SOLUTION: Positive integers x and y have a product of 56 and x < y. Seven times the reciprocal of the smaller integer plus 14 times the reciprocal of the larger integer equals 4. What is the

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: Positive integers x and y have a product of 56 and x < y. Seven times the reciprocal of the smaller integer plus 14 times the reciprocal of the larger integer equals 4. What is the      Log On


   



Question 921289: Positive integers x and y have a product of 56 and x < y. Seven times the reciprocal of the smaller integer plus 14 times the reciprocal of the larger integer equals 4. What is the value of x?
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
xy = 56
7(1/x) + 14(1/y) = 4

Simplify the second equation.
7/x + 14/y = 4
(7y + 14x)/(xy) = 4
7y + 14x = 4 * 56 = 224 ----> 2x + y = 32 ---> y = 32 - 2x

Substitute this into the first equation.

xy = 56
x(32 - 2x) = 56
-2x^2 + 32x - 56 = 0
x^2 - 16x + 28 = 0
(x - 2)(x - 14) = 0

x is the smaller integer, so x = 2 and y = 28.