3/(x-1) - 2/(x+1) = 2/x(^2) -1 Could someone please tell me how you solve that?
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Originally Posted by JQ2009 3/(x-1) - 2/(x+1) = 2/x(^2) -1 Could someone please tell me how you solve that? Hi JQ2009, $\displaystyle \frac{3}{x-1}-\frac{2}{x+1}=\frac{2}{x^2-1}$ $\displaystyle \frac{3}{x-1}-\frac{2}{x+1}=\frac{2}{(x-1)(x+1)}$ Now, multiply each term by (x-1)(x+1) and continue from there.
Thanks but I am still confused. I know that leaves you with 2 on the right hand side but what happens to the left?
Originally Posted by masters Hi JQ2009, $\displaystyle \frac{3}{x-1}-\frac{2}{x+1}=\frac{2}{x^2-1}$ $\displaystyle \frac{3}{x-1}-\frac{2}{x+1}=\frac{2}{(x-1)(x+1)}$ Now, multiply each term by (x-1)(x+1) and continue from there. Originally Posted by JQ2009 Thanks but I am still confused. I know that leaves you with 2 on the right hand side but what happens to the left? $\displaystyle (x-1)(x+1)\left(\frac{3}{(x-1}\right)-(x-1)(x+1)\left(\frac{2}{x+1}\right)=(x-1)(x+1)\left(\frac{2}{(x-1)(x+1)}\right)$ $\displaystyle 3(x+1)-2(x-1)=2$
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