How do you find three consecutive even integers such that three times the largest is 34 more than the sum of the two smaller integers?

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Answer 1 · 2017-07-04T02:52:31

See a solution process below:

Let's call the smallest of the three consecutive integers: $n$ $n$

Because these are consecutive even integers it means we can write the other two integers as:

$n + 2$ $n + 2$

$n + 4$ $n + 4$

Now, we can write "three times the largest" as:

$3 (n + 4)$ $3(n + 4)$

And if this is equal to "34 more than the sum of the two smaller integers" we can write this as the equation:

$3 (n + 4) = 34 + n + n + 2$ $3(n + 4) = 34 + n + n + 2$

We can now solve for $n$ $n$ :

$3 (n + 4) = 34 + 1 n + 1 n + 2$ $3(n + 4) = 34 + 1n + 1n + 2$

$3 (n + 4) = 1 n + 1 n + 2 + 34$ $3(n + 4) = 1n + 1n + 2 + 34$

$3 (n + 4) = (1 + 1) n + 36$ $3(n + 4) = (1 + 1)n + 36$

$3 (n + 4) = 2 n + 36$ $color(red)(3)(n + 4) = 2n + 36$

$(3 \cdot n) + (3 \cdot 4) = 2 n + 36$ $(color(red)(3) * n) + (color(red)(3) * 4) = 2n + 36$

$3 n + 12 = 2 n + 36$ $3n + 12 = 2n + 36$

$- 2 n + 3 n + 12 - 12 = - 2 n + 2 n + 36 - 12$ $-color(blue)(2n) + 3n + 12 - color(red)(12) = -color(blue)(2n) + 2n + 36 - color(red)(12)$

$(- 2 + 3) n + 0 = 0 + 24$ $(-color(blue)(2) + 3)n + 0 = 0 + 24$

$1 n = 24$ $1n = 24$

$n = 24$ $n = 24$

Therefore the three consecutive even integers are:

$n = 24$ $n = 24$

$n + 2 = 26$ $n + 2 = 26$

$n + 4 = 28$ $n + 4 = 28$