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

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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**

We'll remove the brackets from the right side and we'll move the terms to the left:

9x^2 - 16y^2 - 64 = 0

The difference of two squares will be replaced by the product:

(3x - 4y)(3x + 4y) = 64

We'll divide both sides by 3x - 4y:

64/(3x - 4y) = 3x + 4y

**Therefore, the value of the fraction 64/(3x - 4y) is 3x + 4y.**