# How do you combine #(x-2)/(x-6)-(x+2)/(6-x)#?

##### 1 Answer

#### Explanation:

The trick here is to realize that you can write

#6 -x = -1 * (-6 + x) = -1 * (x-6)#

This means that the expression can be rewritten as

#(x-2)/(x-6) - (x+2)/(-1 * (x-6)) = (x-2)/(x-6) - (x+2)/(-(x-6))#

This will be equivalent to

#(x-2)/(x-6) + (x+2)/(x-6)#

You now have two fractions that have the **same denominator**, which means that you can go ahead and combine their *numerators*

#(x-2)/(x-6) + (x+2)/(x-6) = (x - color(red)(cancel(color(black)(2))) + x + color(red)(cancel(color(black)(2))))/(x-6) = color(green)(|bar(ul(color(white)(a/a)color(black)((2x)/(x-6))color(white)(a/a)|)))#

**ALTERNATIVELY**

You can also combine these two fractions by finding their **common denominator**, which in this case would be

Multiply the first fraction by

#(x-2)/(x-6) - (x+2)/(6-x) = (x-2)/(x-6) * (6-x)/(6-x) - (x+2)/(6-x) * (6-x)/(6-x)#

#=((x-2)(6-x))/((x-6)(6-x)) - ((x+2)(6-x))/((x-6)(6-x))#

#= ((x-2)(6-x) - (x+2)(x-6))/((x-6)(6-x))#

Focus on the **numerator** first

#6x - x^2 - color(red)(cancel(color(black)(12))) + color(red)(cancel(color(black)(2x))) - x^2 + 6x - color(red)(cancel(color(black)(2x))) + color(red)(cancel(color(black)(12))) = -2x^2 + 12x#

This can be rewritten as

#-2x^2 + 12x = 12x - 2x^2 = 2x * (6 - x)#

The expression will once again be equal to

#(2x * color(red)(cancel(color(black)((6-x)))))/((x-6)color(red)(cancel(color(black)((6-x))))) = color(green)(|bar(ul(color(white)(a/a)color(black)((2x)/(x-6))color(white)(a/a)|)))#

Keep in mind that you need to have