How do you find three consecutive integers such that the sum of twice the smallest and 3 times the largest is 126?

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Answer 1 · 2018-06-09T15:41:22

$24, 25, 26$ $24, 25, 26$

Let's call the smallest integer $n$ $n$ . Then the next two consecutive integers are $n + 1$ $n+1$ and $n + 2$ $n+2$ .

"Sum of twice the smallest and 3 times the largest is 126"

$2 (n) + 3 (n + 2) = 126$ $2(n) + 3(n+2) = 126$

$2 n + 3 n + 6 = 126$ $2n + 3n + 6 = 126$

$5 n + 6 = 126$ $5n + 6 = 126$

$5 n = 120$ $5n = 120$

$n = 24$ $n = 24$

So the smallest integer is $24$ $24$ .

The three consecutive integers are then: $24$ $24$ , $25$ $25$ , and $26$ $26$ .