Given 9x^2=16(y^2+4), then what is 64/(3x-4y)?

Given 9x^2=16(y^2+4), then what is
64/(3x-4y)?

First, distribute 16 on the right side of the
equation.

9x^2 = 16y^2 +
64

Now move the y-term to the left
side.

9x^2 - 16y^2 =
64

Substitute the polynomial in for 64 in the numerator of
64/(3x-4y).

(9x^2 - 16y^2)/(3x -
4y)

The numerator is an example of the
Difference of Squares.

[(3x +
4y)(3x - 4y)]/(3x - 4y)

The term (3x - 4y) is both the
numerator and denominator, so they cancel out.  You are left with 3x +
4y.

Solution:  3x +
4y