Solve for $$ v $$. $$ \frac{2}{v^{2}-25}-\frac{3}{v-5}=\frac{1}{v+5} $$ If there is more than one solution, separate them with commas. If there is no solution, click on "No solution". $$ v=\square $$

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Solve the equation $$ \frac{2}{v^{2}-25}-\frac{3}{v-5}=\frac{1}{v+5} $$.
Solve the equation by following steps:
- step0: Solve for $$v$$:
  $$\frac{2}{v^{2}-25}-\frac{3}{v-5}=\frac{1}{v+5}$$
- step1: Find the domain:
  $$\frac{2}{v^{2}-25}-\frac{3}{v-5}=\frac{1}{v+5},v \in \left(-\infty,-5ight)\cup \left(-5,5ight)\cup \left(5,+\inftyight)$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{2}{v^{2}-25}-\frac{3}{v-5}ight)\left(v+5ight)\left(v-5ight)=\frac{1}{v+5}	imes \left(v+5ight)\left(v-5ight)$$
- step3: Simplify the equation:
  $$-13-3v=v-5$$
- step4: Move the expression to the left side:
  $$-3v-v=-5+13$$
- step5: Add and subtract:
  $$-4v=-5+13$$
- step6: Add and subtract:
  $$-4v=8$$
- step7: Change the signs:
  $$4v=-8$$
- step8: Divide both sides:
  $$\frac{4v}{4}=\frac{-8}{4}$$
- step9: Divide the numbers:
  $$v=-2$$
- step10: Check if the solution is in the defined range:
  $$v=-2,v \in \left(-\infty,-5ight)\cup \left(-5,5ight)\cup \left(5,+\inftyight)$$
- step11: Find the intersection:
  $$v=-2$$
The solution to the equation is $$ v = -2 $$.
$$ v = -2 $$

Solve for .

If there is more than one solution, 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 solution, separate them with commas. If there is no solution, click on “No solution”.