How do you find three consecutive even integers whose sum is 16 more than two times the middle integer?

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Answer 1 · 2016-10-24T04:52:40

The three consecutive even integers are $14$ $14$ , $16$ $16$ and $18$ $18$ .

We shall consider the three consecutive even integers as $x$ $x$ , $x + 2$ $x+2$ and $x + 4$ $x+4$ .

From the details we know that the sum of all three is $16$ $16$ more than the twice the middle integer.: Expressing this in an equation:

$x + (x + 2) + (x + 4) = 2 (x + 2) + 16$ $x+(x+2)+(x+4)=2(x+2)+16$

Opening the brackets and simplifying:

$x + x + 2 + x + 4 = 2 x + 4 + 16$ $x+x+2+x+4=2x+4+16$

$3 x + 6 = 2 x + 20$ $3x+6=2x+20$

Shifting terms and simplifying:

$3 x - 2 x = 20 - 6$ $3x-2x=20-6$

$x = 14$ $x=14$

$:. x=14, x+2=16, x+4=18$