## Algebra 2 (1st Edition)

$x=3$
Multiplying both sides by $x(x+3)$ we get: $(x-1)(x+3)+(2x-1)x=(x+6)x\\x^2+2x-3+2x^2-x=x^2+6x\\2x^2-5x-3=0\\(x-3)(2x+1)=0$ Thus $x=3$ or $x=-0.5$, and $x=-.5$ is extraneous.