How do you simplify #a/(1-a) + (3a)/(a+1 )- 5/(a^2-1)#?
2 Answers
Explanation:
Note that
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))#
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.