Solve 2/b-3-4/b+3=8/b^2-9

$$\frac{ 2 }{ b-3 } - \frac{ 4 }{ b+3 } = \frac{ 8 }{ { b }^{ 2 } -9 }$$

Solve for b

$$b=5$$

Steps for Solving Linear Equation

Variable b 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(b-3\right)\left(b+3\right), the least common multiple of b-3,b+3,b^{2}-9.

$$\left(b+3\right)\times 2-\left(b-3\right)\times 4=8$$

Use the distributive property to multiply b+3 by 2.

$$2b+6-\left(b-3\right)\times 4=8$$

Use the distributive property to multiply b-3 by 4.

$$2b+6-\left(4b-12\right)=8$$

To find the opposite of 4b-12, find the opposite of each term.

$$2b+6-4b+12=8$$

Combine 2b and -4b to get -2b.

$$-2b+6+12=8$$

Add 6 and 12 to get 18.

$$-2b+18=8$$

Subtract 18 from both sides.

$$-2b=8-18$$

Subtract 18 from 8 to get -10.

$$-2b=-10$$

Divide both sides by -2.

$$b=\frac{-10}{-2}$$

Divide -10 by -2 to get 5.

$$b=5$$