Word problems (solve)

Question

Find three consecutive positive integers such that the product of the first two is 11 more than five times the third.

Answer 1

Three consecutive positive integers

n, (n+1), (n+2)

The product of the first two

n(n+1)

is 11 more than 5 times the third

5(n+2) + 11

11 more than

so we have

n(n+1) = 5(n+2) + 11

n² + n = 5n + 10 + 11

n² + n - 5n - 21 = 0

n² - 4n - 21 = 0

(n - 7)(n + 3) = 0

n is 7 or n = -3. Since n is positive, n = 7.

So we have 7, 8, and 9.

Checking, 7•8 = 5(9) + 11

56 = 45 + 11

56 = 56

Check!