Question

1/(x-2) - 3x/(x-1) = (2x+1)/(x^2-3x+2)

Answer 1

\[\frac{ 1 }{ x-2 }-\frac{ 3x }{ x-1 }=\frac{ 2x+1 }{ x ^{2}-3x+2 }\]

Answer 2

I started by factoring the denominator on the right side:
\[\frac{ 1 }{ x-2 }-\frac{ 3x }{ x-1 }= \frac{ 2x+1 }{ (x-2)(x-1) }\]

Answer 3

And then used this as an LCD:
\[\frac{ 1(x-1) }{ (x-2)(x-1) }-\frac{ 3x(x-2)}{ (x-2)(x-1) } = \frac{ 2x+1 }{ (x-2)(x-1) }\]

Answer 4

I'm not sure what to do after this. It seems I'm messing up signs each time I try this equation.
\[\frac{ x-1 }{ (x-2)(x-1) }-\frac{ 3x ^{2}-6}{ (x-2)(x-1) } = \frac{ 2x+1 }{ (x-2)(x-1) }\]

Answer 5

The answer is x= 2/3, but I'm not sure how to get there.

Answer 6

your last line is wrong when your multiplying through the LCD:

I find it simpler to look at it as:
\[\frac{1(x-1)-3x(x-2)}{(x-2)(x-1)}=\frac{2x+1}{(x-2)(x-1)}\]

multiply both sides by (x-2)(x-1) to get:
\[1(x-1)-3x(x-2)=2x+1\]

simplifiy:
\[x-1-3x^2-6x=2x+1\]
\[-3x^2+5x-2=0\]

Then use quadratic formula to solve:
\[x=\frac{-b +\sqrt{ b^2-4ac}} {2a} \]

I get 2 answers: 2 or \[-\frac{2}{3}\]

Answer 7

however, we know 2 is not a solution, because the equation at 2 is undefined.