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
Subtract 2x from both sides:
Divide both sides by 2:
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
and