Solve for $$ x $$. $$ \frac{x-3}{x-7}+1=\frac{x+7}{x+5} $$ If there is more than one solutional equation that simplifies to quacrauc: separate them with commas, If there is no solution, click on "No solution".

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Solve the equation $$ \frac{x-3}{x-7}+1=\frac{x+7}{x+5} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{x-3}{x-7}+1=\frac{x+7}{x+5}$$
- step1: Find the domain:
  $$\frac{x-3}{x-7}+1=\frac{x+7}{x+5},x \in \left(-\infty,-5ight)\cup \left(-5,7ight)\cup \left(7,+\inftyight)$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{x-3}{x-7}+1ight)\left(x-7ight)\left(x+5ight)=\frac{x+7}{x+5}	imes \left(x-7ight)\left(x+5ight)$$
- step3: Simplify the equation:
  $$2x^{2}-50=x\left(x-7ight)+7\left(x-7ight)$$
- step4: Calculate:
  $$2x^{2}-50=x^{2}-7x+7x-49$$
- step5: Simplify:
  $$2x^{2}-50=x^{2}-49$$
- step6: Move the expression to the left side:
  $$2x^{2}-50-\left(x^{2}-49ight)=0$$
- step7: Subtract the terms:
  $$x^{2}-1=0$$
- step8: Move the constant to the right side:
  $$x^{2}=0+1$$
- step9: Remove 0:
  $$x^{2}=1$$
- step10: Simplify the expression:
  $$x=\pm \sqrt{1}$$
- step11: Simplify:
  $$x=\pm 1$$
- step12: Separate into possible cases:
  $$\begin{align}&x=1\&x=-1\end{align}$$
- step13: Check if the solution is in the defined range:
  $$\begin{align}&x=1\&x=-1\end{align},x \in \left(-\infty,-5ight)\cup \left(-5,7ight)\cup \left(7,+\inftyight)$$
- step14: Find the intersection:
  $$\begin{align}&x=1\&x=-1\end{align}$$
- step15: Rewrite:
  $$x_{1}=-1,x_{2}=1$$
The solutions to the equation $$ \frac{x-3}{x-7}+1=\frac{x+7}{x+5} $$ are $$ x=-1 $$ and $$ x=1 $$.
The solutions are $$ x=-1 $$ and $$ x=1 $$.

Solve for .
If there is more than one solutional equation that simplifies to quacrauc: separate them with commas,
If there is no solution, click on “No solution”.

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Solve for . If there is more than one solutional equation that simplifies to quacrauc: separate them with commas, If there is no solution, click on “No solution”.