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This is it:\[a^2x+(a-2)=(a+2)x\]Not sure where to start..
I have the answer but I don't know how to get it:\[x=\frac{-1}{a+1}\]
\[a ^{2}x+a-2=ax+2x\] multiply them first then put all the a's to one side..

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Well I shouldn't say not sure where to start I guess.. I got: \[a(ax−x+1)=x(x+2)\]This is where I'm actually stuck at.
how about if we do.. \[a ^{2}x+a-2=ax+2x\]
\[a ^{2}x-a-ax=2x+2\] \[a ^{2}x-ax-2x=2-a\] \[x(a ^{2}-a-2)=2-a\]
here solve for x!!
x=2-a/a^2-a-2 -> factor the denominator a^2-a-2 x= - (-2+a)/(a+1)(a-2) x=-1/(a+1)
Yea I had just tried that way before asking on here and didn't think about factoring the a^2-a-2 so thats where I stopped and tried a different way. Thanks though
welcome.. :)

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