## Precalculus (6th Edition)

The solution is $x=-2$
$\dfrac{2x+1}{x-2}+\dfrac{3}{x}=\dfrac{-6}{x^{2}-2x}$ Take out common factor $x$ from the denominator of the fraction on the right side of the equation: $\dfrac{2x+1}{x-2}+\dfrac{3}{x}=\dfrac{-6}{x(x-2)}$ Multiply the whole equation by $x(x-2)$: $x(x-2)\Big[\dfrac{2x+1}{x-2}+\dfrac{3}{x}=\dfrac{-6}{x(x-2)}\Big]$ $(x)(2x+1)+3(x-2)=-6$ Evaluate the indicated operations: $2x^{2}+x+3x-6=-6$ Take $6$ to the left side of the equation and simplify: $2x^{2}+x+3x-6+6=0$ $2x^{2}+4x=0$ Take out common factor $2x$ from the left side: $2x(x+2)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $2x=0$ $x=\dfrac{0}{2}$ $x=0$ $x+2=0$ $x=-2$ Since the original equation is undefined for $x=0$, the solution is just $x=-2$