Question

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

Answer 1

See a solution process below:

Explanation:

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`

Answer 2

`x=-2`

Explanation:

`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`