The sum of three consecutive integers is 31 more than the largest integer. Find the integers

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The sum of three consecutive integers is 31 more than the largest integer. Find the integers

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    0
    2021-09-09T20:48:29+00:00

    To solve with an equation, let’s call the three consecutive integers g, g+1, and g+2.

    So, (g)+(g+1)+(g+2)=(g+2)+31

    3g + 3 = g + 33

    2g = 30

    g = 15

    The integers are 15, 16, and 17.

    To check:

    The sum of 15, 16, and 17 is 48. 17 + 31 is also 48.

    0
    2021-09-09T20:48:46+00:00

    Let’s express this in algebraic terms:

    (x-2)+(x-1)+x=x+31\\x-2+x-1+x=x+31\\3x-3=x+31\\3x=x+31+3\\3x=x+34\\3x-x=34\\2x=34\\x=34/2\\x=17

    The expressions for the integers and respective values are expressed below:

    x=17\\\\Integer 1 => x =17\\Integer 2 => x-1=17-1=16\\Integer 3 => x-2=17-2=15

    Now, to check, we apply the original problem and the values we discovered:

    (x-2)+(x-1)+x=x+31\\((17)-2)+((17)-1)+(17)=(17)+31\\(15)+(16)+17=48\\48=48

    We are right, so:

    x=17\\\\Integer 1 => 17\\Integer 2 => 16\\Integer 3 => 15

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