# How do you simplify #(x-1)/(x^2-1)+2/(5x+5)#?

##### 1 Answer

#### Explanation:

Your starting expression is

#(x-1)/(x^2 -1) + 2/(5x + 5)#

Focus on the first fraction first. Notice that the denominator is the *difference of two squares*, since

#x^2 - 1 = x^2 - 1^2#

As you know, the difference of two squares can be factored as

#color(purple)(|bar(ul(color(white)(a/a)color(black)(a^2 - b^2 = (a-b)(a+b))color(white)(a/a)|)))#

This means that the first fraction is equivalent to

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

Next, focus on the second fraction. In this case, the denominator can be factored as

#2/(5x+5) = 2/(5(x+1))#

The expression can thus be written as

#1/(x+1) + 2/(5(x+1))#

The next thing to do here is multiply the first fraction by **common denominator**, which is equal to

#1/(x+1) * 5/5 + 2/(5(x+1)) = 5/(5(x+1)) + 2/(5(x+1))#

#= (2+5)/(5(x+1)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7/(5(x+1)))color(white)(a/a)|)))#

Notice that you must have