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Find three consecutive even integers such that the sum of 5 and the second is multiplied by -7, the result is 11 greater than 5 times the opposite of the third

User Tashae
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1 Answer

10 votes

Answer:

-20, -18, -16

Explanation:

For consecutive integer problems, it often works well to let a variable represent the middle one (or their average value).

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Here, we can let x represent the second integer. Then the first is (x-2), and the third is (x+2). The given relation can be written as ...

(5 +x)(-7) = 11 +5(-(x+2))

-35 -7x = 11 -5x -10 . . . . . . eliminate parentheses

-36 = 2x . . . . . . . . . add 7x -1

-18 = x . . . . . . . . divide by 2

The three integers are -20, -18, -16.

User Brown Limie
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