Question

How do you simplify `a/(1-a) + (3a)/(a+1 )- 5/(a^2-1)`?

Answer 1

`a/(1-a)+(3a)/(a+1)-5/(a^2-1) = (2a^2-4a-5)/(a^2-1)`

`color(white)(a/(1-a)+(3a)/(a+1)-5/(a^2-1)) = 2-(4a+3)/(a^2-1)`

Explanation:

Note that `a^2-1 = (a-1)(a+1)`

So we find:

`a/(1-a)+(3a)/(a+1)-5/(a^2-1) = (-a(a+1)+3a(a-1)-5)/(a^2-1)`

`color(white)(a/(1-a)+(3a)/(a+1)-5/(a^2-1)) = (-a^2-a+3a^2-3a-5)/(a^2-1)`

`color(white)(a/(1-a)+(3a)/(a+1)-5/(a^2-1)) = color(blue)((2a^2-4a-5)/(a^2-1))`

`color(white)(a/(1-a)+(3a)/(a+1)-5/(a^2-1)) = (2a^2-2-4a-3)/(a^2-1)`

`color(white)(a/(1-a)+(3a)/(a+1)-5/(a^2-1)) = (2(a^2-1)-(4a+3))/(a^2-1)`

`color(white)(a/(1-a)+(3a)/(a+1)-5/(a^2-1)) = color(blue)(2-(4a+3)/(a^2-1))`

Answer 2

`frac{a}{1-a} + frac{3a}{a+1} - frac{5}{a^2-1} = 2 + frac{4a+3}{1-a^2}`

Explanation:

Make the denominator the same for all 3 fractions. After which add up the numerators of the 3 fractions.

`frac{a}{1-a} + frac{3a}{a+1} - frac{5}{a^2-1}`

`= frac{a}{1-a} color(blue)(* frac{a+1}{a+1}) + frac{3a}{a+1} color(blue)(* frac{1-a}{1-a}) + frac{5}{1-a^2}`

`= frac{a^2+a}{1-a^2} + frac{3a-3a^2}{1-a^2} + frac{5}{1-a^2}`

`= frac{(a^2+a)+(3a-3a^2)+5}{1-a^2}`

`= frac{-2a^2+4a+5}{1-a^2}`

To simplify even more, perform long division.

`frac{-2a^2+4a+5}{1-a^2} = 2 + frac{4a+3}{1-a^2}`