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

2 Answers
Sep 4, 2016

#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))#

Sep 4, 2016

#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}#