The product of two positive integers is 546. If thefirst integer is 11 greater than twice the secondinteger, what is the smaller of the two integers?A) 7B) 14C) 39D) 78

Let's denote the two integers as x and y, where y is the smaller integer. According to the problem, we have two equations:

1. x*y = 546 (The product of the two integers is 546)
2. x = 2y + 11 (The first integer is 11 greater than twice the second integer)

We can substitute equation 2) into equation 1) to get:

(2y + 11) * y = 546 2y^2 + 11y - 546 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = 2, b = 11, and c = -546. We can solve this equation for y using the quadratic formula y = [-b ± sqrt(b^2 - 4ac)] / (2a):

y = [-11 ± sqrt((11)^2 - 42(-546))] / (2*2) y = [-11 ± sqrt(121 + 4368)] / 4 y = [-11 ± sqrt(4489)] / 4 y = [-11 ± 67] / 4

We get two solutions: y = 14 and y = -78. However, since y is a positive integer, we discard the negative solution. Therefore, the smaller of the two integers is 14, which corresponds to option B.