SOLUTION: [(6x-x^2-6)/(x-1)]-[(2x-3)/(x-1)]=1 I worked it out and come up with x=2 and x=1 When I plug these numbers in neither one is actually true. What am i doing wrong?

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Question 286352: [(6x-x^2-6)/(x-1)]-[(2x-3)/(x-1)]=1
I worked it out and come up with x=2 and x=1 When I plug these numbers in neither one is actually true. What am i doing wrong?
Answer by subudear(62) (Show Source):
You can put this solution on YOUR website!
[(6x-x^2-6)/(x-1)]-[(2x-3)/(x-1)]=1
[-(x^2-6x+6)/(x-1)]-[(2x-3)/(x-1)]=1
Now (x-1) is common denominator so we can write
[-(x^2-6x+6)-(2x-3)]/(x-1)=1
[-x^2 + 6x - 6 - 2x + 3]/(x-1)=1
[-x^2 +4x -3]/(x-1)=1
[-(x^2 -4x +3)]/(x-1)=1
[-(x^2 -4x +3)]=x-1
-x^2 +4x -3 -x +1 =0
-x^2 +3x -2 =0
-(x^2 -3x +2) =0
do factors of (x^2 -3x +2)
(x^2 -3x +2) = x^2 -2x-x +2 = x(x-2)-(x-2) = (x-2)(x-1)
replacing the values we get
-(x-2)(x-1)=0
so x =1,2 but x =1 cannot be the answer as the equation will be undefined for x=1 so x=2 is the only answer.