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 $-4$

$x-3=6/-4$

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

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

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