The sum of twice an integer and 3 times the next consecutive integer is 48

Discussion in 'Calculator Requests' started by math_celebrity, Oct 21, 2021.

  1. math_celebrity

    math_celebrity Administrator Staff Member

    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 is means equal to, so we set 2n + 3(n + 1) equal to 48:
    2n + 3(n + 1) = 48

    Solve for n 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

    Step 1: Group the n terms on the left hand side:
    (2 + 3)n = 5n

    Step 2: Form modified equation
    5n + 3 = + 48

    Step 3: Group constants:
    We need to group our constants 3 and 48. To do that, we subtract 3 from both sides
    5n + 3 - 3 = 48 - 3

    Step 4: Cancel 3 on the left side:
    5n = 45

    Step 5: Divide each side of the equation by 5
    5n/5 = 45/5

    Cancel the 5's on the left side and we get:
    n = 9
     

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