How do you solve #12/(a+3)+6/(a^2-9)=8/(a+3)#?

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Answer 1 · 2018-03-27T14:55:07

#color(blue)(x=3/2)#

#12/(a+3)+6/(a^2-9)=8/(a+3)#

factor #(a^2-9)#

This is the difference of two squares:

#(a^2-b^2)=(a+b)(a-b)#

#(a^2-3^2)=(a+3)(a-3)#

#12/(a+3)+6/((a+3)(a-3))=8/(a+3)#

Multiply through by #(x+3)#

#(a+3)12/(a+3)+(a+3)6/((a+3)(a-3))=(a+3)8/(a+3)#

Cancel:

#cancel((a+3))12/cancel((a+3))+cancel((a+3))6/(cancel((a+3))(a-3))=cancel((a+3))8/cancel((a+3))#

#12+6/(x-3)=8#

Subtract #12#

#6/(x-3)=-4#

Multiply by #(x-3)#

#(x-3)6/(x-3)=-4(x-3)#

Cancel:

#cancel((x-3))6/(cancel(x-3))=-4(x-3)#

#6=-4(x-3)#

Divide by #-4#

#x-3=6/-4#

#x=6/-4+3=3/2#

#color(blue)(x=3/2)#