How do you solve #\frac { x ^ { 2} + 64} { x ^ { 2} - 64} = \frac { x } { x + 8} - \frac { 8} { x - 8}#?

1 Answer
Steve M
Sep 5, 2017

there are no solutions to the given equation

Explanation:

We have:

# (x^2+64)/(x^2-64) = x/(x+8)-8/(x-8) #

# :. (x^2+64)/(x^2-64) = (x(x-8)-8(x+8)) / ( (x+8)(x-8) ) #

# :. (x^2+64)/(x^2-64) = { x^2-8x-8x-64 } / ( x^2-8x+8x-64 ) #

# :. (x^2+64)/(x^2-64) = { x^2 - 16x -64 } / ( x^2-64 ) #

The denominator is zero, if

# x^2-64 = 0 => x^2=64 => x = +- 8#

Providing that the denominator is nonzero, then we have:

# x^2+64 = x^2 - 16x -64 #
# :. 16x = -128 #
# :. x = -8 #

However because this value would make the denominator zero it has to be excluded, hence there are no solutions to the given equation

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