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2- a-3/a+3= a^2-3/a+3 Solve for A....any help would be great, begining to
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\[2-\frac{a-3}{a+3}=\frac{a^2-3}{a+3}\] like this?
Yes!
i would multiply both sides by \[a+3\] to get \[ 2(a+3)-(a-3)=a^2-3\] \[
then don't forget that a = -3 cannot be a solution. solve the quadratic
\[2a+6-a+3=a^2-3\] \[a+9=a^2-3\] \[a^2-a-12=0\] \[(a-4)(a+3)=0\] \[a=4\] \[a=-3\] but don't forget you had \[x+3\] in your original denominator, so -3 is not a solution and you just have \[x=4\]
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