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?

Question

6582 views around the world

You can reuse this answer
Creative Commons License

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:

$n = 24$ $n = 24$

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

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