Find two consecutive odd integers such that their product is 119 more than 7 times their sum.

Answer - (19 and 21) or (-7 and - 5)

Solution -

let the first odd number is x, then its next odd number will be x+2.

So sum of these numbers = x + (x+2) and product of these numbers x * (x+2)

from the question it can be concluded that

x (x+2) = 7 (x+x+2) + 119

⇒ x² + 2x = 7 (2x+2) + 119

⇒ x² + 2x = 14x + 133

⇒ x² - 12x - 133 = 0

⇒ x² - 19x + 7x - 133 = 0

⇒ x (x-19) + 7 (x-19) = 0

⇒ (x-19) (x+7) = 0

⇒ x = 19, - 7

so here the value of x is either 19 or - 7

if the first odd number is 19, then the next odd number is 21 and if the first number is - 7, then the next number should be - 5.