Work Step by Step
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.
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