Thinking Mathematically (6th Edition)

The solution set is$\left\{ -9,\frac{1}{3} \right\}$.
Consider the expression$\left( x+9 \right)\left( 3x-1 \right)=0$. Then, by the zero product principal either $\left( x+9 \right)=0$or$\left( 3x-1 \right)=0$. Now $\left( x+9 \right)=0$implies that $x=-9$and $\left( 3x-1 \right)=0$ implies that$x=\frac{1}{3}$. Next, check the proposed solution by substituting it in the original equation. Check for$x=-9$. So consider, \begin{align} & \left( x+9 \right)\left( 3x-1 \right)=0 \\ & \left( -9+9 \right)\left( 3\times \left( -9 \right)-1 \right)=0 \\ & 0\left( -28 \right)=0 \\ & 0=0 \end{align} Now check for$x=\frac{1}{3}$. So consider, \begin{align} & \left( x+9 \right)\left( 3x-1 \right)=0 \\ & \left( \frac{1}{3}+9 \right)\left( 3\times \frac{1}{3}-1 \right)=0 \\ & \left( \frac{1}{3}+9 \right)\left( 0 \right)=0 \\ & 0=0 \end{align} Hence, the solution set is$\left\{ -9,\frac{1}{3} \right\}$.