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
