11. \( \frac{2 z}{9}+\frac{1}{3}=\frac{2}{z+4} \)

Solve the equation \( \frac{2z}{9}+\frac{1}{3}=\frac{2}{z+4} \). Solve the equation by following steps: - step0: Solve for \(z\): \(\frac{2z}{9}+\frac{1}{3}=\frac{2}{z+4}\) - step1: Find the domain: \(\frac{2z}{9}+\frac{1}{3}=\frac{2}{z+4},z\neq -4\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{2z}{9}+\frac{1}{3}\right)\times 9\left(z+4\right)=\frac{2}{z+4}\times 9\left(z+4\right)\) - step3: Simplify the equation: \(2z^{2}+11z+12=18\) - step4: Move the expression to the left side: \(2z^{2}+11z+12-18=0\) - step5: Subtract the numbers: \(2z^{2}+11z-6=0\) - step6: Factor the expression: \(\left(z+6\right)\left(2z-1\right)=0\) - step7: Separate into possible cases: \(\begin{align}&z+6=0\\&2z-1=0\end{align}\) - step8: Solve the equation: \(\begin{align}&z=-6\\&z=\frac{1}{2}\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&z=-6\\&z=\frac{1}{2}\end{align},z\neq -4\) - step10: Find the intersection: \(\begin{align}&z=-6\\&z=\frac{1}{2}\end{align}\) - step11: Rewrite: \(z_{1}=-6,z_{2}=\frac{1}{2}\) The solutions to the equation \( \frac{2z}{9}+\frac{1}{3}=\frac{2}{z+4} \) are \( z=-6 \) and \( z=\frac{1}{2} \).