Precalculus (6th Edition)

The solutions are $x=\dfrac{2}{9}$ and $x=-\dfrac{4}{3}$
$\Big|\dfrac{6x+1}{x-1}\Big|=3$ Solving this absolute value equation is equivalent to solving two equations, which are: $\dfrac{6x+1}{x-1}=3$ and $\dfrac{6x+1}{x-1}=-3$ Solve the first equation: $\dfrac{6x+1}{x-1}=3$ Take $x-1$ to multiply the right side: $6x+1=3(x-1)$ $6x+1=3x-3$ Take $3x$ to the left side and $1$ to the right side: $6x-3x=-3-1$ $3x=-4$ Solve for $x$: $x=-\dfrac{4}{3}$ Solve the second equation: $\dfrac{6x+1}{x-1}=-3$ Take $x-1$ to multiply the right side: $6x+1=-3(x-1)$ $6x+1=-3x+3$ Take $3x$ to the left side and $1$ to the right side: $6x+3x=3-1$ $9x=2$ Solve for $x$: $x=\dfrac{2}{9}$ The solutions are $x=\dfrac{2}{9}$ and $x=-\dfrac{4}{3}$