How do you evaluate #\frac{1}{x^{2}-16}=-\frac{1}{x^{2}-4x}#?

First, factor the denominators of each fraction as:

#1/((x + 4)(x - 4)) = -1/(x (x - 4)#

Next, multiply each side of the equation by #color(red)((x - 4))# to eliminate a common factor while keeping the equation balanced:

#color(red)((x - 4)) xx 1/((x + 4)(x - 4)) = color(red)((x - 4)) xx -1/(x(x - 4)#

#cancel(color(red)((x - 4))) xx 1/((x + 4)color(red)(cancel(color(black)((x - 4))))) = cancel(color(red)((x - 4))) xx -1/(xcolor(red)(cancel(color(black)((x - 4)))))#

#1/(x + 4) = -1/x#

Then, because we have a pure fraction on each side of the equation we can flip the fractions giving:

#(x + 4)/1 = -x/1#

#x + 4 = -x#

Next, we can subtract #color(red)(4)# and add #color(blue)(x)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#color(blue)(x) + x + 4 - color(red)(4) = color(blue)(x) - x - color(red)(4)#

#1color(blue)(x) + 1x + 0 = 0 - 4#

#(1 + 1)x = -4#

#2x = -4#

Now, divide both sides of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = -4/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -2#

#x = -2#

#1/(x^2-16)=-1/(x^2-4x)#

We can multiply both sides of the equation by #(x^2-16)(x^2-4x)#

#color(red)(cancel((x^2-16))(x^2-4x))/cancel(x^2-16)=-color(red)((x^2-16)cancel((x^2-4x)))/cancel(x^2-4x)#

#x^2-4x=-(x^2-16)#
#x^2-4x=-x^2+16#

Solve like a quadratic:

#2x^2-4x-16=0#
#x^2-2x-8=0#
#(x-4)(x+2)=0#

#x=-2# or #x=4#

We need to check our answers however.
This is because we can't divide by zero, this is undefined

Because of this, if we let #x=4#, we get denominators of 0. This cannot happen, so #x!=4# This means that the only solution is #x=-2#