) into equation (2) for s: e + e + 24 = 32 To solve this equation for e, we [URL='https://www.mathcelebrity.com/1unk.php?num=e%2Be%2B24%3D32&pl=Solve']type it in our math engine[/URL] and we get: e = [B]4 [/B] Now, we take e = 4 and substitute it into equation (1) to solve for s: s = 4 + 24 s = [B]28[/B]

admission to the school fair is $2.50 for students and $3.75 for others. if 2848 admissions were collected for a total of 10,078.75, how many students attended the fair Let the number of students be s and the others be o. We're given two equations: [LIST=1] [*]o + s = 2848 [*]3.75o + 2.50s = 10078.75 [/LIST] Since we have no coefficients for equation 1, let's solve this the fast way using substitution. Rearrange equation 1 by subtracting o from each side to isolate s [LIST=1] [*]o = 2848 - s [*]3.75o + 2.50s = 10078.75 [/LIST] Now substitute equation 1 into equation 2: 3.75(2848 - s) + 2.50s =10078.75 To solve for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=3.75%282848-s%29%2B2.50s%3D10078.75&pl=Solve']type this equation into our search engine[/URL] and we get: s = [B]481[/B]

Let j be Jacob's age and c be Clinton's age. We have: [LIST=1] [*]j = 4c [*]j - 8 = 9(4c - 8) [/LIST] Substitute (1) into (2) (4c) - 8 = 36c - 72 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=4c-8%3D36c-72&pl=Solve']equation solver,[/URL] we get c = 2 Which means j = 4(2) = 8 8 years ago, Jacob was just born. Which means Clinton wasn't even born yet.

I read it wrong before. Here you go: Jacob is 4 times the age of Clinton. 8 years ago Jacob was 10 times the age of Clinton. How old are they now and how old were they 8 years ago? [LIST=1] [*]j = 4c [*]j - 8 = 10(c - 8) [/LIST] Substitute (1) into (2) (4c) - 8 = 10c - 80 [URL='http://www.mathcelebrity.com/1unk.php?num=4c-8%3D10c-80&pl=Solve']Equation solver[/URL] gives us [B]c = 12[/B] which means j = 4(12) = [B]48[/B]. 8 years ago,[B] j = 40 and c = 4[/B] which holds the 10x rule.

[QUOTE="math_celebrity, post: 1163, member: 1"]I read it wrong before. Here you go: Jacob is 4 times the age of Clinton. 8 years ago Jacob was 10 times the age of Clinton. How old are they now and how old were they 8 years ago? [LIST=1] [*]j = 4c [*]j - 8 = 10(c - 8) [/LIST] Substitute (1) into (2) (4c) - 8 = 10c - 80 [URL='http://www.mathcelebrity.com/1unk.php?num=4c-8%3D10c-80&pl=Solve']Equation solver[/URL] gives us [B]c = 12[/B] which means j = 4(12) = [B]48[/B]. 8 years ago,[B] j = 40 and c = 4[/B] which holds the 10x rule.[/QUOTE] Thank you, I see what I did wrong!

Let f be the age of the father and d be the age of the daughter and s be the age of the son. We have: [LIST=1] [*]f = 3s [*]d = s - 3 [*]d - 3 + f - 3 + s - 3 = 63 [/LIST] Simplify (3) d + f + s - 9 = 63 d + f + s = 72 Now, substitute (1) and (2) into the modified (3) (s - 3) + 3s + s = 72 Combine like terms: 5s - 3 = 72 Add 3 to each side 5s = 75 Divide each side by 5 s = 15 We want f, so we substitute s = 15 into (1) f = 3(15) [B]f = 45[/B]

Alvin is 34 years younger than Elga. Elga is 3 times older than Alvin. What is Elgas age? Let a be Alvin's age. Let e be Elga's age. We're given: [LIST=1] [*]a = e - 34 [*]e = 3a [/LIST] Substitute (2) into (1): a = 3a - 34 [URL='https://www.mathcelebrity.com/1unk.php?num=a%3D3a-34&pl=Solve']Typing this equation into the search engine[/URL], we get a = 17 Subtitute this into Equation (2): e = 3(17) e = [B]51[/B]

An irregular pentagon is a five sided figure. The two longest sides of a pentagon are each three times as long as the shortest side. The remaining two sides are each 8m longer than the shortest side. The perimeter of the Pentagon is 79m. Find the length of each side of the Pentagon. Let long sides be l. Let short sides be s. Let medium sides be m. We have 3 equations: [LIST=1] [*]2l + 2m + s = 79 [*]m = s + 8 [*]l = 3s [/LIST] Substitute (2) and (3) into (1): 2(3s) + 2(s + 8) + s = 79 Multiply through and simplify: 6s + 2s + 16 + s = 79 9s + 16 = 79 [URL='https://www.mathcelebrity.com/1unk.php?num=9s%2B16%3D79&pl=Solve']Using our equation calculator[/URL], we get [B]s = 7[/B]. This means from Equation (2): m = 7 + 8 [B]m = 15 [/B] And from equation (3): l = 3(7) [B]l = 21[/B]

An isosceles triangles non-congruent angle is 16 more than twice the congruent ones. What is the measure of all 3 angles? Let the congruent angles measurement be c. And the non-congruent angle measurement be n. We're given: [LIST=1] [*]n = 2c + 16 <-- Twice means we multiply by 2, and more than means we add 16 [*]2c + n = 180 <-- Since the sum of angles in an isosceles triangle is 180 [/LIST] Substitute (1) into (2): 2c + (2c + 16) = 180 Group like terms: 4c + 16 = 180 [URL='https://www.mathcelebrity.com/1unk.php?num=4c%2B16%3D180&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]c = 41[/B] Substituting this value into Equation 1, we get n = 2(41) + 16 n = 82 + 16 [B]n = 98[/B]

An orchard has 378 orange trees. The number of rows exceeds the number of trees per row by 3. How many trees are there in each row? We have r rows and t trees per row. We're give two equations: [LIST=1] [*]rt = 378 [*]r = t + 3 [/LIST] Substitute equation (2) into equation (1) for r: (t + 3)t = 378 Multiply through: t^2 + 3t = 378 We have a quadratic equation. To solve this equation, we [URL='https://www.mathcelebrity.com/quadratic.php?num=t%5E2%2B3t%3D378&pl=Solve+Quadratic+Equation&hintnum=+0']type it in our search engine [/URL]and we get: t = 18 and t = -21 Since t cannot be negative, we get trees per row (t): [B]t = 18[/B]

An orchard has 816 apple trees. The number of rows exceeds the number of trees per row by 10. How many trees are there in each row? Let the rows be r and the trees per row be t. We're given two equations: [LIST=1] [*]rt = 816 [*]r = t + 10 [/LIST] Substitute equation (2) into equation (1) for r: (t + 10)t = 816 t^2 + 10t = 816 Subtract 816 from each side of the equation: t^2 + 10t - 816 = 816 - 816 t^2 + 10t - 816 = 0 We have a quadratic equation. To solve this, we [URL='https://www.mathcelebrity.com/quadratic.php?num=t%5E2%2B10t-816%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type it in our search engine [/URL]and we get: t = (24, -34) Since the number of trees per row can't be negative, we choose [B]24[/B] as our answer

Andrea has one hour to spend training for an upcoming race she completes her training by running full speed in the distance of the race and walking back the same distance to cool down if she runs at a speed of 9 mph and walks back at a speed of 3 mph how long should she plan on spending to walk back Let r = running time. Let w = walking time We're given two equations [LIST=1] [*]r + w = 1 [*]9r = 3w [/LIST] Rearrange equation (1) by subtract r from each side: [LIST=1] [*]w = 1 - r [*]9r = 3w [/LIST] Now substitute equation (1) into equation (2): 9r = 3(1 - r) 9r = 3 - 3r To solve for r, [URL='https://www.mathcelebrity.com/1unk.php?num=9r%3D3-3r&pl=Solve']we type this equation into our search engine[/URL] and we get: r = 0.25 Plug this into modified equation (1) to solve for w, and we get: w = 1. 0.25 [B]w = 0.75[/B]

Anna has 50 coins in her piggy bank. She notices that she only has dimes and pennies. If she has exactly four times as many pennies as dimes, how many pennies are in her piggy bank? Let d be the number of dimes, and p be the number of pennies. We're given: [LIST=1] [*]d + p = 50 [*]p = 4d [/LIST] Substitute (2) into (1) d + 4d = 50 [URL='https://www.mathcelebrity.com/1unk.php?num=d%2B4d%3D50&pl=Solve']Type that equation into our search engine[/URL]. We get: d = 10 Now substitute this into Equation (2): p = 4(10) [B]p = 40[/B]

Antonio has a change jar that contains $3.65 in half dollars and nickels. He has 7 more nickels than half dollars. How many of each type of coin does he have? Let h be half dollars Let n be nickels We're given two equations: [LIST=1] [*]n = h + 7 [*]0.5h + 0.05n = 3.65 [/LIST] Substitute equation (1) into equation (2) for n: 0.5h + 0.05(h + 7) = 3.65 To solve this equation for h, we[URL='https://www.mathcelebrity.com/1unk.php?num=0.5h%2B0.05%28h%2B7%29%3D3.65&pl=Solve'] type it in our search engine[/URL] and we get: h = [B]6 [/B] To get n, we substitute h = 6 into equation (1) above: n = 6 + 7 n = [B]13[/B]

April, May and June have 90 sweets between them. May has three-quarters of the number of sweets that June has. April has two-thirds of the number of sweets that May has. How many sweets does June have? Let the April sweets be a. Let the May sweets be m. Let the June sweets be j. We're given the following equations: [LIST=1] [*]m = 3j/4 [*]a = 2m/3 [*]a + j + m = 90 [/LIST] Cross multiply #2; 3a = 2m Dividing each side by 2, we get; m = 3a/2 Since m = 3j/4 from equation #1, we have: 3j/4 = 3a/2 Cross multiply: 6j = 12a Divide each side by 12: a = j/2 So we have: [LIST=1] [*]m = 3j/4 [*]a = j/2 [*]a + j + m = 90 [/LIST] Now substitute equation 1 and 2 into equation 3: j/2 + j + 3j/4 = 90 Multiply each side by 4 to eliminate fractions: 2j + 4j + 3j = 360 To solve this equation for j, we [URL='https://www.mathcelebrity.com/1unk.php?num=2j%2B4j%2B3j%3D360&pl=Solve']type it in our search engine[/URL] and we get: j = [B]40[/B]

Ashley age is 2 times Johns age. The sum of their ages is 63. What is Johns age? Let Ashley's age be a. Let John's age be j. We have two equations: [LIST=1] [*]a = 2j [*]a + j = 63 [/LIST] Now substitute (1) into (2) (2j) + j = 63 Combine like terms: 3j = 63 [URL='http://www.mathcelebrity.com/1unk.php?num=3j%3D63&pl=Solve']Typing 3j = 63 into our search engine[/URL], we get [B]j = 21[/B]

At a carnival, the price of an adult ticket is $6 while a child ticket is $4. On a certain day, 30 more child tickets than adult tickets were sold. If a total of $6360 was collected from the total ticket sale that day, how many child tickets were sold? Let the number of adult tickets be a. Let the number of child tickets be c. We're given two equations: [LIST=1] [*]c = a + 30 [*]6a + 4c = 6360 [/LIST] Substitute equation (1) into equation (2): 6a + 4(a + 30) = 6360 Multiply through to remove parentheses: 6a + 4a + 120 = 6360 T[URL='https://www.mathcelebrity.com/1unk.php?num=6a%2B4a%2B120%3D6360&pl=Solve']ype this equation into our search engine[/URL] to solve for a and we get: a = 624 Now substitute a = 624 back into equation (1) to solve for c: c = 124 + 30 c = [B]154[/B]

At a concert there were 25 more women than men. The total number of people at the concert was 139. Find the number of women and the number of men at the concert. Let men be m and women be w. We're given two equations. [LIST=1] [*]w = m + 25 [*]m + w = 139 [/LIST] Substitute equation (1) into equation (2): m + m + 25 = 139 To solve for m, we [URL='https://www.mathcelebrity.com/1unk.php?num=m%2Bm%2B25%3D139&pl=Solve']type this equation into our search engine[/URL] and we get: m = [B]57 [/B] To find w, we substitute m = 57 into equation (1): w = 57 + 25 w = [B]82[/B]

At a football game, a vender sold a combined total of 117 sodas and hot dogs. The number of hot dogs sold was 59 less than the number of sodas sold. Find the number of sodas sold and the number of hot dogs sold. [U]Let h = number of hot dogs and s = number of sodas. Set up our given equations:[/U] [LIST=1] [*]h + s = 117 [*]h = s - 59 [/LIST] [U]Substitute (2) into (1)[/U] (s - 59) + s = 117 [U]Combine s terms[/U] 2s - 59 = 117 [U]Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2s-59%3D117&pl=Solve']equation solver[/URL], we find:[/U] [B]s = 88 [/B] [U]Plug s = 88 into (2)[/U] h = 88 - 59 [B]h = 29[/B]

At a homecoming football game, the senior class sold slices of pizza for $.75 each and hamburgers for $1.35 each. They sold 40 more slices of pizza than hamburgers, and sales totaled $292.5. How many slices of pizza did they sell Let the number of pizza slices be p and the number of hamburgers be h. We're given two equations: [LIST=1] [*]p = h + 40 [*]1.35h + 0.75p = 292.50 [/LIST] [I]Substitute[/I] equation (1) into equation (2) for p: 1.35h + 0.75(h + 40) = 292.50 1.35h + 0.75h + 30 = 292.50 2.10h + 30 = 292.50 To solve for h, we [URL='https://www.mathcelebrity.com/1unk.php?num=2.10h%2B30%3D292.50&pl=Solve']plug this equation into our search engine[/URL] and we get: h = 125 The problem asks for number of pizza slices sold (p). So we substitute our value above of h = 125 into equation (1): p = 125 + 40 p = [B]165[/B]

At the end of the 2021 NBA season, the NY Knicks had 10 more wins than losses. This NBA season the NY Knicks played a total of 72 times. Find a solution to this problem and explain. Let w be the number of wins Let l be the number of losses We're given two equations: [LIST=1] [*]w = l + 10 [*]l + w = 72 [/LIST] To solve this system of equations, substitute equation (1) into equation (2) for w: l + l + 10 = 72 To solve this equation for l, we [URL='https://www.mathcelebrity.com/1unk.php?num=l%2Bl%2B10%3D72&pl=Solve']type it in our math engine[/URL] and we get: l = [B]31 [/B] To solve for w, we substitute l = 31 into equation (1): w = 31 + 10 w = [B]41[/B]

Austin has 15 CDs, which is 3 less than his sister has. How many CDs does his sister have? Let s be the number of CD's his sister has and a be the number Austin has [LIST=1] [*]a = 15 [*]a = s - 3 [/LIST] Substitute (1) into (2) 15 = s - 3 Add 3 to each side [B]s = 18[/B]

Ben is 3 times as old as Daniel and is also 4 years older than Daniel. Let Ben's age be b, let Daniel's age by d. We're given: [LIST=1] [*]b = 3d [*]b = d + 4 [/LIST] Substitute (1) into (2) 3d = d + 4 [URL='https://www.mathcelebrity.com/1unk.php?num=3d%3Dd%2B4&pl=Solve']Type this equation into our search engine[/URL], and we get [B]d = 2[/B]. Substitute this into equation (1), and we get: b = 3(2) [B]b = 6 [/B] So Daniel is 2 years old and Ben is 6 years old.

Ben is 4 times as old as Ishaan and is also 6 years older than Ishaan. Let b be Ben's age and i be Ishaan's age. We're given: [LIST=1] [*]b = 4i [*]b = i + 6 [/LIST] Set (1) and (2) equal to each other: 4i = i + 6 [URL='https://www.mathcelebrity.com/1unk.php?num=4i%3Di%2B6&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]i = 2[/B] Substitute this into equation (1): b = 4(2) [B]b = 8 [/B] [I]Therefore, Ishaan is 2 years old and Ben is 8 years old.[/I]

Beth is 5 years younger than celeste. Next year, their ages will have a sum equal to 57. How old is each now? Let b = Beth's age Let c = Celeste's age We are given: [LIST=1] [*]b = c - 5 [*]b + 1 + c + 1 = 57 [/LIST] Substitute (1) into (2) (c - 5) + 1 + c + 1 = 57 Group like terms: 2c - 3 = 57 [URL='https://www.mathcelebrity.com/1unk.php?num=2c-3%3D57&pl=Solve']Type 2c - 3 = 57 into our search engine[/URL], we get [B]c = 30[/B] Substitute c = 30 into Equation (1), we get: b = 30 - 5 [B]b = 25 [/B] Therefore, Beth is 25 and Celeste is 30.

Bob bought 10 note books and 4 pens for 18$. Bill bought 6 notebooks and 4 pens for 12$. Find the price of one note book and one pen. Let the price of each notebook be n. Let the price of each pen be p. We're given two equations: [LIST=1] [*]10n + 4p = 18 [*]6n + 4p = 12 [/LIST] Since we have matching coefficients for p, we subtract equation 1 from equation 2: (10 - 6)n + (4 - 4)p = 18 - 12 Simplifying and cancelling, we get: 4n = 6 [URL='https://www.mathcelebrity.com/1unk.php?num=4n%3D6&pl=Solve']Type this equation into our search engine[/URL] and we get: [B]n = 1.5[/B] Now, substitute this value for n into equation (2): 6(1.5) + 4p = 12 Multiply through: 9 + 4p = 12 [URL='https://www.mathcelebrity.com/1unk.php?num=9%2B4p%3D12&pl=Solve']Type this equation into our search engine[/URL] and we get: [B]p = 0.75[/B]

Bob has half as many quarters as dimes. He has $3.60. How many of each coin does he have? Let q be the number of quarters. Let d be the number of dimes. We're given: [LIST=1] [*]q = 0.5d [*]0.25q + 0.10d = 3.60 [/LIST] Substitute (1) into (2): 0.25(0.5d) + 0.10d = 3.60 0.125d + 0.1d = 3.6 Combine like terms: 0.225d = 3.6 [URL='https://www.mathcelebrity.com/1unk.php?num=0.225d%3D3.6&pl=Solve']Typing this equation into our search engine[/URL], we're given: [B]d = 16[/B] Substitute d = 16 into Equation (1): q = 0.5(16) [B]q = 8[/B]

Bob is twice as old as Henry. The sum of their ages is 42. How old is Henry? Let Bob's age be b. Let Henry's age be h. We're given two equations: [LIST=1] [*]b = 2h [*]b + h = 42 [/LIST] Substitute b = 2h in equation 1 into equation 2 for b: 2h + h = 42 To solve for h, we [URL='https://www.mathcelebrity.com/1unk.php?num=2h%2Bh%3D42&pl=Solve']type this equation into our search engine[/URL] and we get: h = [B]14[/B]

Bud makes $400 more per month than maxine If their total income is $3600 how much does bud earn per month Let Bud's earnings be b. Let Maxine's earnings be m. We're given two equations: [LIST=1] [*]b = m + 400 [*]b + m = 3600 [/LIST] To solve this system of equations, we substitute equation (1) into equation (2) for b m + 400 + m = 3600 To solve this equation for m, we [URL='https://www.mathcelebrity.com/1unk.php?num=m%2B400%2Bm%3D3600&pl=Solve']type it in our search engine[/URL] and we get: m = 1600 To solve for b, we substitute m = 1600 into equation (1) above: b = 1600 + 400 b = [B]2200[/B]

Building A is 150 feet shorter than Building B. The height of both building is 1530 feet. Find the height of both building A and B. Let a be the height of building A Let b be the height of building B We're given two equations: [LIST=1] [*]a = b - 150 [*]a + b = 1530 [/LIST] To solve this system of equations, we substitute equation (1) into equation (2) for a: (b - 150) + b = 1530 To solve this equation for b, we [URL='https://www.mathcelebrity.com/1unk.php?num=b-150%2Bb%3D1530&pl=Solve']type it in our search engine[/URL] and we get: b = [B]840[/B] To solve for a, we substitute b = 840 into equation (1): a = 840 - 150 a = [B]690[/B]

Let f = dollars spent on flights, h dollars spent on hotels, and p dollars spent on all other purchases. [U]Set up our equations:[/U] (1) 4f + 2h + p = 14660 (2) f + h + p = 9480 (3) f = 2h + 140 [U]First, substitute (3) into (2)[/U] (2h + 140) + h + p = 9480 3h + p + 140 = 9480 3h + p = 9340 [U]Subtract 3h to isolate p to form equation (4)[/U] (4) p = 9340 - 3h [U]Take (3) and (4), and substitute into (1)[/U] 4(2h + 140) + 2h + (9340 - h) = 14660 [U]Multiply through[/U] 8h + 560 + 2h + 9340 - 3h = 14660 [U]Combine h terms and constants[/U] (8 + 2 - 3)h + (560 + 9340) = 14660 7h + 9900 = 14660 [U]Subtract 9900 from both sides:[/U] 7h = 4760 [U]Divide each side by 7[/U] [B]h = 680[/B] [U]Substitute h = 680 into equation (3)[/U] f = 2(680) + 140 f = 1360 + 140 [B]f = 1,500[/B] [U] Substitute h = 680 and f = 1500 into equation (2)[/U] 1500 + 680 + p = 9480 p + 2180 = 9480 [U]Subtract 2180 from each side:[/U] [B]p = 7,300[/B]

Cam is 3 years older than Lara. If their combined age is 63, determine their ages by solving an appropriate pair of equations. Let Cam's age be c. Let Lara's age be l. We're given two equations: [LIST=1] [*]c = l + 3 <-- older means we add [*]c + l = 63 <-- combined ages mean we add [/LIST] Substitute equation (1) into equation (2): l + 3 + l = 63 Combine like terms to simplify our equation: 2l + 3 = 63 To solve for l, [URL='https://www.mathcelebrity.com/1unk.php?num=2l%2B3%3D63&pl=Solve']we type this equation into our search engine[/URL] and we get: l = [B]30[/B] Now, we plug l = 30 into equation (1) to solve for c: c = 30 + 3 c = [B]33[/B]

Cam is 3 years older than Lara. If their combined age is 63, determine their ages by solving an appropriate pair of equations. Let Cam's age be c. Let Lara's age be l. We're given two equations: [LIST=1] [*]c = l + 3 (Since older means we add) [*]c + l = 63 [/LIST] To solve this system of equations, we substitute equation (1) into equation (2) for c: l + 3 + l = 63 To solve this equation for l, we [URL='https://www.mathcelebrity.com/1unk.php?num=l%2B3%2Bl%3D63&pl=Solve']type it in our search engine [/URL]and we get: l = [B]30 [/B] Now, we take l = 30 and substitute it in equation (1) to solve for c: c = 30 + 3 c = [B]33[/B]

Cars and trucks are the most popular vehicles. last year, the number of cars sold was 39,000 more than 3 times the number of trucks sold. There were 216,000 cars sold last year. Write an equation that can be used to find the number of trucks, t, sold last year. Let c be the number of cars. Let t be the number of trucks. We're given two equations: [LIST=1] [*]c = 3t + 39000 [*]c + t = 216000 [/LIST] Substitute equation (1) into equation (2) for c: 3t + 39000 + t = 216000 To solve this equation for t, [URL='https://www.mathcelebrity.com/1unk.php?num=3t%2B39000%2Bt%3D216000&pl=Solve']we type it in our math engine [/URL]and we get: t = [B]44,250[/B]

Connor runs 2 mi more each day than David. The sum of the distances they run each week is 56 mi. How far does David run each day? Let Connor's distance be c Let David's distance be d We're given two equations: [LIST=1] [*]c = d + 2 [*]7(c + d) = 56 [/LIST] Simplifying equation 2 by dividing each side by 7, we get: [LIST=1] [*]c = d + 2 [*]c + d = 8 [/LIST] Substitute equation (1) into equation (2) for c d + 2 + d = 8 To solve for d, we [URL='https://www.mathcelebrity.com/1unk.php?num=d%2B2%2Bd%3D8&pl=Solve']type this equation into our calculation engine[/URL] and we get: d = [B]3[/B]

Country A produces about 7 times the amount of diamonds in carats produce in Country B. If the total produced in both countries is 40,000,000 carats, find the amount produced in each country. Set up our two given equations: [LIST=1] [*]A = 7B [*]A + B = 40,000,000 [/LIST] Substitute (1) into (2) (7B) + B = 40,000,000 Combine like terms 8B = 40,000,000 Divide each side by 8 [B]B = 5,000,000[/B] Substitute this into (1) A = 7(5,000,000) [B]A = 35,000,000[/B]

Daniel is 6cm taller than Kamala. If their total height is 368cm, how tall is Kamala? Let Daniel's height be d. Let Kamala's height be k. We're given two equations: [LIST=1] [*]d = k + 6 [*]d + k = 368 [/LIST] Substitute equation (1) into equation (2) for d: k + 6 + k = 368 To solve for k, we [URL='https://www.mathcelebrity.com/1unk.php?num=k%2B6%2Bk%3D368&pl=Solve']type this equation into our search engine[/URL] and we get: k = [B]181[/B]

devaughn,sageis2timessydneysage,thesumoftheiragesis78,whatissydneysage Let d be Devaughn's age. Let s be Sydney's age. We have two equations: [LIST=1] [*]d = 2s [*]d + s = 78 [/LIST] Substitute (1) into (2) 2s + s = 78 3s = 78 Entering [URL='http://www.mathcelebrity.com/1unk.php?num=3s%3D78&pl=Solve']3x = 78 into the search engine[/URL], we get [B]s = 26[/B].

difference between 2 positive numbers is 3 and the sum of their squares is 117 Declare variables for each of the two numbers: [LIST] [*]Let the first variable be x [*]Let the second variable be y [/LIST] We're given 2 equations: [LIST=1] [*]x - y = 3 [*]x^2 + y^2 = 117 [/LIST] Rewrite equation (1) in terms of x by adding y to each side: [LIST=1] [*]x = y + 3 [*]x^2 + y^2 = 117 [/LIST] Substitute equation (1) into equation (2) for x: (y + 3)^2 + y^2 = 117 Evaluate and simplify: y^2 + 3y + 3y + 9 + y^2 = 117 Combine like terms: 2y^2 + 6y + 9 = 117 Subtract 117 from each side: 2y^2 + 6y + 9 - 117 = 117 - 117 2y^2 + 6y - 108 = 0 This is a quadratic equation: Solve the quadratic equation 2y2+6y-108 = 0 With the standard form of ax2 + bx + c, we have our a, b, and c values: a = 2, b = 6, c = -108 Solve the quadratic equation 2y^2 + 6y - 108 = 0 The quadratic formula is denoted below: y = -b ± sqrt(b^2 - 4ac)/2a [U]Step 1 - calculate negative b:[/U] -b = -(6) -b = -6 [U]Step 2 - calculate the discriminant ?:[/U] ? = b2 - 4ac: ? = 62 - 4 x 2 x -108 ? = 36 - -864 ? = 900 <--- Discriminant Since ? is greater than zero, we can expect two real and unequal roots. [U]Step 3 - take the square root of the discriminant ?:[/U] ?? = ?(900) ?? = 30 [U]Step 4 - find numerator 1 which is -b + the square root of the Discriminant:[/U] Numerator 1 = -b + ?? Numerator 1 = -6 + 30 Numerator 1 = 24 [U]Step 5 - find numerator 2 which is -b - the square root of the Discriminant:[/U] Numerator 2 = -b - ?? Numerator 2 = -6 - 30 Numerator 2 = -36 [U]Step 6 - calculate your denominator which is 2a:[/U] Denominator = 2 * a Denominator = 2 * 2 Denominator = 4 [U]Step 7 - you have everything you need to solve. Find solutions:[/U] Solution 1 = Numerator 1/Denominator Solution 1 = 24/4 Solution 1 = 6 Solution 2 = Numerator 2/Denominator Solution 2 = -36/4 Solution 2 = -9 [U]As a solution set, our answers would be:[/U] (Solution 1, Solution 2) = (6, -9) Since one of the solutions is not positive and the problem asks for 2 positive number, this problem has no solution

Dina is twice as old as Andrea. The sum of their age is 72. Find their present ages. Let d be Dina's age. Let a be Andrea's age. We're given: [LIST=1] [*]d = 2a <-- Twice means multiply by 2 [*]a + d = 72 [/LIST] Substitute equation (1) into equation (2): a + 2a = 72 [URL='https://www.mathcelebrity.com/1unk.php?num=a%2B2a%3D72&pl=Solve']Type this equation into our search engine[/URL] and we get: [B]a = 24[/B] Substitute a = 24 into equation (1): d = 2(24) [B]d = 48 So Andrea is 24 years old and Dina is 48 years old[/B]

does the point (3,0) line on the line y=3x Substitute the x value of (x,y) = (3,0) into y = 3x: y = 3(3) y = 9 Since y = 9 and y <> 0, then no, this point [B]does not[/B] lie on the line

During a recent season Miguel Cabrera and Mike Jacobs hit a combined total of 46 home runs. Cabrera hit 6 more home runs than Jacobs how many home runs did each player hit Let c be Miguel Cabrera's home runs and j be Mike Jacobs home runs. We are given two equations: [LIST=1] [*]c + j = 46 [*]c = j + 6 [/LIST] Substitute (2) into (1) (j + 6) + j = 46 Combine like terms: 2j + 6 = 46 [URL='https://www.mathcelebrity.com/1unk.php?num=2j%2B6%3D46&pl=Solve']Plugging this into our equation calculator[/URL], we get [B]j = 20[/B]. Substitute this into equation (2), we have: c = 20 + 6 [B]c = 26 [/B] Therefore, Mike Jacobs hit 20 home runs and Miguel Cabrera hit 26 home runs.

Emily is three years older than twice her sister Mary’s age. The sum of their ages is less than 30. What is the greatest age Mary could be? Let e = Emily's age and m = Mary's age. We have the equation e = 2m + 3 and the inequality e + m < 30 Substitute the equation for e into the inequality: 2m + 3 + m < 30 Add the m terms 3m + 3 < 30 Subtract 3 from each side of the inequality 3m < 27 Divide each side of the inequality by 3 to isolate m m < 9 Therefore, the [B]greatest age[/B] Mary could be is 8, since less than 9 [U]does not include[/U] 9.

eric is twice as old as Shawn. The sum of their ages is 33. How old is Shawn? Let Eric's age be e. Let Shawn's age be s. We're given two equations: [LIST=1] [*]e = 2s [*]e + s = 33 [/LIST] Substitute equation (1) into equation (2) for e so we can solve for s: 2s + s = 33 To solve for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=2s%2Bs%3D33&pl=Solve']type this equation into our search engine[/URL] and we get: s = [B]11[/B]

Faith is 1/5 her mother's age. Their combined ages are 30. How old is faith? Let Faith's age be f. Let her mother's age be m. We're given: [LIST=1] [*]f = m/5 [*]f + m = 30 [/LIST] Rearrange (1) by cross-multiplying: m = 5f Substitute this into equation (2): f + 5f = 30 [URL='https://www.mathcelebrity.com/1unk.php?num=f%2B5f%3D30&pl=Solve']Type this equation into our search engine[/URL] and we get: f = [B]5[/B]

Find 2 consecutive numbers such that the sum of twice the smaller number and 3 times the larger number is 73. Let x be the smaller number and y be the larger number. We are given: 2x + 3y = 73 Since the numbers are consecutive, we know that y = x + 1. Substitute this into our given equation: 2x + 3(x + 1) = 73 Multiply through: 2x + 3x + 3 = 73 Group like terms: 5x + 3 = 73 [URL='https://www.mathcelebrity.com/1unk.php?num=5x%2B3%3D73&pl=Solve']Type 5x + 3 = 73 into the search engine[/URL], and we get [B]x = 14[/B]. Our larger number is 14 + 1 = [B]15 [/B] Therefore, our consecutive numbers are[B] (14, 15)[/B]

Let g = Gary's pets and a = Abe's pets. We are given two equations: (1) g = a - 3 (2) a + g = 15 Substitute (1) into (2) a + (a - 3) = 15 Combine Like Terms: 2a - 3 = 15 Add 3 to each side: 2a = 18 Divide each side by 2 to isolate a: a = 9 --> Abe has 9 pets Substitute a = 9 into Equation (1) g = 9 - 3 g = 6 --> Gary has 6 pets

George has a certain number of apples, and Sarah has 4 times as many apples as George. They have a total of 25 apples. Let George's apples be g. Let Sarah's apples be s. We're give two equations: [LIST=1] [*]s = 4g [*]g + s = 25 [/LIST] Substitute equation (1) into equation (2) for s: g + 4g = 25 If [URL='https://www.mathcelebrity.com/1unk.php?num=g%2B4g%3D25&pl=Solve']we plug this equation into our search engine[/URL] and solve for g, we get: g = [B]5[/B] Now substitute this into equation 1 for g = 5: s = 4(5) s = [B]20[/B] [B]So George has 5 apples and Sarah has 20 apples[/B]

Germany and Austria have a total of 25 states. Germany has 7 more states than Austria has. Create 2 equations. Let g be the number of German states. Let a be the number of Austrian states. We're given two equations: [LIST=1] [*]a + g = 25 [*]g = a + 7 [/LIST] To solve this system of equations, we substitute equation (2) into equation (1) for g: a + (a + 7) = 25 Combine like terms: 2a + 7 = 25 To solve for a, we[URL='https://www.mathcelebrity.com/1unk.php?num=2a%2B7%3D25&pl=Solve'] type this equation into our search engine[/URL] and we get: [B]a = 9[/B] To solve for g, we plug in a = 9 into equation (2): g = 9 + 7 [B]g = 16[/B]

if a+b=2 and a2-b2=-4, what is the value of a-b? a^2 - b^2 = -4 Factor this: (a + b)(a - b) = -4 We know from above, (a +b) = 2, so substitute: 2(a - b) = -4 Divide each side by 2 [B](a - b) = -2[/B]

If FG=11, GH=x-2, and FH=3x-11, what is FH By segment addition, we have: FG + GH = FH 11 + x - 2 = 3x - 11 To solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=11%2Bx-2%3D3x-11&pl=Solve']type it in or math engine[/URL] and we get: x = 10 FH = 3x - 11. So we substitute x = 10 into this: FH = 3(10) - 11 FH = 30 - 11 FH = [B]19[/B]

If Frank’s age is double of Willis’ age and the sum of their ages is 42. What are their ages? Let Frank's age be f. Let Willis's age be w. We're given two equations: [LIST=1] [*]f = 2w <-- Double means multiply by 2 [*]f + w = 42 [/LIST] Substitute equation (1) into equation (2): 2w + w = 42 To solve for w, [URL='https://www.mathcelebrity.com/1unk.php?num=2w%2Bw%3D42&pl=Solve']type this equation into our search engine[/URL]. We get: w = [B]14 [/B] Now, take w = 14, and substitute it back into equation (1) to solve for f: f = 2(14) f = [B]28[/B]

If JK = PQ and PQ = ST, then JK=ST JK = PQ | Given Substitute ST for PQ since PQ = ST | Substitution [B]JK = ST[/B]

If Jody had $3 more she would have twice as much as Lars together they have $60. Let j be Jody's money and l be Lars's money. We have two equations: [LIST=1] [*]j + l = 60 [*]j + 3 = 2l [/LIST] Rearrange (2) to solve for j by subtracting 3 j = 2l - 3 Now substitute this into (1) (2l - 3) + l = 60 Combine like terms 3l - 3 = 60 Enter this into our [URL='http://www.mathcelebrity.com/1unk.php?num=3l-3%3D60&pl=Solve']equation calculator[/URL], and we get: [B]l = 21[/B] Now plug l = 21 into our rearranged equation above: j = 2(21) - 3 j = 42 - 3 [B]j = 39[/B]

If Mr hernandez has 5 times as many students as Mr daniels and together they have 150 students how many students do each have? Let h = Mr. Hernandez's students and d = Mr. Daniels students. We are given two equations: (1) h = 5d (2) d + h = 150 Substitute equation (1) into equation (2) d + (5d) = 150 Combine like terms: 6d = 150 Divide each side of the equation by 6 to isolate d d = 25 <-- Mr. Daniels Students Now, plug the value for d into equation (1) h = 5(25) h = 125 <-- Mr. Hernandez students

If QR = 16, RS = 4x ? 17, and QS = x + 20, what is RS? From the segment addition rule, we have: QR + RS = QS Plugging our values in for each of these segments, we get: 16 + 4x - 17 = x + 20 To solve this equation for x, [URL='https://www.mathcelebrity.com/1unk.php?num=16%2B4x-17%3Dx%2B20&pl=Solve']we type it in our search engine[/URL] and we get: x = 7 Take x = 7 and substitute it into RS: RS = 4x - 17 RS = 4(7) - 17 RS = 28 - 17 RS = [B]11[/B]

If Quinn has 4 times as many quarters as nickels and they have a combined value of 525 cents, how many of each coin does he have? Using q for quarters and n for nickels, and using 525 cents as $5.25, we're given two equations: [LIST=1] [*]q = 4n [*]0.25q + 0.05n = 5.25 [/LIST] Substitute equation (1) into equation (2) for q: 0.25(4n) + 0.05n = 5.25 Multiply through and simplify: n + 0.05n = 5.25 To solve this equation for n, we [URL='https://www.mathcelebrity.com/1unk.php?num=n%2B0.05n%3D5.25&pl=Solve']type it in our search engine[/URL] and we get: n = [B]5 [/B] To get q, we plug in n = 5 into equation (1) above: q = 4(5) q = [B]20[/B]

if sc = hr and hr=ab then sc=ab sc = hr (given) Since hr = ab, we can substitute ab for hr by substitution: [B]sc = ab[/B]

Let a be the cost of the ball and b be the cost of the bat: We're given 2 equations: [LIST=1] [*]a + b = 1.10 [*]b = a + 1 [/LIST] Substitute equation (2) into equation (1) for b: a + a + 1 = 1.10 Combine like terms: 2a + 1 = 1.10 Subtract 1 from each side: 2a + 1 - 1 = 1.10 - 1 2a = 0.10 Divide each side by 2: 2a/2 = 0.10/2 a = [B]0.05[/B] [MEDIA=youtube]79q346Hy7R8[/MEDIA]

If the perimeter of a rectangular sign is 44cm and the width is 2cm shorter than half the length, then what are the length and width? The perimeter (P) of a rectangle is: 2l + 2w = P We're given P = 44, so we substitute this into the rectangle perimeter equation: 2l + 2w = 44 We're also given w = 0.5l - 2. Substitute the into the Perimeter equation: 2l + 2(0.5l - 2) = 44 Multiply through and simplify: 2l + l - 4 = 44 Combine like terms: 3l - 4 = 44 [URL='https://www.mathcelebrity.com/1unk.php?num=3l-4%3D44&pl=Solve']Type this equation into the search engine[/URL], and we get: [B]l = 16[/B] Substitute this back into the equation w = 0.5l - 2 w = 0.5(16) - 2 w = 8 - 2 [B]w = 6[/B]

If the speed of an aeroplane is reduced by 40km/hr, it takes 20 minutes more to cover 1200m. Find the time taken by aeroplane to cover 1200m initially. We know from the distance formula (d) using rate (r) and time (t) that: d = rt Regular speed: 1200 = rt Divide each side by t, we get: r = 1200/t Reduced speed. 20 minutes = 60/20 = 1/3 of an hour. So we multiply 1,200 by 3 3600 = (r - 40)(t + 1/3) If we multiply 3 by (t + 1/3), we get: 3t + 1 So we have: 3600 = (r - 40)(3t + 1) Substitute r = 1200/t into the reduced speed equation: 3600 = (1200/t - 40)(3t + 1) Multiply through and we get: 3600 = 3600 - 120t + 1200/t - 40 Subtract 3,600 from each side 3600 - 3600 = 3600 - 3600 - 120t + 1200/t - 40 The 3600's cancel, so we get: - 120t + 1200/t - 40 = 0 Multiply each side by t: -120t^2 - 40t + 1200 = 0 We've got a quadratic equation. To solve for t, [URL='https://www.mathcelebrity.com/quadratic.php?num=-120t%5E2-40t%2B1200%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this in our search engine[/URL] and we get: t = -10/3 or t = 3. Since time [I]cannot[/I] be negative, our final answer is: [B]t = 3[/B]

If x = 2y/3 and y = 18, what is the value of 2x - 3? A) 21 B) 15 C) 12 D) 10 Substitute the values into the equation: 2(2y/3) - 3 <-- Given x = 2y/3 Simplifying, we have: 4y/3 - 3 Now substitute y = 18 into this: 4(18)/3 - 3 4(6) - 3 24 - 3 [B]21 or Answer A[/B]

In 2000 a company increased its workforce by 50%. In 2001 it decreased its workforce by 50%. How does the size of its workforce at the end of 2001 compare with the size of the workforce at the beginning of 2000? Let w be the size of the workforce before any changes. We have: [LIST] [*]w(2000) = w(1999) * 1.5 [I](50% increase is the same as multiplying by 1.5)[/I] [*]w(2001) = w(2000)/1.5 [I](50% decrease is the same as dividing by 1.5)[/I] [/LIST] Substitute the first equation back into the second equation w(2001) = w(1999) * 1.5/1.5 Cancel the 1.5 on top and bottom w(2001) = w(1999) This means the workforce had [B]zero net change[/B] from the beginning of 2000 to the end of 2001.

In a class there are 5 more boys than girls. There are 13 students in all. How many boys are there in the class? We start by declaring variables for boys and girls: [LIST] [*]Let b be the number of boys [*]Let g be the number of girls [/LIST] We're given two equations: [LIST=1] [*]b = g + 5 [*]b + g = 13 [/LIST] Substitute equation (1) for b into equation (2): g + 5 + g = 13 Grouping like terms, we get: 2g + 5 = 13 Subtract 5 from each side: 2g + 5 - 5 = 13 - 5 Cancel the 5's on the left side and we get: 2g = 8 Divide each side of the equation by 2 to isolate g: 2g/2 = 8/2 Cancel the 2's on the left side and we get: g = 4 Substitute g = 4 into equation (1) to solve for b: b = 4 + 5 b = [B]9[/B]

In a presidential election ohio had 20 electoral votes. This is 14 less than Texas had. How many electoral votes did Texas have? Let 0 = Ohio votes and t = Texas votes. We have: [LIST=1] [*]o = 20 [*]0 = t - 14 [/LIST] [U]Substitute (1) into (2)[/U] 20 = t - 14 [U]Add 14 to each side[/U] [B]t = 34[/B]

In an algebra test, the highest grade was 50 points higher than the lowest grade. The sum of the two grades was 180. Let the high grade be h and the low grade be l. We're given: [LIST=1] [*]h = l + 50 [*]h + l = 180 [/LIST] Substitute equation (1) into equation (2) for h l + 50 + l = 180 To solve this equation, [URL='https://www.mathcelebrity.com/1unk.php?num=l%2B50%2Bl%3D180&pl=Solve']we type it in our search engine[/URL] and we get: l = [B]65 [/B] Now, we take l = 65 and substitute it into equation (1) to solve for h: h = 65 + 50 h = [B]115[/B]

In Super Bowl XXXV, the total number of points scored was 41. The winning team outscored the losing team by 27 points. What was the final score of the game? In Super Bowl XXXV, the total number of points scored was 41. The winning team outscored the losing team by 27 points. What was the final score of the game? Let w be the winning team's points, and l be the losing team's points. We have two equations: [LIST=1] [*]w + l = 41 [*]w - l = 27 [/LIST] Add the two equations: 2w = 68 Divide each side by 2 [B]w = 34[/B] Substitute this into (1) 34 + l = 41 Subtract 34 from each side [B]l = 7[/B] Check your work: [LIST=1] [*]34 + 7 = 41 <-- check [*]34 - 7 = 27 <-- check [/LIST] The final score of the game was [B]34 to 7[/B]. You could also use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=w+%2B+l+%3D+41&term2=w+-+l+%3D+27&pl=Cramers+Method']simultaneous equation solver[/URL].

Jack bought 7 tickets for a movie. He paid $7 for each adult ticket and $4 for each child ticket. Jack spent $40 for the tickets Let a = Number of adult tickets and c be the number of child tickets. [LIST=1] [*]7a + 4c = 40 [*]a + c = 7 [*]Rearrange (2): a = 7 - c [/LIST] Now substitute a in (3) into (1): 7(7 - c) + 4c = 40 49 - 7c + 4c = 40 49 - 3c = 40 Add 3c to each side and subtract 40: 3c = 9 Divide each side by 3: [B]c = 3 [/B] Substitute c = 3 into Equation (2) a + 3 = 7 Subtract 3 from each side: [B]a = 4[/B]

James is four time as old as peter if their combined age is 30 how old is James. Let j be Jame's age. Let p be Peter's age. We're given: [LIST=1] [*]j = 4p [*]j + p = 30 [/LIST] Substitute (1) into (2) 4p + p = 30 Combine like terms: 5p = 30 [URL='https://www.mathcelebrity.com/1unk.php?num=5p%3D30&pl=Solve']Type 5p = 30 into our search engine[/URL], and we get p = 6. Plug p = 6 into equation (1) to get James's age, we get: j = 4(6) j = [B]24[/B]

Jason has an equal number of nickels and dimes. The total value of his nickels and dimes is $2.25. How many nickels does Jason have? Let the number of nickels be n Let the number of dimes be d We're given two equations: [LIST=1] [*]d = n [*]0.05n + 0.1d = 2.25 [/LIST] Substitute equation (1) for d into equation (2): 0.05n + 0.1n = 2.25 Solve for [I]n[/I] in the equation 0.05n + 0.1n = 2.25 [SIZE=5][B]Step 1: Group the n terms on the left hand side:[/B][/SIZE] (0.05 + 0.1)n = 0.15n [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 0.15n = + 2.25 [SIZE=5][B]Step 3: Divide each side of the equation by 0.15[/B][/SIZE] 0.15n/0.15 = 2.25/0.15 n = [B]15[/B] [URL='https://www.mathcelebrity.com/1unk.php?num=0.05n%2B0.1n%3D2.25&pl=Solve']Source[/URL]

Jennifer is twice as old as Peter. The difference between their ages is 15. What is Peters age Let j be Jennifer's age Let p be Peter's age We're given two equations: [LIST=1] [*]j = 2p [*]j - p = 15 [/LIST] Substitute equation (1) into equation (2) for j 2p - p = 15 To solve for p, we [URL='https://www.mathcelebrity.com/1unk.php?num=2p-p%3D15&pl=Solve']type this equation into our calculation engine[/URL] and we get: p = [B]15[/B]

Jenny threw the javelin 4 metres further than Angus but 5 metres less than Cameron. if the combined distance thrown by the 3 friends is 124 metres, how far did Angus throw the javelin? Assumptions and givens: [LIST] [*]Let a be the distance Angus threw the javelin [*]Let c be the distance Cameron threw the javelin [*]Let j be the distance Jenny threw the javelin [/LIST] We're given 3 equations: [LIST=1] [*]j = a + 4 [*]j = c - 5 [*]a + c + j = 124 [/LIST] Since j is the common variable in all 3 equations, let's rearrange equation (1) and equation (2) in terms of j as the dependent variable: [LIST=1] [*]a = j - 4 [*]c = j + 5 [*]a + c + j = 124 [/LIST] Now substitute equation (1) and equation (2) into equation (3) for a and c: j - 4 + j + 5 + j = 124 To solve this equation for j, we [URL='https://www.mathcelebrity.com/1unk.php?num=j-4%2Bj%2B5%2Bj%3D124&pl=Solve']type it in our math engine[/URL] and we get: j = 41 The question asks how far Angus (a) threw the javelin. Since we have Jenny's distance j = 41 and equation (1) has j and a together, let's substitute j = 41 into equation (1): a = 41 - 4 a = [B]37 meters[/B]

Jenny went shoe shopping. Now she has 5 more pairs than her brother. Together they have 25 pairs. How many pairs does Jenny have and how many pairs does her brother have? [U]Let j be the number of shoes Jenny has and b be the number of s hoes her brother has. Set up 2 equations:[/U] (1) b + j = 25 (2) j = b + 5 [U]Substitute (2) into (1)[/U] b + (b + 5) = 25 [U]Group the b terms[/U] 2b + 5 = 25 [U]Subtract 5 from each side[/U] 2b = 20 [U]Divide each side by b[/U] [B]b = 10 [/B] [U]Substitute b = 10 into (2)[/U] j = 10 + 5 [B]j = 15[/B]

Jim is 9 years older than June. Alex is 8 years younger than June. If the total of their ages is 82, how old is the eldest of them Let j be Jim's age, a be Alex's age, and u be June's age. We have 3 given equations: [LIST=1] [*]j + a + u = 82 [*]j = u + 9 [*]a = u - 8 [/LIST] Substitute (2) and (3) into (1) (u + 9) + (u - 8) + u = 82 Combine Like Terms: 3u + 1 = 82 [URL='https://www.mathcelebrity.com/1unk.php?num=3u%2B1%3D82&pl=Solve']Type this equation into the search engine[/URL], and we get u = 27. The eldest (oldest) of the 3 is Jim. So we have from equation (2) j = u + 9 j = 27 + 9 [B]j = 36[/B]

Joey and Romnick play in the same soccer team. Last Saturday, Romnick scored 3 more goals than Joey,but together they scored less than 9 goals. What are the possible number of goal Romnick scored? Let j be Joey's goals Let r by Romnick's goals We're given 1 equation and 1 inequality: [LIST=1] [*]r = j + 3 [*]r + j < 9 [/LIST] Rearranging equation 1 for j, we have: [LIST=1] [*]j = r - 3 [*]r + j < 9 [/LIST] Substitute equation (1) into inequality (2) for j: r + r - 3 < 9 2r - 3 < 9 [URL='https://www.mathcelebrity.com/1unk.php?num=2r-3%3C9&pl=Solve']Typing this inequality into our math engine[/URL], we get: [B]r < 6[/B]

JP's age is twice the age of Reyna. The sum of their ages does not exceed 51 Let JP's age be j. Let Reyna's age be r. We're given two expressions: [LIST=1] [*]w = 2r [*]r + w <= 51. ([I]Does not exceed means less than or equal to)[/I] [/LIST] We substitute (1) into (2) for w to get the inequality: r + 2r <= 51 To solve this inequality, we type it in our search engine and we get: [B]r <= 17[/B]

Juan spent at most $2.50 on apples and oranges. He bought 5 apples at $0.36 each. What is the most he spent on oranges? Let a be spending apples and o be spending on oranges, we have: [LIST=1] [*]a + o <= 2.36 <-- At most means less than or equal to [*]a = 5 * 0.36 = 1.8 [/LIST] Substitute (2) into (1) 1.8 + o <= 2.36 Subtract 1.8 from each side [B]o <= 0.56[/B]

kate is twice as old as her sister mars. the sum of their ages is 24. find their ages. Let k be Kate's age Let m be Mars's age We're given two equations: [LIST=1] [*]k = 2m. (Because twice means multiply by 2) [*]k + m = 24 [/LIST] Substitute equation (1) for k into equation (2): 2m + m = 24 T o solve for m, we [URL='https://www.mathcelebrity.com/1unk.php?num=2m%2Bm%3D24&pl=Solve']type this equation into our math engine[/URL]: m = [B]8 [/B] We want to solve for k using m= 8. Substitute this into equation 1 k = 2(8) k = [B]16 [/B] Check our work for equation 1 16 = 2 * 8 16 = 16 Check our work for equation 2 16 + 8 ? 24 24 = 24 [MEDIA=youtube]TJMTRYP-Ct8[/MEDIA]

Kate spent 1 more than Lauren, and together they spent 5. Let k be the amount Kate spent, and l be the amount Lauren spent. We're given: [LIST=1] [*]k = l + 1 [*]k + l = 5 [/LIST] Substitute (1) into (2): (l + 1) + l = 5 Group like terms 2l + 1 = 5 [URL='https://www.mathcelebrity.com/1unk.php?num=2l%2B1%3D5&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]l = 2[/B] Plug this into Equation (1), we get: k = 2 + 1 [B]k = 3 [/B] Kate Spent 3, and Lauren spent 2

Let k = Katie's age and m = Mara's age. We have 2 equations: (1) k = 2m (2) k + m = 24 Substitute (1) into (2) (2m) + m = 24 Combine like terms: 3m = 24 Divide each side of the equation by 3 to isolate m m = 8 If m = 8, substituting into (1) or (2), we get k = 16. [MEDIA=youtube]Cu7gSgNkQPg[/MEDIA]

Kendra has $5.70 in quarters and nickels. If she has 12 more quarters than nickels, how many of each coin does she have? Let n be the number of nickels and q be the number of quarters. We have: [LIST=1] [*]q = n + 12 [*]0.05n + 0.25q = 5.70 [/LIST] Substitute (1) into (2) 0.05n + 0.25(n + 12) = 5.70 0.05n + 0.25n + 3 = 5.70 Combine like terms: 0.3n + 3 = 5.70 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=0.3n%2B3%3D5.70&pl=Solve']equation calculator[/URL], we get [B]n = 9[/B]. Substituting that back into (1), we get: q = 9 + 12 [B]q = 21[/B]

Kendra is half as old as Morgan and 3 years younger than Lizzie. The total of their ages is 39. How old are they? Let k be Kendra's age, m be Morgan's age, and l be Lizzie's age. We're given: [LIST=1] [*]k = 0.5m [*]k = l - 3 [*]k + l + m = 39 [/LIST] Rearranging (1) by multiplying each side by 2, we have: m = 2k Rearranging (2) by adding 3 to each side, we have: l = k + 3 Substituting these new values into (3), we have: k + (k + 3) + (2k) = 39 Group like terms: (k + k + 2k) + 3 = 39 4k + 3 = 39 [URL='https://www.mathcelebrity.com/1unk.php?num=4k%2B3%3D39&pl=Solve']Type this equation into the search engine[/URL], and we get: [B]k = 9 [/B] Substitute this back into (1), we have: m = 2(9) [B]m = 18 [/B] Substitute this back into (2), we have: l = (9) + 3 [B][B]l = 12[/B][/B]

Kerry asked a bank teller to cash 390 check using 20 bills and 50 bills. If the teller gave her a total of 15 bills, how many of each type of bill did she receive? Let t = number of 20 bills and f = number of 50 bills. We have two equations. (1) 20t + 50f = 390 (2) t + f = 15 [U]Rearrange (2) into (3) for t, by subtracting f from each side:[/U] (3) t = 15 - f [U]Now substitute (3) into (1)[/U] 20(15 - f) + 50f = 390 300 - 20f + 50f = 390 [U]Combine f terms[/U] 300 + 30f = 390 [U]Subtract 300 from each side[/U] 30f = 90 [U]Divide each side by 30[/U] [B]f = 3[/B] [U]Substitute f = 3 into (3)[/U] t = 15 - 3 [B]t = 12[/B]

Kevin and randy have a jar containing 41 coins, all of which are either quarters or nickels. The total value of the jar is $7.85. How many of each type? Let d be dimes and q be quarters. Set up two equations from our givens: [LIST=1] [*]d + q = 41 [*]0.1d + 0.25q = 7.85 [/LIST] [U]Rearrange (1) by subtracting q from each side:[/U] (3) d = 41 - q [U]Now, substitute (3) into (2)[/U] 0.1(41 - q) + 0.25q = 7.85 4.1 - 0.1q + 0.25q = 7.85 [U]Combine q terms[/U] 0.15q + 4.1 = 7.85 [U]Using our [URL='http://www.mathcelebrity.com/1unk.php?num=0.15q%2B4.1%3D7.85&pl=Solve']equation calculator[/URL], we get:[/U] [B]q = 25[/B] [U]Substitute q = 25 into (3)[/U] d = 41 - 25 [B]d = 16[/B]

Kevin ran 4 miles more than Steve ran. The sum of their distances is 26 miles. How far did Steve run? The domain of the solution is: Let k be Kevin's miles ran Let s be Steve's miles ran We have 2 given equtaions: [LIST=1] [*]k = s + 4 [*]k + s = 26 [/LIST] Substitute (1) into (2) (s + 4) + s = 26 2s + 4 = 26 Plug this into our [URL='http://www.mathcelebrity.com/1unk.php?num=2s%2B4%3D26&pl=Solve']equation calculator[/URL] and we get s = 11

Kiko is now 6 times as old as his sister. In 6 years, he will be 3 times as old as his sister. What is their present age? Let k be Kiko's present age Let s be Kiko's sisters age. We're given two equations: [LIST=1] [*]k = 6s [*]k + 6 = 3(s + 6) [/LIST] To solve this system of equations, we substitute equation (1) into equation (2) for k: 6s + 6 = 3(s + 6) [URL='https://www.mathcelebrity.com/1unk.php?num=6s%2B6%3D3%28s%2B6%29&pl=Solve']Typing this equation into our math engine[/URL] to solve for s, we get: s = [B]4[/B] To solve for k, we substitute s = 4 into equation (1) above: k = 6 * 4 k = [B]24[/B]

larger of 2 numbers is 12 more than the smaller number. if the sum of the 2 numbers is 74 find the 2 numbers Declare Variables for each number: [LIST] [*]Let l be the larger number [*]Let s be the smaller number [/LIST] We're given two equations: [LIST=1] [*]l = s + 12 [*]l + s = 74 [/LIST] Equation (1) already has l solved for. Substitute equation (1) into equation (2) for l: s + 12 + s = 74 Solve for [I]s[/I] in the equation s + 12 + s = 74 [SIZE=5][B]Step 1: Group the s terms on the left hand side:[/B][/SIZE] (1 + 1)s = 2s [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 2s + 12 = + 74 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 12 and 74. To do that, we subtract 12 from both sides 2s + 12 - 12 = 74 - 12 [SIZE=5][B]Step 4: Cancel 12 on the left side:[/B][/SIZE] 2s = 62 [SIZE=5][B]Step 5: Divide each side of the equation by 2[/B][/SIZE] 2s/2 = 62/2 s = [B]31[/B] To solve for l, we substitute in s = 31 into equation (1): l = 31 + 12 l = [B]43[/B]

larger of 2 numbers is 4 more than the smaller. the sum of the 2 is 40. what is the larger number? Declare variables for the 2 numbers: [LIST] [*]Let l be the larger number [*]Let s be the smaller number [/LIST] We're given two equations: [LIST=1] [*]l = s + 4 [*]l + s = 40 [/LIST] To get this problem in terms of the larger number l, we rearrange equation (1) in terms of l. Subtract 4 from each side in equation (1) l - 4 = s + 4 - 4 Cancel the 4's and we get: s = l - 4 Our given equations are now: [LIST=1] [*]s = l - 4 [*]l + s = 40 [/LIST] Substitute equation (1) into equation (2) for s: l + l - 4 = 40 Grouping like terms for l, we get: 2l - 4 = 40 Add 4 to each side: 2l - 4 + 4 = 40 + 4 Cancelling the 4's on the left side, we get 2l = 44 Divide each side of the equation by 2 to isolate l: 2l/2 = 44/2 Cancel the 2's on the left side and we get: l = [B]22[/B]

Laura weighs 45 pounds more than her pet dog. When they are on the scale together, they weigh 85 pounds. How much does Laura weigh? Let Laura weigh l and her dog weigh d. WE have: [LIST=1] [*]l = d + 45 [*]d + l = 85 [/LIST] Substitute equation (1) into Equation (2) for l: d + d + 45 = 85 Solve for [I]d[/I] in the equation d + d + 45 = 85 [SIZE=5][B]Step 1: Group the d terms on the left hand side:[/B][/SIZE] (1 + 1)d = 2d [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 2d + 45 = + 85 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 45 and 85. To do that, we subtract 45 from both sides 2d + 45 - 45 = 85 - 45 [SIZE=5][B]Step 4: Cancel 45 on the left side:[/B][/SIZE] 2d = 40 [SIZE=5][B]Step 5: Divide each side of the equation by 2[/B][/SIZE] 2d/2 = 40/2 d = 20 From equation (1), we substitute d = 20: l = d + 45 l = 20 + 45 l = [B]65 pounds [URL='https://www.mathcelebrity.com/1unk.php?num=d%2Bd%2B45%3D85&pl=Solve']Source[/URL][/B]

Length (l) is the same as width (w) and their product is 64. We're given 2 equations: [LIST=1] [*]lw = 64 [*]l = w [/LIST] Substitute equation (2) into equation (1): w * w = 64 w^2 = 64 [B]w = 8[/B] Since l = w, then [B]l = 8[/B]

Liz harold has a jar in her office that contains 47 coins. Some are pennies and the rest are dimes. If the total value of the coins is 2.18, how many of each denomination does she have? [U]Set up two equations where p is the number of pennies and d is the number of dimes:[/U] (1) d + p = 47 (2) 0.1d + 0.01p = 2.18 [U]Rearrange (1) into (3) by solving for d[/U] (3) d = 47 - p [U]Substitute (3) into (2)[/U] 0.1(47 - p) + 0.01p = 2.18 4.7 - 0.1p + 0.01p = 2.18 [U]Group p terms[/U] 4.7 - 0.09p = 2.18 [U]Add 0.09p to both sides[/U] 0.09p + 2.18 = 4.7 [U]Subtract 2.18 from both sides[/U] 0.09p = 2.52 [U]Divide each side by 0.09[/U] [B]p = 28[/B] [U]Now substitute that back into (3)[/U] d =47 - 28 [B]d = 19[/B]

Lorda is older than Kate. The sum of their ages is 30. The difference in their ages is 6. What are their ages? Let Lorda's age be l. Let Kate's age be k. We're given two equations: [LIST=1] [*]l + k = 30 [*]l - k = 6 <-- Since Lorda is older [/LIST] Add the 2 equations together and we eliminate k: 2l = 36 [URL='https://www.mathcelebrity.com/1unk.php?num=2l%3D36&pl=Solve']Typing this equation into our search engine[/URL] and solving for l, we get: l = [B]18[/B] Now substitute l = 18 into equation 1: 18 + k = 30 [URL='https://www.mathcelebrity.com/1unk.php?num=18%2Bk%3D30&pl=Solve']Type this equation into our search engine[/URL] and solving for k, we get: k = [B]12[/B]

Luke and Dan's total debt is $72. If Luke's debt is three times Dan's debt, what is Dan's debt? Let Dan's debt be d. Let Luke's debt be l. We're given two equations: [LIST=1] [*]d + l = 72 [*]l = 3d [/LIST] Substitute equation (2) for l into equation (1): d + 3d = 72 Solve for [I]d[/I] in the equation d + 3d = 72 [SIZE=5][B]Step 1: Group the d terms on the left hand side:[/B][/SIZE] (1 + 3)d = 4d [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 4d = + 72 [SIZE=5][B]Step 3: Divide each side of the equation by 4[/B][/SIZE] 4d/4 = 72/4 d = [B]18[/B]

Mark and Jennie are bowling. Jennie’s score is double Mark’s score. If the sum of their score is 171, find each person’s score by writing out an equation. Let Mark's score be m. Let Jennie's score be j. We're given two equations: [LIST=1] [*]j = 2m [*]j + m = 171 [/LIST] Substitute equation (1) into equation (2): 2m + m = 171 [URL='https://www.mathcelebrity.com/1unk.php?num=2m%2Bm%3D171&pl=Solve']Type this equation into our search engine[/URL] to solve for m: m = [B]57 [/B] To solve for j, we substitute m = 57 in equation (1) above: j = 2(57) j = [B]114[/B]

Martha is 18 years older than Harry. Their ages add to 106. Write an equation and solve it to find the ages of Martha and Harry. Let m be Martha's age. Let h be Harry's age. We're given two equations: [LIST=1] [*]m = h + 18 [I](older means we add)[/I] [*]h + m = 106 [/LIST] Substitute equation (1) into equation (2) for m: h + h + 18 = 106 To solve for h, [URL='https://www.mathcelebrity.com/1unk.php?num=h%2Bh%2B18%3D106&pl=Solve']we type this equation into our search engine[/URL] and we get: h = [B]44[/B]

Max and Bob went to the store. Max bought 2 burgers and 2 drinks for $5.00. Bob bought 3 burgers and 1 drink for $5.50. How much is each burger and drink? [U]Set up the givens where b is the cost of a burger and d is the cost of a drink:[/U] Max: 2b + 2d = 5 Bob: 3b + d = 5.50 [U]Rearrange Bob's equation by subtracting 3b from each side[/U] (3) d = 5.50 - 3b [U]Now substitute that d equation back into Max's Equation[/U] 2b + 2(5.50 - 3b) = 5 2b + 11 - 6b = 5 [U]Combine b terms:[/U] -4b + 11 = 5 [U]Subtract 11 from each side[/U] -4b = -6 [U]Divide each side by -4[/U] b = 3/2 [B]b = $1.50[/B] [U]Now plug that back into equation (3):[/U] d = 5.50 - 3(1.50) d = 5.50 - 4.50 [B]d = $1.00[/B]

Max is 23 years younger than his father.Together their ages add up to 81. Let Max's age be m, and his fathers' age be f. We're given: [LIST=1] [*]m = f - 23 <-- younger means less [*]m + f = 81 [/LIST] Substitute Equation (1) into (2): (f - 23) + f = 81 Combine like terms to form the equation below: 2f - 23 = 81 [URL='https://www.mathcelebrity.com/1unk.php?num=2f-23%3D81&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]f = 52[/B] Substitute this into Equation (1): m = 52 - 23 [B]m = 29[/B]

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water as juice. How many ounces of strawberry juice and how many ounces of water does she need to make 40 ounces of strawberry infused water? Let j be the ounces of strawberry juice and w be the ounces of water. We're given: [LIST=1] [*]j + w = 40 [*]w = 3j [/LIST] Substitute (2) into (1): j + 3j = 40 Combine like terms: 4j = 40 [URL='https://www.mathcelebrity.com/1unk.php?num=4j%3D40&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]j = 10[/B] From equation (2), we substitute j = 2: w = 3(10) [B]w = 30 [/B] This means we have [B]10 ounces of juice[/B] and [B]30 ounces of water[/B] for a 40 ounce mix.

Mr turner sent his car to the workshop for repair work as well as to change 4 tires. Mr turner paid $1035 in all. The repair work cost 5 times the price of each tire. The mechanic told Mr. turner that the repair work cost $500. Explain the mechanic’s mistake Let the cost for work be w. Let the cost for each tire be t. We're given; [LIST=1] [*]w = 5t [*]w + 4t = 1035 [/LIST] Substitute equation 1 into equation 2: (5t) + 4t = 1035 [URL='https://www.mathcelebrity.com/1unk.php?num=%285t%29%2B4t%3D1035&pl=Solve']Type this equation into our search engine[/URL], and we get: t = 115 Substitute this into equation (1): w = 5(115) w = [B]575[/B] The mechanic underestimated the work cost.

Mr. Wilson wants to park his carin a parking garage that charges 3 per hour along with a flat fee of 6. If Mr. Wilson paid 54 to park in the garage, for how many hours did he park there? [U]Set up an equation, where f is the flat fee, and h is the number of hours parked:[/U] 3h + f = 54 [U]Substitute f = 6 into the equation:[/U] 3h + 6 = 54 [U]Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3h%2B6%3D54&pl=Solve']equation solver[/URL], we get[/U] [B]h = 16[/B]

Mr. Winkle downloaded 34 more songs than Mrs. Winkle downloaded. Together they downloaded 220 songs. How many songs did each download? Let x = Mr. Winkle downloads and y = Mrs. Winkle downloads. We then have x = y + 34 and x + y = 220. Substitute equation 1 into equation 2, we have: (y + 34) + y = 220 2y + 34 = 220 Subtract 34 from each side: 2y = 186 Divide each side by 2: y = 93 (Mrs. Winkle) x = 93 + 34 x = 127 (Mr. Winkle)

Ms. Jeffers is splitting $975 among her three sons. If the oldest gets twice as much as the youngest and the middle son gets $35 more than the youngest, how much does each boy get? Let 0 be the oldest son, m be the middle sun, and y be the youngest son. Set up our given equations [LIST] [*]o = 2y [*]m = y + 35 [*]o + m + y = 975 [/LIST] [U]Substitute the first and second equations into Equation 3[/U] 2y + y + 35 + y = 975 [U]Combine the y terms[/U] 4y + 35 = 975 Subtract 35 using our [URL='http://www.mathcelebrity.com/1unk.php?num=4y%2B35%3D975&pl=Solve']equation calculator[/URL] to solve and get [B]y = 235[/B] [U]Plug y = 235 into equation 2[/U] m = 235 + 35 [B]m = 270[/B] [U]Plug y = 235 into equation 2[/U] o = 2(235) [B]o = 470[/B]

n and m are congruent and supplementary. prove n and m are right angles Given: [LIST] [*]n and m are congruent [*]n and m are supplementary [/LIST] If n and m are supplementary, that means we have the equation: m + n = 180 We're also given n and m are congruent, meaning they are equal. So we can substitute n = m into the supplementary equation: m + m = 180 To solve this equation for m, [URL='https://www.mathcelebrity.com/1unk.php?num=m%2Bm%3D180&pl=Solve']we type it in our search engine[/URL] and we get: m = 90 This means m = 90, n = 90, which means they are both right angles since by definition, a right angle is 90 degrees.

Nava is 17 years older than Edward. the sum of Navas age and Edwards ages id 29. How old is Nava? Let Nava's age be n and Edward's age be e. We have 2 equations: [LIST=1] [*]n = e + 17 [*]n + e = 29 [/LIST] Substitute (1) into (2) (e + 17) + e = 29 Group like terms: 2e + 17 = 29 Running this equation [URL='http://www.mathcelebrity.com/1unk.php?num=2e%2B17%3D29&pl=Solve']through our search engine[/URL], we get: e = 6 Substitute this into equation (1) n = 6 + 17 [B]n = 23[/B]

Nicole is half as old as Donald. The sum of their ages is 72. How old is Nicole in years? Let n be Nicole's age. Let d be Donald's age. We're given two equations: [LIST=1] [*]n = 0.5d [*]n + d = 72 [/LIST] Substitute equation (1) into (2): 0.5d + d = 72 1.5d = 72 [URL='https://www.mathcelebrity.com/1unk.php?num=1.5d%3D72&pl=Solve']Typing this equation into the search engine and solving for d[/URL], we get: d = [B]48[/B]

numerator of a fraction is 5 less than its denominator. if 1 is added to the numerator and to the denominator the new fraction is 2/3. find the fraction. Let n be the numerator. Let d be the denominator. We're given 2 equations: [LIST=1] [*]n = d - 5 [*](n + 1)/(d + 1) = 2/3 [/LIST] Substitute equation (1) into equation (2) for n: (d - 5 + 1) / (d + 1) = 2/3 (d - 4) / (d + 1) = 2/3 Cross multiply: 3(d - 4) = 2(d + 1) To solve this equation for d, we type it in our search engine and we get: d = 14 Substitute d = 14 into equation (1) to solve for n: n = 14 - 5 n = 9 Therefore, our fraction n/d is: [B]9/14[/B]

On an algebra test, the highest grade was 42 points higher than the lowest grade. The sum of the two grades was 138. Find the lowest grade. [U]Let h be the highest grade and l be the lowest grade. Set up the given equations:[/U] (1) h = l + 42 (2) h + l = 138 [U]Substitute (1) into (2)[/U] l + 42 + l = 138 [U]Combine l terms[/U] 2l + 42 = 138 [U]Enter that equation into our [URL='http://www.mathcelebrity.com/1unk.php?num=2l%2B42%3D138&pl=Solve']equation calculator[/URL] to get[/U] [B]l = 48 [/B] [U]Substitute l = 48 into (1)[/U] h = 48 + 42 [B]h = 90[/B]

One number is 1/4 of another number. The sum of the two numbers is 25. Find the two numbers. Use a comma to separate your answers. Let the first number be x and the second number be y. We're given: [LIST=1] [*]x = 1/4y [*]x + y = 25 [/LIST] Substitute (1) into (2) 1/4y + y = 25 Since 1/4 = 0.25, we have: 0.25y + y = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=0.25y%2By%3D25&pl=Solve']Type this equation into the search engine[/URL] to get: [B]y = 20 [/B] Now, substitute this into (1) to solve for x: x = 1/4y x = 1/4(20) [B]x = 5 [/B] The problem asks us to separate the answers by a comma. So we write this as: [B](x, y) = (5, 20)[/B]

One number is 1/5 of another number. The sum of the two numbers is 18. Find the two numbers. Let the two numbers be x and y. We're given: [LIST=1] [*]x = 1/5y [*]x + y = 18 [/LIST] Substitute (1) into (2): 1/5y + y = 18 1/5 = 0.2, so we have: 1.2y = 18 [URL='https://www.mathcelebrity.com/1unk.php?num=1.2y%3D18&pl=Solve']Type 1.2y = 18 into the search engine[/URL], and we get [B]y = 15[/B]. Which means from equation (1) that: x = 15/5 [B]x = 3 [/B] Our final answer is [B](x, y) = (3, 15)[/B]

One number is 3 times another. Their sum is 44. Let the first number be x, and the second number be y. We're given: [LIST=1] [*]x = 3y [*]x + y = 44 [/LIST] Substitute (1) into (2): 3y + y = 44 [URL='https://www.mathcelebrity.com/1unk.php?num=3y%2By%3D44&pl=Solve']Type this equation into the search engine[/URL], and we get: [B]y = 11[/B] Plug this into equation (1): x = 3(11) [B]x = 33[/B]

one number is 3 times as large as another. Their sum is 48. Find the numbers Let the first number be x. Let the second number be y. We're given two equations: [LIST=1] [*]x = 3y [*]x + y = 48 [/LIST] Substitute equation (1) into equation (2): 3y + y = 48 To solve for y, [URL='https://www.mathcelebrity.com/1unk.php?num=3y%2By%3D48&pl=Solve']we type this equation into the search engine[/URL] and we get: [B]y = 12[/B] Now, plug y = 12 into equation (1) to solve for x: x = 3(12) [B]x = 36[/B]

One number is 8 times another number. The numbers are both positive and have a difference of 70. Let the first number be x, the second number be y. We're given: [LIST=1] [*]x = 8y [*]x - y = 70 [/LIST] Substitute(1) into (2) 8y - y = 70 [URL='https://www.mathcelebrity.com/1unk.php?num=8y-y%3D70&pl=Solve']Plugging this equation into our search engine[/URL], we get: [B]y = 10[/B] <-- This is the smaller number Plug this into Equation (1), we get: x = 8(10) [B]x = 80 [/B] <-- This is the larger number

one number is twice a second number. the sum of those numbers is 45. Let the first number be x and the second number be y. We're given: [LIST=1] [*]x = 2y [*]x + y = 45 [/LIST] Substitute Equation (1) into Equation (2): 2y + y = 45 [URL='https://www.mathcelebrity.com/1unk.php?num=2y%2By%3D45&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]y = 15[/B] Plug this into equation (1) to solve for x, and we get: x = 2(15) [B]x = 30[/B]

One positive number is one-fifth of another number. The difference between the two numbers is 192, find the numbers. Let the first number be x and the second number be y. We're given two equations: [LIST=1] [*]x = y/5 [*]x + y = 192 [/LIST] Substitute equation 1 into equation 2: y/5 + y = 192 Since 1 equals 5/5, we rewrite our equation like this: y/5 = 5y/5 = 192 We have fractions with like denominators, so we add the numerators: (1 + 5)y/5 = 192 6y/5 = 192 [URL='https://www.mathcelebrity.com/prop.php?num1=6y&num2=192&den1=5&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']Type this equation into our search engine[/URL], and we get: [B]y = 160[/B] Substitute this value into equation 1: x = 160/5 x = [B]32[/B]

Let x be the first number, y be the second number, and z be the number. We have the following equations: [LIST=1] [*]x + y + z = 305 [*]x = y - 5 [*]z = 3y [/LIST] Substitute (2) and (3) into (1) (y - 5) + y + (3y) = 305 Combine like terms: 5y - 5 = 305 Use our [URL='http://www.mathcelebrity.com/1unk.php?num=5y-5%3D305&pl=Solve']equation solver[/URL] [B]y = 62 [/B] Substitute y = 62 into (3) z = 3(62) [B]z = 186 [/B] x = (62) - 5 [B]x = 57[/B]

A Web music store offers two versions of a popular song. The size of the standard version is 2.7 megabytes (MB). The size of the high-quality version is 4.7 MB. Yesterday, the high-quality version was downloaded three times as often as the standard version. The total size downloaded for the two versions was 4200 MB. How many downloads of the standard version were there? Let s be the standard version downloads and h be the high quality downloads. We have two equations: [LIST=1] [*]h = 3s [*]2.7s + 4.7h = 4200 [/LIST] Substitute (1) into (2) 2.7s + 4.7(3s) = 4200 2.7s + 14.1s = 4200 Combine like terms: 16.8s = 4200 Divide each side by 16.8 [B]s = 250[/B]

Prove 0! = 1 Let n be a whole number, where n! represents the product of n and all integers below it through 1. The factorial formula for n is: n! = n · (n - 1) * (n - 2) * ... * 3 * 2 * 1 Written in partially expanded form, n! is: n! = n * (n - 1)! [U]Substitute n = 1 into this expression:[/U] n! = n * (n - 1)! 1! = 1 * (1 - 1)! 1! = 1 * (0)! For the expression to be true, 0! [U]must[/U] equal 1. Otherwise, 1! <> 1 which contradicts the equation above

[URL='https://www.mathcelebrity.com/proofs.php?num=prove0%21%3D1&pl=Prove']Prove 0! = 1[/URL] Let n be a whole number, where n! represents: The product of n and all integers below it through 1. The factorial formula for n is n! = n · (n - 1) · (n - 2) · ... · 3 · 2 · 1 Written in partially expanded form, n! is: n! = n · (n - 1)! [SIZE=5][B]Substitute n = 1 into this expression:[/B][/SIZE] n! = n · (n - 1)! 1! = 1 · (1 - 1)! 1! = 1 · (0)! For the expression to be true, 0! [U]must[/U] equal 1. Otherwise, 1! ? 1 which contradicts the equation above [MEDIA=youtube]wDgRgfj1cIs[/MEDIA]

Prove the following statement for non-zero integers a, b, c, If a divides b and b divides c, then a divides c. If an integer a divides an integer b, then we have: b = ax for some non-zero integer x If an integer b divides an integer c, then we have: c = by for some non-zero integer y Since b = ax, we substitute this into c = by for b: c = axy We can write this as: c = a(xy) [LIST] [*]Since x and y are integers, then xy is also an integer. [*]Therefore, c is the product of some integer multiplied by a [*]This means a divides c [/LIST] [MEDIA=youtube]VUIUFAFFVU4[/MEDIA]

Richard is thrice as old as Alvin. The sum of their ages is 52 years. Find their ages. Let r be Richard's age. And a be Alvin's age. We have: [LIST=1] [*]r = 3a [*]a + r = 52 [/LIST] Substitute (1) into (2) a + 3a = 52 Group like terms: 4a = 52 [URL='https://www.mathcelebrity.com/1unk.php?num=4a%3D52&pl=Solve']Typing this into the search engine[/URL], we get [B]a = 13[/B]. This means Richard is 3(13) = [B]39[/B]

Rigby and Eleanor's combined score on the systems of equations test was 181. Rigby scored 9 more points than Eleanor. What were Eleanor and Rigby's scores? Let Rigby's score be r Let Eleanor's score be e We're given two equations: [LIST=1] [*]r = e + 9 [*]e + r = 181 [/LIST] Substitute equation (1) into equation (2): e + (e + 9) = 181 Group like terms: 2e + 9 = 181 To solve this equation for e, we [URL='https://www.mathcelebrity.com/1unk.php?num=2e%2B9%3D181&pl=Solve']type it in our search engine[/URL] and we get: e = [B]86[/B]

Sadie and Connor both play soccer. Connor scored 2 times as many goals as Sadie. Together they scored 9 goals. Could Sadie have scored 4 goals? Why or why not? [U]Assumptions:[/U] [LIST] [*]Let Connor's goals be c [*]Let Sadie's goals be s [/LIST] We're given the following simultaneous equations: [LIST=1] [*]c = 2s [*]c + s = 9 [/LIST] We substitute equation (1) into equation (2) for c: 2s + s = 9 To solve the equation for s, we type it in our search equation and we get: s = [B]3[/B] So [U][B]no[/B][/U], Sadie could not have scored 4 goals since s = 3

Sally and Adam works a different job. Sally makes $5 per hour and Adam makes $4 per hour. They each earn the same amount per week but Adam works 2 more hours. How many hours a week does Adam work? [LIST] [*]Let [I]s[/I] be the number of hours Sally works every week. [*]Let [I]a[/I] be the number of hours Adam works every week. [*]We are given: a = s + 2 [/LIST] Sally's weekly earnings: 5s Adam's weekly earnings: 4a Since they both earn the same amount each week, we set Sally's earnings equal to Adam's earnings: 5s = 4a But remember, we're given a = s + 2, so we substitute this into Adam's earnings: 5s = 4(s + 2) Multiply through on the right side: 5s = 4s + 8 <-- [URL='https://www.mathcelebrity.com/expand.php?term1=4%28s%2B2%29&pl=Expand']multiplying 4(s + 2)[/URL] [URL='https://www.mathcelebrity.com/1unk.php?num=5s%3D4s%2B8&pl=Solve']Typing this equation into the search engine[/URL], we get s = 8. The problem asks for Adam's earnings (a). We plug s = 8 into Adam's weekly hours: a = s + 2 a = 8 + 2 [B]a = 10[/B]

Sally is 4 years older than Mark. Twice Sally's age plus 5 times Mark's age is equal to 64. Let Sally's age be s. Let Mark's age be m. We're given two equations: [LIST=1] [*]s = m + 4 [*]2s + 5m = 64 <-- [I]Since Twice means we multiply by 2[/I] [/LIST] Substitute equation (1) into equation (2): 2(m + 4) + 5m = 64 Multiply through: 2m + 8 + 5m = 64 Group like terms: (2 + 5)m + 8 = 64 7m + 8 = 64 [URL='https://www.mathcelebrity.com/1unk.php?num=7m%2B8%3D64&pl=Solve']Type this equation into the search engine[/URL] and we get: m = [B]8[/B]

Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which equation could be used to find Jeremy's age, j, if he is the younger man. Let Sam's age be s. Let' Jeremy's age be j. We're given: [LIST=1] [*]s = j + 2 <-- consecutive odd integers [*]sj = 783 [/LIST] Substitute (1) into (2): (j + 2)j = 783 j^2 + 2j = 783 Subtract 783 from each side: j^2 + 2j - 783 = 0 <-- This is the equation to find Jeremy's age. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=j%5E2%2B2j-783%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this quadratic equation into the search engine[/URL] and get: j = 27, j = -29. Since ages cannot be negative, we have: [B]j = 27[/B]

She earns $20 per hour as a carpenter and $25 per hour as a blacksmith, last week Giselle worked both jobs for a total of 30 hours, and a total of $690. How long did Giselle work as a carpenter and how long did she work as a blacksmith? Assumptions: [LIST] [*]Let b be the number of hours Giselle worked as a blacksmith [*]Let c be the number of hours Giselle worked as a carpenter [/LIST] Givens: [LIST=1] [*]b + c = 30 [*]25b + 20c = 690 [/LIST] Rearrange equation (1) to solve for b by subtracting c from each side: [LIST=1] [*]b = 30 - c [*]25b + 20c = 690 [/LIST] Substitute equation (1) into equation (2) for b 25(30 - c) + 20c = 690 Multiply through: 750 - 25c + 20c = 690 To solve for c, we [URL='https://www.mathcelebrity.com/1unk.php?num=750-25c%2B20c%3D690&pl=Solve']type this equation into our search engine[/URL] and we get: c = [B]12 [/B] Now, we plug in c = 12 into modified equation (1) to solve for b: b = 30 - 12 b = [B]18[/B]

Sheila wants build a rectangular play space for her dog. She has 100 feet of fencing and she wants it to be 5 times as long as it is wide. What dimensions should the play area be? Sheila wants: [LIST=1] [*]l =5w [*]2l + 2w = 100 <-- Perimeter [/LIST] Substitute (1) into (2) 2(5w) + 2w = 100 10w + 2w = 100 12w = 100 Divide each side by 12 [B]w = 8.3333[/B] Which means l = 5(8.3333) -->[B] l = 41.6667[/B]

Sherry is 31 years younger than her mom. The sum of their ages is 61. How old is Sherry? Let Sherry's age be s. Let the mom's age be m. We're given two equations: [LIST=1] [*]s = m - 31 [*]m + s = 61 [/LIST] Substitute equation (1) into equation (2) for s: m + m - 31 = 61 To solve for m, [URL='https://www.mathcelebrity.com/1unk.php?num=m%2Bm-31%3D61&pl=Solve']we type this equation into our search engine[/URL] and we get: m = 46 Now, we plug m = 46 into equation (1) to find Sherry's age s: s = 46 - 31 s = [B]15[/B]

Stanley bought a ruler and a yardstick for $1.25. If the yardstick cost 45 cents more than the ruler, what was the cost of the yardstick? Let r be the cost of the ruler Let y be the cost of the yardstick We're given 2 equations: [LIST=1] [*]r + y = 1.25 [*]y = r + 0.45 [/LIST] Substitute equation (2) into equation (1) for y r + r + 0.45 = 1.25 Solve for [I]r[/I] in the equation r + r + 0.45 = 1.25 [SIZE=5][B]Step 1: Group the r terms on the left hand side:[/B][/SIZE] (1 + 1)r = 2r [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 2r + 0.45 = + 1.25 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 0.45 and 1.25. To do that, we subtract 0.45 from both sides 2r + 0.45 - 0.45 = 1.25 - 0.45 [SIZE=5][B]Step 4: Cancel 0.45 on the left side:[/B][/SIZE] 2r = 0.8 [SIZE=5][B]Step 5: Divide each side of the equation by 2[/B][/SIZE] 2r/2 = 0.8/2 r = 0.4 Substitute r = 0.4 into equation (2) above: y = r + 0.45 y = 0.4 + 0.45 r = [B]0.85 [URL='https://www.mathcelebrity.com/1unk.php?num=r%2Br%2B0.45%3D1.25&pl=Solve']Source[/URL][/B]

Substitute the given values into given formula and solve for the unknown variable. S = 4LW + 2 WH; S= 144, L= 8, W= 4. H= S = 4LW + 2 WH Substituting our given values, we have: 144 = 4(8)(4) + 2(4)H 144 = 128 + 8H Using our [URL='http://www.mathcelebrity.com/1unk.php?num=128%2B8h%3D144&pl=Solve']equation calculator[/URL], we get: [B]H = 2[/B]

The basketball team is selling candy as a fundraiser. A regular candy bar cost 0.75 and a king sized candy bar costs 1.50. In the first week of the sales the team made 36.00. Exactly 12 regular sized bars were sold that week. How many king size are left? Let r be the number of regular bars and k be the number of king size bars. Set up our equations: [LIST=1] [*]0.75r + 1.5k = 36 [*]r = 12 [/LIST] [U]Substitute (2) into (1)[/U] 0.75(12) + 1.5k = 36 9 + 1.5k = 36 [U]Use our equation solver, we get:[/U] [B]k = 18[/B]

The bigger of 2 numbers in 5 larger than the smaller. Twice the smaller, increased by, twice the larger, is equal to 50. Find each number. Let the big number be b. Let the small number be s. We're given two equations: [LIST=1] [*]b = s + 5 [*]2s + 2b = 50 [/LIST] Substitute equation (1) into equation (2) 2s + 2(s + 5) = 50 [URL='https://www.mathcelebrity.com/1unk.php?num=2s%2B2%28s%2B5%29%3D50&pl=Solve']Type this equation into our search engine[/URL], and we get: [B]s = 10[/B] Now substitute s = 10 into equation (1) to solve for b: b = 10 + 5 [B]b = 15[/B]

The denominator of a fraction is 4 more than the numerator. If 4 is added to the numerator and 7 is added to the denominator, the value of the fraction is 1/2. Find the original fraction. Let the original fraction be n/d. We're given: [LIST=1] [*]d = n + 4 [*](n + 4) / (d + 7) = 1/2 [/LIST] Cross multiply Equation 2: 2(n + 4) = d + 7 2n + 8 = d + 7 Now substitute equation (1) into tihs: 2n + 8 = (n + 4) + 7 2n + 8 = n + 11 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B8%3Dn%2B11&pl=Solve']Type this equation into our search engine[/URL], and we get: n = 3 This means from equation (1), that: d = 3 + 4 d = 7 So our original fraction n/d = [B]3/7[/B]

The difference between 2 numbers is 108. 6 times the smaller is equal to 2 more than the larger. What are the numbers? Let the smaller number be x. Let the larger number be y. We're given: [LIST=1] [*]y - x = 108 [*]6x = y + 2 [/LIST] Rearrange (1) by adding x to each side: [LIST=1] [*]y = x + 108 [/LIST] Substitute this into (2): 6x = x + 108 + 2 Combine like terms 6x = x +110 Subtract x from each side: 5x = 110 [URL='https://www.mathcelebrity.com/1unk.php?num=5x%3D110&pl=Solve']Plugging this equation into our search engine[/URL], we get: x = [B]22[/B]

The difference between two numbers is 96. One number is 9 times the other. What are the numbers? Let x be the first number Let y be the second number We're given two equations: [LIST=1] [*]x - y = 96 [*]x = 9y [/LIST] Substitute equation (2) into equation (1) for x 9y - y = 96 [URL='https://www.mathcelebrity.com/1unk.php?num=9y-y%3D96&pl=Solve']Plugging this equation into our math engine[/URL], we get: y = [B]12 [/B] If y = 12, then we plug this into equation 2: x = 9(12) x = [B]108[/B]

The difference between two positive numbers is 5 and the square of their sum is 169. Let the two positive numbers be a and b. We have the following equations: [LIST=1] [*]a - b = 5 [*](a + b)^2 = 169 [*]Rearrange (1) by adding b to each side. We have a = b + 5 [/LIST] Now substitute (3) into (2): (b + 5 + b)^2 = 169 (2b + 5)^2 = 169 [URL='https://www.mathcelebrity.com/community/forums/calculator-requests.7/create-thread']Run (2b + 5)^2 through our search engine[/URL], and you get: 4b^2 + 20b + 25 Set this equal to 169 above: 4b^2 + 20b + 25 = 169 [URL='https://www.mathcelebrity.com/quadratic.php?num=4b%5E2%2B20b%2B25%3D169&pl=Solve+Quadratic+Equation&hintnum=+0']Run that quadratic equation in our search engine[/URL], and you get: b = (-9, 4) But the problem asks for [I]positive[/I] numbers. So [B]b = 4[/B] is one of our solutions. Substitute b = 4 into equation (1) above, and we get: a - [I]b[/I] = 5 [URL='https://www.mathcelebrity.com/1unk.php?num=a-4%3D5&pl=Solve']a - 4 = 5[/URL] [B]a = 9 [/B] Therefore, we have [B](a, b) = (9, 4)[/B]

The difference of 2 positive numbers is 54. The quotient obtained on dividing the 1 by the other is 4. Find the numbers. Let the numbers be x and y. We have: [LIST] [*]x - y = 54 [*]x/y = 4 [*]Cross multiply x/y = 4 to get x = 4y [*]Now substitute x = 4y into the first equation [*](4y) - y = 54 [*]3y = 54 [*]Divide each side by 3 [*][B]y = 18[/B] [*]If x = 4y, then x = 4(18) [*][B]x = 72[/B] [/LIST]

The difference of two numbers is 12 and their mean is 15. Find the two numbers. Let the two numbers be x and y. We're given: [LIST=1] [*]x - y = 12 [*](x + y)/2 = 15. <-- Mean is an average [/LIST] Rearrange equation 1 by adding y to each side: x - y + y = y + 12 Cancelling the y's on the left side, we get: x = y + 12 Now substitute this into equation 2: (y + 12 + y)/2 = 15 Cross multiply: y + 12 + y = 30 Group like terms for y: 2y + 12 = 30 [URL='https://www.mathcelebrity.com/1unk.php?num=2y%2B12%3D30&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]y = 9[/B] Now substitute this into modified equation 1: x = y + 12 x = 9 + 12 [B]x = 21[/B]

The fraction has a value of 3/5. The sum of the numerator and the denominator was 40. What was the fraction? We're given two equations with a fraction with numerator (n) and denominator (d): [LIST=1] [*]n + d = 40 [*]n/d = 3/5 [/LIST] Cross multiply equation 2, we get: 5n = 3d Divide each side by 5: 5n/5 = 3d/5 n = 3d/5 Substitute this into equation 1: 3d/5 + d = 40 Multiply through both sides of the equation by 5: 5(3d/5) = 5d = 40 * 5 3d + 5d =200 To solve this equation for d, we [URL='https://www.mathcelebrity.com/1unk.php?num=3d%2B5d%3D200&pl=Solve']type it in our search engine and we get[/URL]: d = [B]25 [/B] Now substitute that back into equation 1: n + 25 = 40 Using [URL='https://www.mathcelebrity.com/1unk.php?num=n%2B25%3D40&pl=Solve']our equation solver again[/URL], we get: n = [B]15[/B]

The function P(x) = -30x^2 + 360x + 785 models the profit, P(x), earned by a theatre owner on the basis of a ticket price, x. Both the profit and the ticket price are in dollars. What is the maximum profit, and how much should the tickets cost? Take the [URL='http://www.mathcelebrity.com/dfii.php?term1=-30x%5E2+%2B+360x+%2B+785&fpt=0&ptarget1=0&ptarget2=0&itarget=0%2C1&starget=0%2C1&nsimp=8&pl=1st+Derivative']derivative of the profit function[/URL]: P'(x) = -60x + 360 We find the maximum when we set the profit derivative equal to 0 -60x + 360 = 0 Subtract 360 from both sides: -60x = -360 Divide each side by -60 [B]x = 6 <-- This is the ticket price to maximize profit[/B] Substitute x = 6 into the profit equation: P(6) = -30(6)^2 + 360(6) + 785 P(6) = -1080 + 2160 + 785 [B]P(6) = 1865[/B]

The length of a rectangle is 6 less than twice the width. If the perimeter is 60 inches, what are the dimensions? Set up 2 equations given P = 2l + 2w: [LIST=1] [*]l = 2w - 6 [*]2l + 2w = 60 [/LIST] Substitute (1) into (2) for l: 2(2w - 6) + 2w = 60 4w - 12 + 2w = 60 6w - 12 = 60 To solve for w, [URL='https://www.mathcelebrity.com/1unk.php?num=6w-12%3D60&pl=Solve']type this into our math solver [/URL]and we get: w = [B]12 [/B] To solve for l, substitute w = 12 into (1) l = 2(12) - 6 l = 24 - 6 l = [B]18[/B]

The length of a rectangle is equal to triple the width. Find the length of the rectangle if the perimeter is 80 inches. The perimeter (P) of a rectangle is: 2l + 2w = P We're given two equations: [LIST=1] [*]l = 3w [*]2l + 2w = 80 [/LIST] We substitute equation 1 into equation 2 for l: 2(3w) + 2w = 80 6w + 2w = 80 To solve this equation for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B2w%3D80&pl=Solve']type it in our search engine[/URL] and we get: w = 10 To solve for the length (l), we substitute w = 10 into equation 1 above: l = 3(10) l = [B]30[/B]

The length of a rectangle is three times its width.If the perimeter is 80 feet, what are the dimensions? We're given 2 equations: [LIST=1] [*]l = 3w [*]P = 80 = 2l + 2w = 80 [/LIST] Substitute (1) into (2) for l: 2(3w) + 2w = 80 6w + 2w = 80 8w = 80 Divide each side by 8: 8w/8 = 80/8 w = [B]10 [/B] Substitute w = 10 into (1) l = 3(10) l = [B]30[/B]

The length of a rectangular building is 6 feet less than 3 times the width. The perimeter is 120 feet. Find the width and length of the building. P = 2l + 2w Since P = 120, we have: (1) 2l + 2w = 120 We are also given: (2) l = 3w - 6 Substitute equation (2) into equation (1) 2(3w - 6) + 2w = 120 Multiply through: 6w - 12 + 2w = 120 Combine like terms: 8w - 12 = 120 Add 12 to each side: 8w = 132 Divide each side by 8 to isolate w: w =16.5 Now substitute w into equation (2) l = 3(16.5) - 6 l = 49.5 - 6 l = 43.5 So (l, w) = (43.5, 16.5)

The length of a wooden frame is 1 foot longer than its width and its area is equal to 12ft² The frame is a rectangle. The area of a rectangle is A = lw. So were given: [LIST=1] [*]l = w + 1 [*]lw = 12 [/LIST] Substitute equation (1) into equation (2) for l: (w + 1) * w = 12 Multiply through and simplify: w^2 + w = 12 We have a quadratic equation. To solve for w, we type this equation into our search engine and we get two solutions: w = 3 w = -4 Since width cannot be negative, we choose the positive result and have: w = [B]3[/B] To solve for length, we plug w = 3 into equation (1) above and get: l = 3 + 1 l = [B]4[/B]

The length of Sally’s garden is 4 meters greater than 3 times the width. The perimeter of her garden is 72 meters. Find the dimensions of Sally’s garden. Gardens have a rectangle shape. Perimeter of a rectangle is 2l + 2w. We're given: [LIST=1] [*]l = 3w + 4 [I](3 times the width Plus 4 since greater means add)[/I] [*]2l + 2w = 72 [/LIST] We substitute equation (1) into equation (2) for l: 2(3w + 4) + 2w = 72 Multiply through and simplify: 6w + 8 + 2w = 72 (6 +2)w + 8 = 72 8w + 8 = 72 To solve this equation for w, we [URL='https://www.mathcelebrity.com/1unk.php?num=8w%2B8%3D72&pl=Solve']type it in our search engine[/URL] and we get: w = [B]8 [/B] To solve for l, we substitute w = 8 above into Equation (1): l = 3(8) + 4 l = 24 + 4 l = [B]28[/B]

The length of Sally’s garden is 4 meters greater than 3 times the width. The perimeter of her garden is 72 meters A garden is a rectangle, which has perimeter P of: P = 2l + 2w With P = 72, we have: 2l + 2w = 72 We're also given: l = 3w + 4 We substitute this into the perimeter equation for l: 2(3w + 4) + 2w = 72 6w + 8 + 2w = 72 To solve this equation for w, we t[URL='https://www.mathcelebrity.com/1unk.php?num=6w%2B8%2B2w%3D72&pl=Solve']ype it in our search engine[/URL] and we get: w =[B] 8[/B] Now, to solve for l, we substitute w = 8 into our length equation above: l = 3(8) + 4 l = 24 + 4 l = [B]28[/B]

The length of the flag is 2 cm less than 7 times the width. The perimeter is 60cm. Find the length and width. A flag is a rectangle shape. So we have the following equations Since P = 2l + 2w, we have 2l + 2w = 60 l = 7w - 2 Substitute Equation 1 into Equation 2: 2(7w -2) + 2w = 60 14w - 4 + 2w = 60 16w - 4 = 60 Add 4 to each side 16w = 64 Divide each side by 16 to isolate w w = 4 Which means l = 7(4) - 2 = 28 - 2 = 26

The perimeter of a rectangle is 400 meters. The length is 15 meters less than 4 times the width. Find the length and the width of the rectangle. l = 4w - 15 Perimeter = 2l + 2w Substitute, we get: 400 = 2(4w - 15) + 2w 400 = 8w - 30 + 2w 10w - 30 = 400 Add 30 to each side 10w = 370 Divide each side by 10 to isolate w w = 37 Plug that back into our original equation to find l l = 4(37) - 15 l = 148 - 15 l = 133 So we have (l, w) = (37, 133)

The perpendicular height of a right-angled triangle is 70 mm longer than the base. Find the perimeter of the triangle if its area is 3000. [LIST] [*]h = b + 70 [*]A = 1/2bh = 3000 [/LIST] Substitute the height equation into the area equation 1/2b(b + 70) = 3000 Multiply each side by 2 b^2 + 70b = 6000 Subtract 6000 from each side: b^2 + 70b - 6000 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=b%5E2%2B70b-6000%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: b = 50 and b = -120 Since the base cannot be negative, we use b = 50. If b = 50, then h = 50 + 70 = 120 The perimeter is b + h + hypotenuse Using the [URL='http://www.mathcelebrity.com/righttriangle.php?angle_a=&a=70&angle_b=&b=50&c=&pl=Calculate+Right+Triangle']right-triangle calculator[/URL], we get hypotenuse = 86.02 Adding up all 3 for the perimeter: 50 + 70 + 86.02 = [B]206.02[/B]

The product of two positive numbers is 96. Determine the two numbers if one is 4 more than the other. Let the 2 numbers be x and y. We have: [LIST=1] [*]xy = 96 [*]x = y - 4 [/LIST] [U]Substitute (2) into (1)[/U] (y - 4)y = 96 y^2 - 4y = 96 [U]Subtract 96 from both sides:[/U] y^2 - 4y - 96 = 0 [U]Factoring using our quadratic calculator, we get:[/U] (y - 12)(y + 8) So y = 12 and y = -8. Since the problem states positive numbers, we use [B]y = 12[/B]. Substituting y = 12 into (2), we get: x = 12 - 4 [B]x = 8[/B] [B]We have (x, y) = (8, 12)[/B]

The sum of 2 consecutive numbers is 3 less than 3 times the first number. What are the numbers? Let the first number be x. And the second number be y. We're given: [LIST=1] [*]y = x + 1 [*]x + y = 3x - 3 (less 3 means subtract 3) [/LIST] Substitute (1) into (2): x + x + 1 = 3x - 3 Combine like terms: 2x + 1 = 3x - 3 [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B1%3D3x-3&pl=Solve']Type this equation into the search engine[/URL], we get: x = 4 Substituting x = 4 into equation 1: y = 4 + 1 y = 5 So (x, y) = [B](4, 5)[/B]

The sum of 2 numbers is 60. The larger number is thrice the smaller. Let the 2 numbers be x and y, where x is the smaller number and y is the larger number. We are given: [LIST=1] [*]x + y = 60 [*]y = 3x [/LIST] Substitute (2) into (1): x + (3x) = 60 Combine like terms: 4x = 60 [URL='https://www.mathcelebrity.com/1unk.php?num=4x%3D60&pl=Solve']Type 4x = 60 into our search engine[/URL], and you get [B]x = 15[/B]. Substituting x = 15 into Equation (2) above, we get: y = 3(15) [B]y = 45 [/B] Check our work in Equation (1): 15 + 45 ? 60 60 = 60 Check our work in Equation (2): 45 ? 15(3) 45 = 45 The numbers check out, so our answer is [B](x, y) = (15, 45)[/B]

The sum of Jocelyn and Joseph's age is 40. In 5 years, Joseph will be twice as Jocelyn's present age. How old are they now? Let Jocelyn's age be a Let Joseph's age be b. We're given two equations: [LIST=1] [*]a + b = 40 [*]2(a + 5) = b + 5 [/LIST] We rearrange equation (1) in terms of a to get: [LIST=1] [*]a = 40 - b [*]2a = b + 5 [/LIST] Substitute equation (1) into equation (2) for a: 2(40 - b) = b + 5 80 - 2b = b + 5 To solve this equation for b, we [URL='https://www.mathcelebrity.com/1unk.php?num=80-2b%3Db%2B5&pl=Solve']type it in our search engine[/URL] and we get: [B]b (Joseph's age) = 25[/B] Now, substitute b = 25 into equation (1) to solve for a: a = 40 - 25 [B]a (Jocelyn's age) = 15[/B]

The sum of Juan’s age and Sara’s age is 33 yrs. If Sara is 15 yrs old, how old is Juan? Let j be Juan's age and s be Sara's age. We have the following equations: [LIST=1] [*]j + s = 33 [*]s = 15 [/LIST] Substitute (2) into (1) j + 15 = 33 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=j%2B15%3D33&pl=Solve']equation solver[/URL], we get[B] j = 18[/B]

The sum of the ages of levi and renee is 89 years. 7 years ago levi's age was 4 times renees age. How old is Levi now? Let Levi's current age be l. Let Renee's current age be r. Were given two equations: [LIST=1] [*]l + r = 89 [*]l - 7 = 4(r - 7) [/LIST] Simplify equation 2 by multiplying through: [LIST=1] [*]l + r = 89 [*]l - 7 = 4r - 28 [/LIST] Rearrange equation 1 to solve for r and isolate l by subtracting l from each side: [LIST=1] [*]r = 89 - l [*]l - 7 = 4r - 28 [/LIST] Now substitute equation (1) into equation (2): l - 7 = 4(89 - l) - 28 l - 7 = 356 - 4l - 28 l - 7 = 328 - 4l To solve for l, we [URL='https://www.mathcelebrity.com/1unk.php?num=l-7%3D328-4l&pl=Solve']type the equation into our search engine[/URL] and we get: l = [B]67[/B]

The sum of the digits of a 2 digit number is 10. The value of the number is four more than 15 times the unit digit. Find the number. Let the digits be (x)(y) where t is the tens digit, and o is the ones digit. We're given: [LIST=1] [*]x + y = 10 [*]10x+ y = 15y + 4 [/LIST] Simplifying Equation (2) by subtracting y from each side, we get: 10x = 14y + 4 Rearranging equation (1), we get: x = 10 - y Substitute this into simplified equation (2): 10(10 - y) = 14y + 4 100 - 10y = 14y + 4 [URL='https://www.mathcelebrity.com/1unk.php?num=100-10y%3D14y%2B4&pl=Solve']Typing this equation into our search engine[/URL], we get: y = 4 Plug this into rearranged equation (1), we get: x = 10 - 4 x = 6 So our number xy is [B]64[/B]. Let's check our work against equation (1): 6 + 4 ? 10 10 = 10 Let's check our work against equation (2): 10(6)+ 4 ? 15(4) + 4 60 + 4 ? 60 + 4 64 = 64

The sum of three numbers is 171. The second number is 1/2 of the first and the third is 3/4 of the first. Find the numbers. We have three numbers, x, y, and z. [LIST=1] [*]x + y + z = 171 [*]y = 1/2x [*]z = 3/4x [/LIST] Substitute (2) and (3) into (1) x + 1/2x + 3/4x = 171 Use a common denominator of 4 for each x term 4x/4 + 2x/4 + 3x/4 = 171 (4 + 2 + 3)x/4 = 171 9x/4 = 171 [URL='https://www.mathcelebrity.com/prop.php?num1=9x&num2=171&den1=4&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']Plug this equation into our search engine[/URL], and we get [B]x = 76[/B] So y = 1/2(76) --> [B]y = 38[/B] Then z = 3/4(76) --> [B]z = 57[/B]

Let x be the larger number. Let y be the smaller number. We're given two equations: [LIST=1] [*]x + y = 231 [*]x = 2y [/LIST] Substitute (2) into (1) for x: 2y + y = 231 3y = 231 [URL='https://www.mathcelebrity.com/1unk.php?num=3y%3D231&pl=Solve']Type this into our math solver[/URL] and get y = 77 This means x is: x = 2(77) x = [B]154[/B]

The value of all the quarters and dimes in a parking meter is $18. There are twice as many quarters as dimes. What is the total number of dimes in the parking meter? Let q be the number of quarters. Let d be the number of dimes. We're given: [LIST=1] [*]q = 2d [*]0.10d + 0.25q = 18 [/LIST] Substitute (1) into (2): 0.10d + 0.25(2d) = 18 0.10d + 0.5d = 18 [URL='https://www.mathcelebrity.com/1unk.php?num=0.10d%2B0.5d%3D18&pl=Solve']Type this equation into our search engine[/URL], and we get [B]d = 30[/B].

there are $4.20 in nickel and quarters. There are 6 more nickels than quarters there. How many coins of each are there We're given two equations: [LIST=1] [*]n = q + 6 [*]0.05n + 0.25q = 4.2 [/LIST] Substitute equation (1) into equation (2): 0.05(q + 6) + 0.25q = 4.2 Multiply through and simplify: 0.05q + 0.3 + 0.25q 0.3q + 0.3 = 4.2 To solve for q, we [URL='https://www.mathcelebrity.com/1unk.php?num=0.3q%2B0.3%3D4.2&pl=Solve']type this equation into the search engine[/URL] and we get: q = [B]13 [/B] To solve for n, we plug in q = 13 into equation (1): n = 13 + 6 n = [B]19[/B]

There are 13 animals in the barn. some are chickens and some are pigs. there are 40 legs in all. How many of each animal are there? Chickens have 2 legs, pigs have 4 legs. Let c be the number of chickens and p be the number of pigs. Set up our givens: (1) c + p = 13 (2) 2c + 4p = 40 [U]Rearrange (1) to solve for c by subtracting p from both sides:[/U] (3) c = 13 - p [U]Substitute (3) into (2)[/U] 2(13 - p) + 4p = 40 26 - 2p + 4p = 40 [U]Combine p terms[/U] 2p + 26 = 40 [U]Subtract 26 from each side:[/U] 2p = 14 [U]Divide each side by 2[/U] [B]p = 7[/B] [U]Substitute p = 7 into (3)[/U] c = 13 - 7 [B]c = 6[/B] For a shortcut, you could use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+p+%3D+13&term2=2c+%2B+4p+%3D+40&pl=Cramers+Method']simultaneous equations calculator[/URL]

There are 2 consecutive integers. Twice the first increased by the second yields 16. What are the numbers? Let x be the first integer. y = x + 1 is the next integer. We have the following givens: [LIST=1] [*]2x + y = 16 [*]y = x + 1 [/LIST] Substitute (2) into (1) 2x + (x + 1) = 16 Combine x terms 3x + 1 = 16 Subtract 1 from each side 3x = 15 Divide each side by 3 [B]x = 5[/B] So the other integer is 5 + 1 = [B]6[/B]

Let b = boys and g = girls. We have two equations: (1) b + g = 320 (2) g = b + 24 Substitute (2) into (1) for g b + (b + 24) = 320 Combine b terms: 2b + 24 = 320 Use our [URL='http://www.mathcelebrity.com/1unk.php?num=2b%2B24%3D320&pl=Solve']equation calculator[/URL]: [B]b = 148 [/B] Substitute b = 148 into (2) g = 148 + 24 [B]g = 172[/B]

There are 33 students in an Algebra I class. There are 7 fewer girls than boys. How many girls are in the class? Let b be the number of boys and g be the number of girls. We are given 2 equations: [LIST=1] [*]g = b - 7 [*]b + g = 33 [/LIST] Substitute (1) into (2): b + (b - 7) = 33 Combine like terms: 2b - 7 = 33 [URL='https://www.mathcelebrity.com/1unk.php?num=2b-7%3D33&pl=Solve']Typing this equation into our search engine[/URL], we get b = 20. Since the problem asks for how many girls (g) we have, we substitute b = 20 into Equation (1): g = 20 - 7 [B]g = 13[/B]

There are 63 students in middle school chorus. There are 11 more boys than girls. How many boys x and girls y are in the chorus? Set up equations: [LIST=1] [*]x + y = 63 [*]x = y + 11 [/LIST] Substitute (1) into (2) y + 11 + y = 63 2y + 11 = 63 Use our [URL='http://www.mathcelebrity.com/1unk.php?num=2y%2B11%3D63&pl=Solve']equation solver[/URL]: [B]y = 26[/B]

There are 64 members in the history club. 11 less than half of the members are girls. How many members are boys? Set up two equations where b = the number of boys and g = the number of girls [LIST=1] [*]b + g = 64 [*]1/2(b + g) - 11 = g [/LIST] Substitute (1) for b + g into (2) 1/2(64) - 11 = g 32 - 11 = g [B]g = 21[/B] Substitute g = 21 into (1) b + 21 = 64 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=b%2B21%3D64&pl=Solve']equation calculator[/URL], we get: [B]b = 43[/B]

There are two numbers. The sum of 4 times the first number and 3 times the second number is 24. The difference between 2 times the first number and 3 times the second number is 24. Find the two numbers. Let the first number be x and the second number be y. We have 2 equations: [LIST=1] [*]4x + 3y = 24 [*]2x - 3y = 24 [/LIST] Without doing anything else, we can add the 2 equations together to eliminate the y term: (4x + 2x) + (3y - 3y) = (24 + 24) 6x = 48 Divide each side by 6: [B]x = 8 [/B] Substitute this into equation (1) 4(8) + 3y = 24 32 + 3y = 24 [URL='https://www.mathcelebrity.com/1unk.php?num=32%2B3y%3D24&pl=Solve']Type 32 + 3y = 24 into our search engine[/URL] and we get [B]y = 2.6667[/B].

Set up two equations: (1) b = g + 16 (2) b + g = 150 Substitute equation (1) into (2) (g + 16) + g = 150 Combine like terms 2g + 16 = 150 Subtract 16 from each side 2g = 134 Divide each side by 2 to isolate g g = 67 Substitute this into equation (1) b = 67 + 16 [B]b = 83[/B]

To make an international telephone call, you need the code for the country you are calling. The code for country A, country B, and C are three consecutive integers whose sum is 90. Find the code for each country. If they are three consecutive integers, then we have: [LIST=1] [*]B = A + 1 [*]C = B + 1, which means C = A + 2 [*]A + B + C = 90 [/LIST] Substitute (1) and (2) into (3) A + (A + 1) + (A + 2) = 90 Combine like terms 3A + 3 = 90 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3a%2B3%3D90&pl=Solve']equation calculator[/URL], we get: [B]A = 29[/B] Which means: [LIST] [*]B = A + 1 [*]B = 29 + 1 [*][B]B = 30[/B] [*]C = A + 2 [*]C = 29 + 2 [*][B]C = 31[/B] [/LIST] So we have [B](A, B, C) = (29, 30, 31)[/B]

Tom is 2 years older than Sue and Bill is twice as old as Tom. If you add all their ages and subtract 2, the sum is 20. How old is Bill? Let t be Tom's age., s be Sue's age, and b be Bill's age. We have the following equations: [LIST=1] [*]t = s + 2 [*]b = 2t [*]s + t + b - 2 = 20 [/LIST] Get (2) in terms of s (2) b = 2(s + 2) <-- using (1), substitute for t So we have (3) rewritten with substitution as: s + (s + 2) + 2(s + 2) - 2 = 20 s + (s + 2) + 2s + 4 - 2 = 20 Group like terms: (s + s + 2s) + (2 + 4 - 2) = 20 4s + 4 = 20 Run this through our [URL='https://www.mathcelebrity.com/1unk.php?num=4s%2B4%3D20&pl=Solve']equation calculator [/URL]to get s = 4 Above, we had b = 2(s + 2) Substituting s = 4, we get: 2(4 + 2) = 2(6) = [B]12 Bill is 12 years old[/B]

two mechanics worked on a car. the first mechanic worked for 10 hours, and the second mechanic worked for 5 hours. together they charged a total of 1225. what was the rate charged per hour by each mechanic if the sum of the two rates was 170 per hour? Set up two equations: (1) 10x + 5y = 1225 (2) x + y = 170 Rearrange (2) x = 170 - y Substitute that into (1) 10(170 - y) + 5y = 1225 1700 - 10y + 5y = 1225 1700 - 5y = 1225 Move 5y to the other side 5y + 1225 = 1700 Subtract 1225 from each side 5y =475 Divide each side by 5 [B]y = 95[/B] Which means x = 170 - 95, [B]x = 75[/B]

Two numbers total 12, and their differences is 20. Find the two numbers. Let the first number be x. Let the second number be y. We're given two equations: [LIST=1] [*]x + y = 12 [*]x - y = 20 [/LIST] Since we have y coefficients of (-1 and 1) that cancel, we add the two equations together: (x + x) + (y - y) = 12 + 20 The y terms cancel, so we have: 2x = 32 [URL='https://www.mathcelebrity.com/1unk.php?num=2x%3D32&pl=Solve']Type this equation into our search engine[/URL] and we get: x = [B]16[/B] Substitute this value of x = 16 back into equation 1: 16 + y = 12 [URL='https://www.mathcelebrity.com/1unk.php?num=16%2By%3D12&pl=Solve']Typing this equation into our search engine[/URL], we get: y = [B]-4 [/B] Now, let's check our work for both equations: [LIST=1] [*]16 - 4 = 12 [*]16 - -4 --> 16 + 4 = 20 [/LIST] So these both check out. (x, y) = ([B]16, -4)[/B]

Victoria is 4 years older than her neighbor. The sum of their ages is no more than 14 years. Let Victoria's age be v. And her neighbor's age be n. We're given: [LIST=1] [*]v = n + 4 [*]v + n <=14 <-- no more than means less than or equal to [/LIST] Substitute Equation (1) into Inequality (2): (n + 4) + n <= 14 Combine like terms: 2n + 4 <= 14 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B4%3C%3D14&pl=Solve']Typing this inequality into our search engine[/URL], we get: n <= 5 Substituting this into inequality (2): v + 5 <= 14 [URL='https://www.mathcelebrity.com/1unk.php?num=v%2B5%3C%3D14&pl=Solve']Typing this inequality into our search engine[/URL], we get: [B]v <= 9[/B]

You earned $141 last week babysitting and cleaning. You earned $5 per hour babysitting and $7 per hour cleaning. You worked 9 more hours babysitting than cleaning. How many hours did you work last week? Let b be the hours of babysitting and c be the hours of cleaning. We're given two equations: [LIST=1] [*]b = c + 9 [*]5b + 7c = 141 [/LIST] Substitute equation (1) into (2): 5(c + 9) + 7c = 141 Multiply through: 5c + 45 + 7c = 141 Combine like terms: 12c + 45 = 141 [URL='https://www.mathcelebrity.com/1unk.php?num=12c%2B45%3D141&pl=Solve']Typing this equation into our search engine[/URL], we get: c = 8 Now substitute this value of c back into Equation (1): b = 8 + 9 b = 17 The total hours worked (t) is: t = b + c t = 17 + 8 t = [B]25[/B]

You have a total of 42 math and science problems for homework. You have 10 more math problems than science problems. How many problems do you have in each subject? Let m be the math problems and s be the science problems. We have two equations: (1) m + s = 42 (2) m = s + 10 Substitute (2) into (1) (s + 10) + s = 42 Combine like terms 2s + 10 = 42 Subtract 10 from each side 2s = 32 Divide each side by 2 [B]s = 16[/B] So that means m = 16 + 10 --> [B]m = 26 (m, s) = (26, 16)[/B]

Your friends in class want you to make a run to the vending machine for the whole group. Everyone pitched in to make a total of $12.50 to buy snacks. The fruit drinks are $1.50 and the chips are $1.00. Your friends want you to buy a total of 10 items. How many drinks and how many chips were you able to purchase? Let c be the number of chips. Let f be the number of fruit drinks. We're given two equations: [LIST=1] [*]c + f = 10 [*]c + 1.5f = 12.50 [/LIST] Rearrange equation 1 by subtracting f from both sides: [LIST=1] [*]c = 10 - f [*]c + 1.5f = 12.50 [/LIST] Substitute equation (1) into equation (2): 10 - f + 1.5f = 12.50 To solve for f, we [URL='https://www.mathcelebrity.com/1unk.php?num=10-f%2B1.5f%3D12.50&pl=Solve']type this equation into our search engine[/URL] and we get: [B]f = 5[/B] Now, substitute this f = 5 value back into modified equation (1) above: c = 10 - 5 [B]c = 5[/B]