SOLUTION: I could really use your help today, I am stuck.
x/x-2=1/2-x+2/x
2x(x)/2x(x-2)=x(x-2)/2x(x-2)-2(x-2)(x+2)/2x(x-2)
2x^2/2x(x-2)=x^2-2x/2x(x-2)-x^2+8/2x(x-2)
I am lost from h
Algebra.Com
Question 285289: I could really use your help today, I am stuck.
x/x-2=1/2-x+2/x
2x(x)/2x(x-2)=x(x-2)/2x(x-2)-2(x-2)(x+2)/2x(x-2)
2x^2/2x(x-2)=x^2-2x/2x(x-2)-x^2+8/2x(x-2)
I am lost from here as the last term numerator is negative. I must have made a wrong turn along the way. I appreciate your help.
Thank you,
Tina
Answer by user_dude2008(1862) (Show Source): You can put this solution on YOUR website!
x/(x-2)=1/2-(x+2)/x
2x(x-2)[x/(x-2)]=2x(x-2)[1/2]-2x(x-2)[(x+2)/x]
2x^2=x(x-2)-2(x-2)(x+2)
2x^2=x^2-2x-2x^2+8
3x^2+2x-8=0
(x+2)(3x-4)=0
x=-2 or x=4/3 <--------------- Answers
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