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find three consecutive even intergers such that the sum of the smallest number and twice the middle number is 20 more than the largest number

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Answer:

10, 12, 14

Explanation:

Hi there!

Let x be equal to the smallest integer.

Let (x+2) be equal to the next even integer.

Let (x+4) be equal to the largest even integer out of the three.

We're given:

⇒ smallest number + (2 × middle number) = 20 + largest number

Construct an equation:


x+2(x+2)=20+(x+4)

Combine like terms:


x+2x+4=20+x+4\\3x+4=x+24\\2x=20\\x=10

Therefore, the smallest integer is 10, making the next two even integers 12 and 14.

I hope this helps!

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User Bklyn
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