## Precalculus (6th Edition)

The solutions are $x=4$ and $x=-\dfrac{3}{2}$
$\dfrac{x+4}{2x}=\dfrac{x-1}{3}$ Take $2x$ to multiply the numerator of the right side and $3$ to multiply the numerator of the left side: $(x+4)(3)=(x-1)(2x)$ Evaluate the indicated operations: $3x+12=2x^{2}-2x$ Take all terms to the right side and simplify: $0=2x^{2}-2x-3x-12$ $0=2x^{2}-5x-12$ Rearrange: $2x^{2}-5x-12=0$ Solve by factoring: $(2x+3)(x-4)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $2x+3=0$ $2x=-3$ $x=-\dfrac{3}{2}$ $x-4=0$ $x=4$ The solutions are $x=4$ and $x=-\dfrac{3}{2}$