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
[tex]4x = 2x+3+18\iff 4x=2x+21[/tex]
Subtract 2x from both sides:
[tex]2x=21[/tex]
Divide both sides by 2:
[tex]x=10.5[/tex]
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
[tex]4\cdot 10.5 = 42[/tex]
and
[tex]11.5+12.5+18 = 24+18=42[/tex]
