One positive number is 5 less than twice a second number, and their product is 117. Find the two numbers.

x, y - the numbers

one positive number is 5 less than twice a second number

(1) x - 5 = 2y

their product is 117

(2) xy = 117

(1) x - 5 = 2y add 5 to both sides

x = 2y + 5 substitute it to (2)

(2y + 5) y = 117 use distributive property

(2y) (y) + (5) (y) = 117

2y² + 5y = 117 subtract 117 from both sides

2y² + 5y - 117 = 0

2y² + 18y - 13y - 117 = 0

2y (y + 9) - 13 (y + 9) = 0

(y + 9) (2y - 13) = 0 ↔ y + 9 = 0 ∨ 2y - 13 = 0

y + 9 = 0 subtract 9 from both sides

y = - 9

2y - 13 = 0 add 13 to both sides

2y = 13 divide both sides by 2

y = 6.5

substitute the values of y to (1)

x = 2 (-9) + 5 = - 18 + 5 = - 13 < 0

x = 2 (6.5) + 5 = 13 + 5 = 18

Answer: x = 18 and y = 6.5