How do you solve $\frac{x}{x + 4} - \frac{4}{x - 4} = \frac{x^{2} + 16}{x^{2} - 16}$ $x/(x+4) - 4/(x-4) = (x^2+16)/(x^2-16)$ ?

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How do you solve $\frac{x}{x + 4} - \frac{4}{x - 4} = \frac{x^{2} + 16}{x^{2} - 16}$ $x/(x+4) - 4/(x-4) = (x^2+16)/(x^2-16)$ ?

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Answer 1 · 2015-06-16T23:55:00

NO SOLUTIONS!!!!

Explanation:

Consider as common denominator $x^{2} - 16 = (x - 4) (x + 4)$ $x^2-16=(x-4)(x+4)$ :
$(x(x-4)-4(x+4))/cancel((x^2-16))=(x^2+16)/cancel((x^2-16))$
$cancel(x^2)-4x-4x-16=cancel(x^2)+16$
$-8x=32$
$x=-4$
BUT this value cannot be accpeted because it makes the denominator of my original expression equal to ZERO!!!
So...NO real solutions!!!!

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How do you solve $\frac{x}{x + 4} - \frac{4}{x - 4} = \frac{x^{2} + 16}{x^{2} - 16}$ $x/(x+4) - 4/(x-4) = (x^2+16)/(x^2-16)$ ?

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How do you solve x/(x+4) - 4/(x-4) = (x^2+16)/(x^2-16)xx+4−4x−4=x2+16x2−16x/(x+4) - 4/(x-4) = (x^2+16)/(x^2-16)?

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How do you solve $\frac{x}{x + 4} - \frac{4}{x - 4} = \frac{x^{2} + 16}{x^{2} - 16}$ $x/(x+4) - 4/(x-4) = (x^2+16)/(x^2-16)$ ?