Find three consecutive integers such that 4 times the first is 13 more than the third

let the first integer be n

the second is (n+1), the third (n+2)

4n = (n+2) + 13

4n = n + 15

3n = 15

n = 5

4 x 5 = 20 = 7 + 13

the 3 integers are 5, 6, 7

Let the integers be x, x + 1 and x + 2

So we have

4x = x+ 2 + 13 simplify

4x = x + 15 subtract x from both sides

3x = 15 divide both sides by 3

x = 5

So...the integers are 5, 6, 7