Question

Find the 2nd of three consecutive positive even integers such that the product o the second and third integers is twenty more than ten times the first integer.

Answer 1

10

Explanation:

Let the three consecutive positive even integers be x-1, x and x+1

First = x-1

Second = x

Third = x+1

If the product of the second and third integers is twenty more than ten times the first integer, then;

x(x+1) = 20 + 10(x-1)

Open the bracket

x² + x = 20 + 10x - 10

x² + x = 10x + 10

x² + x - 10x - 10 = 0

x² - 9x - 10 = 0

x² -10x + x - 10 = 0

x(x-10)+ 1(x-10) = 0

(x+1)(x-10) = 0

x+1 = 0 and x-10 = 0

x = -1 and 10

Hence the 2nd of three consecutive positive even integers is 10