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

Answer 1

#(x-1)/(x^2-1)+2/(5x+5)=7/(5x+5)#

#(x-1)/(x^2-1)+2/(5x+5)#
= #(x-1)/((x-1)(x+1))+2/(5(x+1))#
= #(1cancel((x-1)))/(cancel((x-1))(x+1))+2/(5(x+1))#
= #1/((x+1))+2/(5(x+1))#
= #(5xx1)/(5(x+1))+2/(5(x+1))#
= #5/(5(x+1))+2/(5(x+1))#
= #7/(5(x+1))#
= #7/(5x+5)#
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Answer 2

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)).

Combining like terms in the numerator, we have (5x^2 - 5 + 2x^2 - 2)/((x-1)(x+1)(5x+5)).

Simplifying further, we get (7x^2 - 7)/((x-1)(x+1)(5x+5)).

Finally, we can factor out a 7 from the numerator, resulting in 7(x^2 - 1)/((x-1)(x+1)(5x+5)).

Therefore, the simplified expression is 7(x^2 - 1)/((x-1)(x+1)(5x+5)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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