Question

How do I solve for X?

`(x^2 - 16) / (3x) * 12 / (2x - 8)`

Answer 1

without out an equal sign you can't solve for x, but you can simplify it to `(2( x +4))/ x`

Explanation:

I scan the problem to see if I can simplify any terms. Two things jumped out at me.

first on the bottom of the second fraction I can factor out a 2 from `2x -8` which would become `2(x -4)`

Then I noticed the terms on top of the first fraction can be written as the difference of two squares. http://www.purplemath.com/modules/specfact.htm) .

So I rewrote `x^1 - 16` as `(x-4)(x+4)`

The problem now looks like: `((x-4)(x+4) )/ (3x) * 12 / (2(x - 4)`

Start by canceling out anything that the same on any part of top and bottom.

`(cancel((x-4))(x+4) )/ (3x) * 12 / (2(cancel(x - 4))`

Next I see there is a 12 on the top and a 3 on the bottom. I can divide 12 by 3 to get 4. The problem now look like this

`((x+4) )/ (x) * 4 / 2`

Now I divide the 4 by 2 and get

`((x+4) )/ (x) * 2`

Write the equation and you get

`(2( x +4))/ x`

If you need any more explanation please just let me know.

Answer 2

Because this is not an equation you cannot solve for `X`.

See a process below for how to simplify this expression if this is what you are looking to do:

Explanation:

First, factor the denominator of the fraction on the left with the numerator of the fraction on the right:

`(x^2 - 16)/(3x) * 12/(2x - 8) => (x^2 - 16)/(color(red)(cancel(color(black)(3)))x) * (color(red)(cancel(color(black)(12)))4)/(2x - 8) =>`

`(x^2 - 16)/x * 4/(2x - 8)`

Next, factor and cancel the common terms for the numerator of the fraction on the left and the denominator of the fraction on the right:

`(x^2 - 16)/x * 4/(2x - 8) => ((x + 4)(x - 4))/x * 4/(2(x - 4)) =>`

`((x + 4)color(red)(cancel(color(black)((x - 4)))))/x * 4/(2color(red)(cancel(color(black)((x - 4))))) =>`

`(x + 4)/x * 4/2 => (x + 4)/x * 2 => (2x + 8)/x`

If necessary, we can split this result into two fractions as:

`(2x + 8)/x => (2x)/x + 8/x => 2 + 8/x`