## Algebra: A Combined Approach (4th Edition)

$x=-6$ and $x=1$
$x+5=\dfrac{6}{x}$ Multiply the whole equation by $x$: $x\Big(x+5=\dfrac{6}{x}\Big)$ $x^{2}+5x=6$ Take the $6$ to the left side of the equation: $x^{2}+5x-6=0$ Solve this equation by factoring: $(x+6)(x-1)=0$ Set each factor equal to $0$ and solve each individual equation: $x+6=0$ $x=-6$ $x-1=0$ $x=1$ The original equation is not undefined for neither of the values of $x$ found, so the solutions are: $x=-6$ and $x=1$