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

Question

Eva Abramson · Accepted Answer

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

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#.

Kennedy Adams · Answer

undefined To simplify the expression (x-1)/(x^2-1)+2/(5x+5), we first factor the denominators. The denominator x^2-1 can be factored as (x-1)(x+1), and the denominator 5x+5 can be factored as 5(x+1).

Next, we find the least common denominator (LCD) of the two fractions, which is (x-1)(x+1)(5x+5).

To make the fractions have a common denominator, we multiply the first fraction by (5x+5)/(5x+5), and the second fraction by (x-1)(x+1)/((x-1)(x+1)).

After simplifying, we get ((x-1)(5x+5)+2(x-1)(x+1))/((x-1)(x+1)(5x+5)).

Expanding and combining like terms in the numerator, we have (5x^2-5+2x^2-2)/(5x^2-5).

Simplifying further, we get (7x^2-7)/(5x^2-5).

This expression cannot be simplified any further.