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!

Let x be the first even integer. That means the next consecutive even integer must be x + 2. Set up our equation: x + (x + 2) = 118 Group x terms 2x + 2 = 118 Subtract 2 from each side 2x = 116 Divide each side by 2 x = 58 Which means the next consecutive even integer is 58 + 2 = 60 So our two consecutive even integers are [B]58, 60[/B] Check our work: 58 + 60 = 118

2 consecutive odd integers such that their product is 15 more than 3 times their sum. Let the first integer be n. The next odd, consecutive integer is n + 2. We are given the product is 15 more than 3 times their sum: n(n + 2) = 3(n + n + 2) + 15 Simplify each side: n^2 + 2n = 6n + 6 + 15 n^2 + 2n = 6n + 21 Subtract 6n from each side: n^2 - 4n - 21 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-4n-21%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: n = (-3, 7) If we use -3, then the next consecutive odd integer is -3 + 2 = -1. So we have [B](-3, -1)[/B] If we use 7, then the next consecutive odd integer is 7 + 2 = 9. So we have [B](7, 9)[/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)

3 consecutive odd integers such that thrice the middle is 15 more than the sum of the other 2. [LIST] [*]Let the first integer be n [*]The next odd one (middle) is n + 2. [*]The next odd one is n + 4 [/LIST] We are given 3(n + 2) = n + n + 4 + 15. Simplifying, we get: 3n + 6 = 2n + 19 [URL='http://www.mathcelebrity.com/1unk.php?num=3n%2B6%3D2n%2B19&pl=Solve']Plugging that problem[/URL] into our search engine, we get n = 13. So the next odd integer is 13 + 2 = 15 The next odd integer is 15 + 2 = 17

4 consecutive integers such that the sum of the first 3 integers is equal to the 4th Let n be our first consecutive integer. [LIST=1] [*]n [*]n + 1 [*]n + 2 [*]n + 3 [/LIST] The sum of the first 3 integers is equal to the 4th: n + n + 1 + n + 2 = n + 3 Simplify by grouping like terms: (n + n + n) + (1 + 2) = n + 3 3n + 3 = n + 3 3n = n n = 0 n = 0 n + 1 = 1 n + 2 = 2 n + 3 = 3 Check our work: 0 + 1 +2 ? 3 3 = 3 Our final answer is [B](0, 1, 2, 3}[/B]

a baseball player has 9 hits in his first 60 at bats. how many consecutive hits would he need to bring his average up to 0.400? Let the amount of consecutive hits needed be h. We have: hits / at bats = Batting Average Plugging in our numbers, we get: (9 + h)/60 = 0.400 Cross multiply: 9 + h = 60 * 0.4 9 + h = 24 To solve this equation for h, [URL='https://www.mathcelebrity.com/1unk.php?num=9%2Bh%3D24&pl=Solve']we type it in our search engine[/URL] and we get: h = [B]15[/B]

A football team lost 7 yards each play for four consecutive plays. Represent the team’s total change in position for the four plays as an integer. A net loss in yardage for 7 yards is written as -7 4 plays * -7 yards equals [B]-28[/B]

A set of 4 consecutive integers adds up to 314. What is the least of the 4 integers? First integer is x. The next 3 are x + 1, x + 2, and x + 3. Set up our equation: x + (x + 1) + (x + 2) + (x + 3) = 314 Group x terms and group constnats (x + x + x + x) + (1 + 2 + 3) = 314 Simplify and combine 4x + 6 = 314 [URL='http://www.mathcelebrity.com/1unk.php?num=4x%2B6%3D314&pl=Solve']Enter this in the equation solver[/URL] [B]x = 77[/B]

A software company, in 3 consecutive years, makes profits of -3 million dollars, 10 million dollars, and -2 million dollars. What was its profit over the 3 year period? Profit = -3,o00,000 + 10,000,000 - 2,000,000 Profit = [B]5,000,000[/B]

Ages are consecutive integers. The sum of ages are 111. What are the ages In the search engine, we type [I][URL='http://www.mathcelebrity.com/consecintwp.php?num=111&pl=Sum']sum of 2 consecutive integers is 111[/URL][/I]. We get [B]55 and 56[/B].

Balls numbered 1 to 10 are placed in a bag. Two of the balls are drawn out at random. Find the probability that the numbers on the balls are consecutive. Build our sample set: [LIST] [*](1, 2) [*](2, 3) [*](3, 4) [*](4, 5) [*](5, 6) [*](6, 7) [*](7, 8) [*](8, 9) [*](9, 10) [/LIST] Each of these 9 possibilities has a probability of: 1/10 * 1/9 This is because we draw without replacement. To start, the bag has 10 balls. On the second draw, it only has 9. We multiply each event because each draw is independent. We have 9 possibilities, so we have: 9 * 1/10 * 1/9 Cancelling, the 9's, we have [B]1/10[/B]

Before Barry Bonds, Mark McGwire, and Sammy Sosa, Roger Maris held the record for the most home runs in one season. Just behind Maris was Babe Ruth. The numbers of home runs hit by these two athletes in their record-breaking seasons form consecutive integers. Combined, the two athletes hit 121 home runs. Determine the number of home runs hit by Maris and Ruth in their record-breaking seasons. We want [URL='https://www.mathcelebrity.com/consecintwp.php?num=121&pl=Sum']the sum of 2 consecutive integers equals 121[/URL]. [B]We get Maris at 61 and Ruth at 60[/B]

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[INDENT]Consecutive odd integers are odd integers that differ by ______ , such as ______ and 13. Consecutive even integers are even integers that differ by ______ , such as 12 and ______ . Consecutive odd integers are odd integers that differ by ___2___ , such as ___11___ and 13. Consecutive even integers are even integers that differ by ___2___ , such as 12 and ___14___ .[/INDENT]

Customers arrive at the claims counter at the rate of 20 per hour (Poisson distributed). What is the probability that the arrival time between consecutive customers is less than five minutes? Use the [I]exponential distribution[/I] 20 per 60 minutes is 1 every 3 minutes 1/λ = 3 so λ = 0.333333333 Using the [URL='http://www.mathcelebrity.com/expodist.php?x=+5&l=0.333333333&pl=CDF']exponential distribution calculator[/URL], we get F(5,0.333333333) = [B]0.811124396848[/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]

Find 3 consecutive integers such that the sum of twice the smallest and 3 times the largest is 126. Let the first integer be n, the second integer be n + 1, and the third integer be n + 2. We have: Sum of the smallest and 3 times the largest is 126: n + 3(n + 2) = 126 Multiply through: n + 3n + 6 = 126 Group like terms: 4n + 6 = 126 [URL='https://www.mathcelebrity.com/1unk.php?num=4n%2B6%3D126&pl=Solve']Type 4n + 6 = 126 into our calculator[/URL], we get n = 30. Which means the next two integers are 31 and 32. [B]{30, 31, 32}[/B]

Find four consecutive odd numbers which add to 64. Let the first number be x. The next three numbers are: x + 2 x + 4 x + 6 Add them together to get 64: x + (x + 2) + (x + 4) + (x + 6) = 64 Group like terms: 4x + 12 = 64 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=4x%2B12%3D64&pl=Solve']equation calculator[/URL], we get: [B]x = 13[/B] The next 3 odd numbers are: x + 2 = 13 + 2 = 15 x + 4 = 13 + 4 = 17 x + 6 = 13 + 6 = 19 So the 4 consecutive odd numbers which add to 64 are: [B](13, 15, 17, 19)[/B]

Find the largest of three consecutive even integers when six times the first integers is equal to five times the middle integer. Let the first of the 3 consecutive even integers be n. The second consecutive even integer is n + 2. The third (largest) consecutive even integer is n + 4. We are given 6n = 5(n + 2). Multiply through on the right side, and we get: 6n = 5n + 10 [URL='https://www.mathcelebrity.com/1unk.php?num=6n%3D5n%2B10&pl=Solve']Typing 6n = 5n + 10 into the search engine[/URL], we get n = 10. Remember, n was our smallest of 3 consecutive even integers. So the largest is: n + 4 10 + 4 [B]14[/B]

Find two consecutive integers if the sum of their squares is 1513 Let the first integer be n. The next consecutive integer is (n + 1). The sum of their squares is: n^2 + (n + 1)^2 = 1513 n^2 + n^2 + 2n + 1 = 1513 2n^2 + 2n + 1 = 1513 Subtract 1513 from each side: 2n^2 + 2n - 1512 = 0 We have a quadratic equation. We [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B2n-1512%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this into our search engine[/URL] and get: n = (-27, 28) Let's take the positive solution. The second integer is: n + 1 28 + 1 = 29

Find two consecutive intergers whose sum is 15 Use our [URL='http://www.mathcelebrity.com/consecintwp.php?pl=Sum&num=+15']consecutive integer calculator[/URL], we get: (7, 8)

Find two consecutive odd integers such that the sum of their squares is 290. Let the first odd integer be n. The next odd integer is n + 2 Square them both: n^2 (n + 2)^2 = n^2 + 4n + 4 from our [URL='https://www.mathcelebrity.com/expand.php?term1=%28n%2B2%29%5E2&pl=Expand']expansion calculator[/URL] The sum of the squares equals 290 n^2 + n^2 + 4n + 4 = 290 Group like terms: 2n^2 + 4n + 4 = 290 [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B4n%2B4%3D290&pl=Solve+Quadratic+Equation&hintnum=+0']Enter this quadratic into our search engine[/URL], and we get: n = 11, n = -13 Which means the two consecutive odd integer are: 11 and 11 + 2 = 13. [B](11, 13)[/B] -13 and -13 + 2 = -11 [B](-13, -11)[/B]

Find two consecutive positive integers such that the difference of their square is 25. Let the first integer be n. This means the next integer is (n + 1). Square n: n^2 Square the next consecutive integer: (n + 1)^2 = n^2 + 2n + 1 Now, we take the difference of their squares and set it equal to 25: (n^2 + 2n + 1) - n^2 = 25 Cancelling the n^2, we get: 2n + 1 = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B1%3D25&pl=Solve']Typing this equation into our search engine[/URL], we get: n = [B]12[/B]

Find two consecutive positive integers such that the sum of their squares is 25. Let the first integer be x. The next consecutive positive integer is x + 1. The sum of their squares equals 25. We write this as:: x^2 + (x + 1)^2 Expanding, we get: x^2 + x^2 + 2x + 1 = 25 Group like terms: 2x^2 + 2x + 1 = 25 Subtract 25 from each side: 2x^2 + 2x - 24 = 0 Simplify by dividing each side by 2: x^2 + x - 12 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3. This means, our next positive integer is 3 + 1 = 4. So we have [B](3, 4) [/B]as our answers. Let's check our work: 3^2 + 4^2 = 9 + 16 = 25

Four consecutive integers beginning with n consecutive meaning one after another. So we have: [LIST] [*][B]n[/B] [*][B]n + 1[/B] [*][B]n + 2[/B] [*][B]n + 3[/B] [/LIST]

Four cousins were born at two-year intervals. The sum of their ages is 36. What are their ages? So the last cousin is n years old. this means consecutive cousins are n + 2 years older than the next. whether their ages are even or odd, we have the sum of 4 consecutive (odd|even) integers equal to 36. We [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof4consecutiveevenintegersis36&pl=Calculate']type this into our search engine[/URL] and we get the ages of: [B]6, 8, 10, 12[/B]

If first integer is 5y, then the next two consecutive integers are integers increase by 1, so we have: [B]5y + 1[/B] 5y + 1 + 1 = [B]5y + 2[/B]

If the third of 6 consecutive numbers is 12, what is their sum? If 12 is the third of 6 consecutive numbers: First consecutive number is 12 - 2 = 10 Second consecutive number = 12 - 1 = 11 Third consecutive number = 12 Fourth consecutive number = 12 + 1 = 13 Fifth consecutive number = 13 + 1 = 14 Sixth consecutive number = 14 + 1 = 15 The sum of all consecutive numbers is: 10 + 11 + 12 + 13 + 14 + 15 =[B] 75[/B]

If two consecutive even numbers are added, the sum is equal to 226. What is the smaller of the two numbers? Let the smaller number be n. The next consecutive even number is n + 2. Add them together to equal 226: n + n + 2 = 226 Solve for [I]n[/I] in the equation n + n + 2 = 226 [SIZE=5][B]Step 1: Group the n terms on the left hand side:[/B][/SIZE] (1 + 1)n = 2n [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 2n + 2 = + 226 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 2 and 226. To do that, we subtract 2 from both sides 2n + 2 - 2 = 226 - 2 [SIZE=5][B]Step 4: Cancel 2 on the left side:[/B][/SIZE] 2n = 224 [SIZE=5][B]Step 5: Divide each side of the equation by 2[/B][/SIZE] 2n/2 = 224/2 n = [B]112 [URL='https://www.mathcelebrity.com/1unk.php?num=n%2Bn%2B2%3D226&pl=Solve']Source[/URL][/B]

If x represents the first, or the smaller, of two consecutive odd integers, express the sum of the two integers in terms of x If x is the first of two consecutive odd integers, then we find the next consecutive odd integer by adding 2 to x: x + 2 The sum of the two consecutive odd integers is expressed by x + (x + 2) Simplify by grouping like terms, we get: [B]2x + 2[/B]

Let n be the middle number of three consecutive integers This means: [LIST] [*]n is the second of three consecutive integers [*]The first consecutive integer is n - 1 [*]The third consecutive integer is n + 1 [/LIST] The sum is found by: n - 1 + n + n + 1 Simplifying, we get: (n + n + n) + 1 - 1 [B]3n[/B]

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Take an integer n. The next alternate consecutive integer is n + 2 Subtract the difference of the squares: (n + 2)^2 - n^2 n^2 + 4n + 4 - n^2 n^2 terms cancel, we get: 4n + 4 Factor out a 4: 4(n + 1) If n is odd, n + 1 is even. 4 * even is always even If n is even, n + 1 is odd. 4 * odd is always odd Since both cases are even, we've proven our statement. [MEDIA=youtube]J_E9lR5qFY0[/MEDIA]

Take two consecutive integers: n, n + 1 The difference of their cubes is: (n + 1)^3 - n^3 n^3 + 3n^2 + 3n + 1 - n^3 Cancel the n^3 3n^2 + 3n + 1 Factor out a 3 from the first 2 terms: 3(n^2 + n) + 1 The first two terms are always divisible by 3 but then the + 1 makes this expression not divisible by 3: 3(n^2 + n) + 1 = 1 (mod 3) [MEDIA=youtube]hFvJ3epqmyE[/MEDIA]

Take an integer n. The next consecutive integer is n + 1 Subtract the difference of the squares: (n + 1)^2 - n^2 n^2 + 2n + 1 - n^2 n^2 terms cancel, we get: 2n + 1 2 is even. For n, if we use an even: we have even * even = Even Add 1 we have Odd 2 is even. For n, if we use an odd: we have even * odd = Even Add 1 we have Odd Since both cases are odd, we've proven our statement. [MEDIA=youtube]RAi0HbH5bqc[/MEDIA]

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]

sum of 3 consecutive odd integers equals 1 hundred 17 The sum of 3 consecutive odd numbers equals 117. What are the 3 odd numbers? 1) Set up an equation where our [I]odd numbers[/I] are n, n + 2, n + 4 2) We increment by 2 for each number since we have [I]odd numbers[/I]. 3) We set this sum of consecutive [I]odd numbers[/I] equal to 117 n + (n + 2) + (n + 4) = 117 [SIZE=5][B]Simplify this equation by grouping variables and constants together:[/B][/SIZE] (n + n + n) + 2 + 4 = 117 3n + 6 = 117 [SIZE=5][B]Subtract 6 from each side to isolate 3n:[/B][/SIZE] 3n + 6 - 6 = 117 - 6 [SIZE=5][B]Cancel the 6 on the left side and we get:[/B][/SIZE] 3n + [S]6[/S] - [S]6[/S] = 117 - 6 3n = 111 [SIZE=5][B]Divide each side of the equation by 3 to isolate n:[/B][/SIZE] 3n/3 = 111/3 [SIZE=5][B]Cancel the 3 on the left side:[/B][/SIZE] [S]3[/S]n/[S]3 [/S]= 111/3 n = 37 Call this n1, so we find our other 2 numbers n2 = n1 + 2 n2 = 37 + 2 n2 = 39 n3 = n2 + 2 n3 = 39 + 2 n3 = 41 [SIZE=5][B]List out the 3 consecutive odd numbers[/B][/SIZE] ([B]37, 39, 41[/B]) 37 ? 1st number, or the Smallest, Minimum, Least Value 39 ? 2nd number 41 ? 3rd or the Largest, Maximum, Highest Value

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Free Sum of Five Consecutive Integers Calculator - Finds five consecutive integers, if applicable, who have a sum equal to a number. Sum of 5 consecutive integers

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Sum of N and its next consecutive even integer is 65 Next even consecutive integer is N + 2. We have N + (N + 2) = 65. Combine like terms, we have 2N + 2 = 65 [URL='http://www.mathcelebrity.com/1unk.php?num=2n%2B2%3D65&pl=Solve']Running this problem through the search engine[/URL], we get n = 31.5. Meaning this problem is impossible, it cannot be done. n is not an integer, and neither is the next consecutive even integer.

Free Sum of Three Consecutive Integers Calculator - Finds three consecutive integers, if applicable, who have a sum equal to a number. Sum of 3 consecutive integers

Sum of two consecutive numbers is always odd Definition: [LIST] [*]A number which can be written in the form of 2 m where m is an integer, is called an even integer. [*]A number which can be written in the form of 2 m + 1 where m is an integer, is called an odd integer. [/LIST] Take two consecutive integers, one even, and one odd: 2n and 2n + 1 Now add them 2n + (2n+ 1) = 4n + 1 = 2(2 n) + 1 The sum is of the form 2n + 1 (2n is an integer because the product of two integers is an integer) Therefore, the sum of two consecutive integers is an odd number.

The age of three sister are consecutive intergers the sum of their age is 45 what is their ages Type this into the search engine: [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=thesumofthreeconsecutivenumbersis45&pl=Calculate']The sum of three consecutive numbers is 45[/URL]. We get [B]14, 15, 16[/B].

The ages of three siblings are all consecutive integers. The sum of of their ages is 39. Let the age of the youngest sibling be n. This means the second sibling is n + 1. This means the oldest/third sibling is n + 2. So what we want is the[URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof3consecutiveintegersequalto39&pl=Calculate'] sum of 3 consecutive integers equal to 39[/URL]. We type this command into our search engine. We get: n = 12. So the youngest sibling is [B]12[/B]. The next sibling is 12 + 1 = [B]13[/B] The oldest/third sibling is 12 + 2 = [B]14[/B]

The difference between the squares of two consecutive numbers is 141. Find the numbers Take two consecutive numbers: n- 1 and n Given a difference (d) between the squares of two consecutive numbers, the shortcut for this is: 2n - 1 = d Proof of this: n^2- (n - 1)^2 = d n^2 - (n^2 - 2n + 1) = d n^2 - n^2 + 2n - 1 = d 2n - 1 = d Given d = 141, we have 2n - 1 = 141 Add 1 to each side: 2n = 142 Divide each side by 2: 2n/2 = 142/2 n = [B]71[/B] Therefore, n - 1 = [B]70 Our two consecutive numbers are (70, 71)[/B] Check your work 70^2 = 4900 71^2 = 5041 Difference = 5041 - 4900 Difference = 141 [MEDIA=youtube]vZJtZyYWIFQ[/MEDIA]

The distance between consecutive bases is 90 feet. An outfielder catches the ball on the third base line about 40 feet behind third base. How far would the outfielder have to throw the ball to first base? We have a right triangle. From home base to third base is 90 feet. We add another 40 feet to the outfielder behind third base to get: 90 + 40 = 130 The distance from home to first is 90 feet. Our hypotenuse is the distance from the outfielder to first base. [URL='https://www.mathcelebrity.com/pythag.php?side1input=130&side2input=90&hypinput=&pl=Solve+Missing+Side']Using our Pythagorean theorem calculator[/URL], we get: d = [B]158.11 feet[/B]

the left and right page numbers of an open book are two consecutive integers whose number is 235 find the page numbers Using our [URL='https://www.mathcelebrity.com/consecintwp.php?pl=Sum&num=+235']consecutive integer calculator[/URL], we get: [B]117, 118[/B]

The left and right page numbers of an open book are two consecutive integers whose sum is 403. Find these page numbers. Page numbers left and right are consecutive integers. So we want to find a number n and n + 1 where: n + n + 1 = 403 Combining like terms, we get: 2n + 1 = 403 Typing that equation into our search engine, we get: [B]n = 201[/B] This is our left hand page. Our right hand page is: 201 + 1 = [B]202[/B]

The sides of a triangle are consecutive numbers. If the perimeter of the triangle is 240 m, find the length of each side Let the first side be n. Next side which is consecutive is n + 1 Next side which is consecutive is n + 1 + 1 = n + 2 So we have the sum of 3 consecutive numbers is 240. We type in [I][URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof3consecutivenumbersis240&pl=Calculate']sum of 3 consecutive numbers is 240[/URL][/I] into our search engine and we get: [B]79, 80, 81[/B]

The square of a positive integer minus twice its consecutive integer is equal to 22. Find the integers. Let x = the original positive integer. We have: [LIST] [*]Consecutive integer is x + 1 [*]x^2 - 2(x + 1) = 22 [/LIST] Multiply through: x^2 - 2x - 2 = 22 Subtract 22 from each side: x^2 - 2x - 24 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2-2x-24%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: x = 6 and x = -4 Since the problem states [U]positive integers[/U], we use: x = 6 and x + 1 = 7 [B](6, 7)[/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 3 consecutive integers is greater than 30. Let the first consecutive integer be n The second consecutive integer is n + 1 The third consecutive integer is n + 2 The sum is written as: n + n + 1 + n + 2 Combine like terms: (n + n + n) + (1 + 2) 3n + 3 The phrase [I]greater than[/I] means an inequality, which we write as: [B]3n + 3 > 30[/B]

the sum of 3 consecutive natural numbers, the first of which is n Natural numbers are counting numbers, so we the following expression: n + (n + 1) + (n + 2) Combine n terms and constants: (n + n + n) + (1 + 2) [B]3n + 3 Also expressed as 3(n + 1)[/B]

the sum of 3 consecutive natural numbers, the first of which is n We have: n + (n + 1) + (n + 2) Grouping like terms, we have: [B]3n + 3[/B]

The sum of 3 consecutive natural numbers, the first of which is n. We have 3 numbers: n, n + 1, and n + 2 Add them up: n + (n + 1) + (n + 2) Group like terms: [B]3n + 3[/B]

The sum of 5 odd consecutive numbers is 145. Let the first odd number be n. We have the other 4 odd numbers denoted as: [LIST] [*]n + 2 [*]n + 4 [*]n + 6 [*]n + 8 [/LIST] Add them all together n + (n + 2) + (n + 4) + (n + 6) + (n + 8) The sum of the 5 odd consecutive numbers equals 145 n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 145 Combine like terms: 5n + 20 = 145 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=5n%2B20%3D145&pl=Solve']equation solver[/URL], we get [B]n = 25[/B]. Using our other 4 consecutive odd numbers above, we get: [LIST] [*]27 [*]29 [*]31 [*]33 [/LIST] Adding the sum up, we get: 25 + 27 + 29 + 31 + 33 = 145. So our 5 odd consecutive number added to get 145 are [B]{25, 27, 29, 31, 33}[/B]. [MEDIA=youtube]0T2PDuQIIwI[/MEDIA]

The sum of the squares of two consecutive positive integers is 61. Find these two numbers. Let the 2 consecutive integers be x and x + 1. We have: x^2 + (x + 1)^2 = 61 Simplify: x^2 + x^2 + 2x + 1 = 61 2x^2 + 2x + 1 = 61 Subtract 61 from each side: 2x^2 + 2x - 60 = 0 Divide each side by 2 x^2 + x - 30 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL], we get: x = 5 and x = -6 The question asks for [I]positive integers[/I], so we use [B]x = 5. [/B]This means the other number is [B]6[/B].

The sum of twice an integer and 3 times the next consecutive integer is 48 Let the first integer be n This means the next consecutive integer is n + 1 Twice an integer means we multiply n by 2: 2n 3 times the next consecutive integer means we multiply (n + 1) by 3 3(n + 1) The sum of these is: 2n + 3(n + 1) The word [I]is[/I] means equal to, so we set 2n + 3(n + 1) equal to 48: 2n + 3(n + 1) = 48 Solve for [I]n[/I] in the equation 2n + 3(n + 1) = 48 We first need to simplify the expression removing parentheses Simplify 3(n + 1): Distribute the 3 to each term in (n+1) 3 * n = (3 * 1)n = 3n 3 * 1 = (3 * 1) = 3 Our Total expanded term is 3n + 3 Our updated term to work with is 2n + 3n + 3 = 48 We first need to simplify the expression removing parentheses Our updated term to work with is 2n + 3n + 3 = 48 [SIZE=5][B]Step 1: Group the n terms on the left hand side:[/B][/SIZE] (2 + 3)n = 5n [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 5n + 3 = + 48 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 3 and 48. To do that, we subtract 3 from both sides 5n + 3 - 3 = 48 - 3 [SIZE=5][B]Step 4: Cancel 3 on the left side:[/B][/SIZE] 5n = 45 [SIZE=5][B]Step 5: Divide each side of the equation by 5[/B][/SIZE] 5n/5 = 45/5 Cancel the 5's on the left side and we get: n = [B]9[/B]

The sum of two consecutive integers if n is the first integer. consecutive means immediately after, so we have: n n + 1 [U]The sum is written as:[/U] n + n + 1 [U]Grouping like terms, we have:[/U] (n + n) + 1 [B]2n + 1[/B]

The sum of two consecutive integers plus 18 is 123. Let our first integer be n and our next integer be n + 1. We have: n + (n + 1) + 18 = 123 Group like terms to get our algebraic expression: 2n + 19 = 123 If we want to solve the algebraic expression using our [URL='http://www.mathcelebrity.com/1unk.php?num=2n%2B19%3D123&pl=Solve']equation solver[/URL], we get n = 52. This means the next integer is 52 + 1 = 53

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]

There are five consecutive numbers and the smallest is called n. What is the largest number called? List out consecutive numbers. Each consecutive number is found by adding 1 to the prior number [LIST=1] [*]n [*]n + 1 [*]n + 2 [*]n + 3 [*][B]n + 4[/B] [/LIST]

Three good friends are in the same algebra class, their scores on a recent test are three consecutive odd integers whose sum is 273. Find the score In our search engine, we type in [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=3consecutiveintegerswhosesumis273&pl=Calculate']3 consecutive integers whose sum is 273[/URL] and we get: [B]90, 91, 92[/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]

Two consecutive even integers that equal 126 Let the first integer equal x. So the next even integer must be x + 2. The sum which is equal to 126 is written as x + (x + 2) = 126 Simplify: 2x + 2 = 126 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2x%2B2%3D126&pl=Solve']equation calculator,[/URL] we get: x = 62 This means the next consecutive even integer is 62 = 2 = 64. So our two even consecutive integers with a sum of 126 are [B](62, 64)[/B]

two pages that face each other in a book have a sum of 569 Pages that face each other are consecutive. Let the first page be p. The second page is p + 1. Their sum is: p + p + 1 = 569 [URL='https://www.mathcelebrity.com/1unk.php?num=p%2Bp%2B1%3D569&pl=Solve']Type this equation into our search engine to solve for p[/URL], and we get: p = 284 This means p + 1 = 284 + 1 = 285 So the pages that face each other having a sum of 569 are: [B]284, 285[/B]

What is the average of 7 consecutive numbers if the smallest number is called n? [LIST] [*]First number = n [*]Second number = n + 1 [*]Third number = n + 2 [*]Fourth number = n + 3 [*]Fifth number = n + 4 [*]Sixth number = n + 5 [*]Seventh number = n + 6 [/LIST] Average = Sum of all numbers / Total numbers Average = (n + n + 1 + n + 2 + n + 3 + n + 4 + n + 5 + n + 6)/7 Average = 7n + 21/7 Factor out a 7 from the top: 7(n + 3)/7 Cancel the 7's: [B]n + 3[/B]

What is the sum of four consecutive multiples of 5? First number = n Second number = n + 5 Third number = n + 10 Fourth number = n + 15 Add them together: (n + n + n + n) + (5 + 10 + 15) [B]4n + 30[/B]

What pair of consecutive integers gives the following: 7 times the smaller is less than 6 times the larger? Let x and y be consecutive integers, where y = x + 1 We have 7x < 6y as our inequality. Substituting x, y = x + 1, we have: 7x < 6(x + 1) 7x < 6x + 6 Subtracting x from each side, we have: x < 6, so y = 6 + 1 = 7 (x, y) = (6, 7)

When 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 integers. What is the difference of the product minus the sum? Let the 3 consecutive positive integers be: [LIST=1] [*]x [*]x + 1 [*]x + 2 [/LIST] The product is: x(x + 1)(x + 2) The sum is: x + x + 1 + x + 2 = 3x + 3 We're told the product is equivalent to: x(x + 1)(x + 2) = 16(3x + 3) x(x + 1)(x + 2) = 16 * 3(x + 1) Divide each side by (x + 1) x(x + 2) = 48 x^2 + 2x = 48 x^2 + 2x - 48 = 0 Now subtract the sum from the product: x^2 + 2x - 48 - (3x + 3) [B]x^2 - x - 51[/B]