What is the domain of #(4/(a+1))-(5/a) = (20/(a^2+a))#?

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Answer 1 · 2015-10-26T03:26:33

#a={-25}#

#[1]" "4/(a+1)-5/a=20/(a^2+a)#

Combine #4/(a+1)# and #-5/a# by making them have the same denominator.

#[2]" "(4(a))/(a(a+1))-(5(a+1))/(a(a+1))=20/(a^2+a)#

#[3]" "(4(a)-5(a+1))/(a(a+1))=20/(a^2+a)#

#[4]" "(4a-5a-5)/(a^2+a)=20/(a^2+a)#

#[5]" "(-a-5)/(a^2+a)=20/(a^2+a)#

Multiply both sides by #(a^2+a)# to remove the denominator.

#[6]" "cancel(a^2+a)((-a-5)/cancel(a^2+a))=cancel(a^2+a)(20/cancel(a^2+a))#

#[7]" "-a-5=20#

#[8]" "-a=25#

#[9]" "color(blue)(a=-25)#