Find two consecutive even integers such that the sum of the larger and twice the smaller is 62?

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Answer 1 · 2018-06-15T04:16:46

$20 and 22$ $20 and 22$

Let the two consecutive even numbers be: $2 n and 2 (n + 1)$ $2n and 2(n+1)$ for some $n in NN$

Since $n in NN -> 2(n+1) > 2n$

We are told that the sum of the larger plus twice the smaller equals 62.

Hence, $2(n+1) + 2xx 2n =62$

$2n+2+4n =62$

$6n = 60 -> n=10$

Replacing $n=10$ into $2n and 2(n+1)$

$->$ Smaller number $=20$ and larger number $=22$

Check:

$20 and 22$ are consecutive even numbers.

$22+2xx20 = 62$