Four times the first of three consecutive even integers is six more than the product of two and the third integer. find the integers.

First integer = x

Second integer = x + 2

Third integer = x + 4

Since four times the first integer equals six more than the product of two and the third integer.

=> 4x = 6 + 2(x + 4)

=> 4x = 6 + 2x + 8

=> 2x = 14

=> x = 7.

Hence,

First integer = x = 7

Second integer = x + 2 = 7 + 2 = 9

Third integer = x + 4 = 7 + 4 = 11.

Hope it helps