How do you simplify #(3a+9)/a^2-:(a+3)/(a-1)#?

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Answer 1 · 2018-02-27T09:51:57

#(3a+9)/(a^2)-:(a+3)/(a-1)=color(blue)((3(a-1))/a^2#

#(3a+9)/(a^2)-:(a+3)/(a-1)#

Invert the second fraction and multiply.

#(3a+9)/(a^2)xx(a-1)/(a+3)#

#((3a+9)(a-1))/((a^2)(a+3))#

Factor out the common #3# from #(3a+9)#.

#(3(a+3)(a-1))/((a^2)(a+3)#

Cancel #(a+3)# from the numerator and denominator.

#(3color(red)cancel(color(black)((a+3)))(a-1))/((a^2)color(red)cancel(color(black)((a+3)))#

Simplify.

#(3(a-1))/a^2#