Question

there are 3 consecutive even integers .if twice the first integer added to the third is 268,222, find all three integers

Answer 1

Answer:

89406, 89408 and 89410.

Step-by-step explanation:

Let n be the any even integer.

So n+2 will be 2nd even integer and n+4 will be 3rd even integer.

We are told that twice the first integer added to the third is 268,222.

Twice the first integer will be 2*n.

We can represent this information in an equation as:

$2n+(n+4)=268222$

$2n+n+4=268222$

$3n+4=268222$

Let us subtract 4 from both sides of our equation.

$3n+4-4=268222-4$

$3n=268218$

Let us divide both sides of our equation by 3.

$\frac{3n}{3}=\frac{268218}{3}$

$n=89406$

Therefore, our 1st even integer will be 89406.

2nd even integer = n+2

$n+2=89406+2=89408$

3rd even integer = n+4

$n+4=89406+4=89410$

Therefore, our three consecutive even integers will be 89406, 89408 and 89410.