Solve 1/v-3-5/v+3=2/v^2-9

$$\frac { 1 } { v - 3 } - \frac { 5 } { v + 3 } = \frac { 2 } { v ^ { 2 } - 9 }$$

Solve for v

$$v=4$$

Steps for Solving Linear Equation

Variable v cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(v-3\right)\left(v+3\right), the least common multiple of v-3,v+3,v^{2}-9.

$$v+3-\left(v-3\right)\times 5=2$$

Use the distributive property to multiply v-3 by 5.

$$v+3-\left(5v-15\right)=2$$

To find the opposite of 5v-15, find the opposite of each term.

$$v+3-5v+15=2$$

Combine v and -5v to get -4v.

$$-4v+3+15=2$$

Add 3 and 15 to get 18.

$$-4v+18=2$$

Subtract 18 from both sides.

$$-4v=2-18$$

Subtract 18 from 2 to get -16.

$$-4v=-16$$

Divide both sides by -4.

$$v=\frac{-16}{-4}$$

Divide -16 by -4 to get 4.

$$v=4$$