Originally Posted by

**harish21** No it does not. You are subtracting x and 1 from both sides of the equation.

$\displaystyle (x+1)=(6x^2-5x+1)$

subtract 1 from both the sides:

$\displaystyle (x+1)-1=(6x^2-5x+1)-1$

that leaves you with:

$\displaystyle x = 6x^2-5x$

now you subtract x from both the sides. this gives you:

$\displaystyle x-x=(6x^2-5x)-x$

which gives you:

$\displaystyle 0=6x^2-5x-x$

therefore $\displaystyle 0=6x^2-6x$

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Note: $\displaystyle \boxed{-5x-x=-6x} \;\;and \;\;\boxed{-5x+x=-4x}$

In the first box, you are subtracting x from -5x. In the second box, you are adding x to -5x. This is where your confusion lies.

If you still dont understand it, please refer to your books/notes/instructor to avoid trouble in the future.

Please refer to post Number 2 by **Unknown008**

your question was $\displaystyle \dfrac{3}{2x - 1} - \dfrac{4}{3x-1}=1$

then **Unknown008**'s method (which you've been told multiple number of times) takes you :

$\displaystyle \dfrac{x+1}{6x^2 - 5x + 1}=1$

multiplying both sides of this equation by $\displaystyle 6x^2 - 5x + 1$.gives you:

$\displaystyle x+1=6x^2 - 5x + 1$.

Have fun!