Question

Find four consecutive even integers so that the sum of the first three is 2 more than twice the fourth

Answer 1

Answer:

The numbers are : -20,-18,-16,-14

Step-by-step explanation:

Let 1st even integer be $=x$

So, next consecutive even integer would be $=x+2$

3rd consecutive even integer would be $=x+4$

4th consecutive even integer would be $=x+6$

Sum of first three integers would be $=x+x+2+x+4=3x+6$

It says sum of the first three is 2 more than twice the fourth,

So we have

$3x+6=2+4(x+6)$

using distribution to simplify.

$3x+6=2+(4\times x)+(4\times6)$

$3x+6=2+4x+24$

$3x+6=4x+26$

Subtracting $4x$ from both sides

$3x+6-4x=4x+26-4x$

$-x+6=26$

Subtracting 6 from both sides

$-x+6-6=26-6$

$-x=20$

∴ $x=-20$

So first integer =-20

2nd integer  $=-20+2=-18$

3rd integer $=-18+2=-16$

4th integer  $=-16+2=-14$

So, the numbers are : -20,-18,-16,-14