What is the solution(s) of the equation ? (1 point)
3
.
and 3
no solution

Ask by Bartlett Higgins. in the United States

Nov 21,2024

Question

What is the solution(s) of the equation $$ \frac{9}{z}+\frac{9}{z-2}=12 $$ ? (1 point) 3 $$ \frac{1}{2} $$. $$ \frac{1}{2} $$ and 3 no solution

UpStudy AI · Accepted Answer

Solve the equation $$ \frac{9}{z}+\frac{9}{z-2}=12 $$.
Solve the equation by following steps:
- step0: Solve for $$z$$:
  $$\frac{9}{z}+\frac{9}{z-2}=12$$
- step1: Find the domain:
  $$\frac{9}{z}+\frac{9}{z-2}=12,z \in \left(-\infty,0ight)\cup \left(0,2ight)\cup \left(2,+\inftyight)$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{9}{z}+\frac{9}{z-2}ight)z\left(z-2ight)=12z\left(z-2ight)$$
- step3: Simplify the equation:
  $$18z-18=12z^{2}-24z$$
- step4: Move the expression to the left side:
  $$18z-18-\left(12z^{2}-24zight)=0$$
- step5: Subtract the terms:
  $$42z-18-12z^{2}=0$$
- step6: Factor the expression:
  $$-6\left(3-zight)\left(1-2zight)=0$$
- step7: Divide the terms:
  $$\left(3-zight)\left(1-2zight)=0$$
- step8: Separate into possible cases:
  $$\begin{align}&3-z=0\&1-2z=0\end{align}$$
- step9: Solve the equation:
  $$\begin{align}&z=3\&z=\frac{1}{2}\end{align}$$
- step10: Check if the solution is in the defined range:
  $$\begin{align}&z=3\&z=\frac{1}{2}\end{align},z \in \left(-\infty,0ight)\cup \left(0,2ight)\cup \left(2,+\inftyight)$$
- step11: Find the intersection:
  $$\begin{align}&z=3\&z=\frac{1}{2}\end{align}$$
- step12: Rewrite:
  $$z_{1}=\frac{1}{2},z_{2}=3$$
The solutions to the equation $$ \frac{9}{z}+\frac{9}{z-2}=12 $$ are $$ z=\frac{1}{2} $$ and $$ z=3 $$.
The solutions are $$ z=\frac{1}{2} $$ and $$ z=3 $$.

What is the solution(s) of the equation ? (1 point)
3
.
and 3
no solution

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What is the solution(s) of the equation ? (1 point) 3 . and 3 no solution