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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.
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