Solve x in the rational equation: $$ \frac{x^{2}-4 x}{x-2}=\frac{14-9 x}{x-2} $$ Answer:

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Solve the equation $$ \frac{x^{2}-4 x}{x-2}=\frac{14-9 x}{x-2} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{x^{2}-4x}{x-2}=\frac{14-9x}{x-2}$$
- step1: Find the domain:
  $$\frac{x^{2}-4x}{x-2}=\frac{14-9x}{x-2},x
eq 2$$
- step2: Cross multiply:
  $$\left(x^{2}-4xight)\left(x-2ight)=\left(x-2ight)\left(14-9xight)$$
- step3: Calculate:
  $$x^{3}-6x^{2}+8x=\left(x-2ight)\left(14-9xight)$$
- step4: Calculate:
  $$x^{3}-6x^{2}+8x=32x-9x^{2}-28$$
- step5: Move the expression to the left side:
  $$x^{3}-6x^{2}+8x-\left(32x-9x^{2}-28ight)=0$$
- step6: Calculate:
  $$x^{3}+3x^{2}-24x+28=0$$
- step7: Factor the expression:
  $$\left(x-2ight)^{2}\left(x+7ight)=0$$
- step8: Separate into possible cases:
  $$\begin{align}&\left(x-2ight)^{2}=0\&x+7=0\end{align}$$
- step9: Solve the equation:
  $$\begin{align}&x=2\&x=-7\end{align}$$
- step10: Check if the solution is in the defined range:
  $$\begin{align}&x=2\&x=-7\end{align},x
eq 2$$
- step11: Find the intersection:
  $$x=-7$$
The solution to the rational equation $$ \frac{x^{2}-4 x}{x-2}=\frac{14-9 x}{x-2} $$ is $$ x=-7 $$.
$$ x=-7 $$.

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