Question

find three consecutive integers. such that 4 times of first integer is 18 more than the sum of the second and third integer

Answer 1

Three consecutive numbers are x, x+1 and x+2.

Four times the first integer is 4x

The sum of the second and third is (x+1)+(x+2)=2x+3.

So, we have

$4x = 2x+3+18\iff 4x=2x+21$

Subtract 2x from both sides:

$2x=21$

Divide both sides by 2:

$x=10.5$

So, you can't have three consecutive integers such that four times the first is 18 more than the sum of the other two: the three numbers would be 10.5, 11.5, 12.5.

In fact, you have

$4\cdot 10.5 = 42$

and

$11.5+12.5+18 = 24+18=42$