Question

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

Answer 1

`7/(5(x+1))`

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 `1=5/5`. This will allow you to express both fractions in terms of their common denominator, which is equal to `5(x+1)`

`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 `x!= +-1`.