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

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

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step-by-step explanation: