Answer
The solution set is\[\left\{ -9,\frac{1}{3} \right\}\].
Work Step by Step
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\[\left\{ -9,\frac{1}{3} \right\}\].
