## College Algebra 7th Edition

$x=0$
We solve the equation: $\displaystyle \frac{x+1}{x-1}=\frac{2x-1}{2x+1}$ First we cross-multiply: $(x+1)(2x+1)=(2x-1)(x-1)$ Next we distribute: $2x^{2}+3x+1=2x^{2}-3x+1$ And simplify: $2x^{2}-2x^{2}+3x+3x+1-1=0$ $6x=0$ $x=0$