Question

Factor the denominator on the right side of the equation

$$\frac{b}{b-5} - \frac{2}{b+5} = \frac{50}{(b-5)(b+5)}$$




Multiply both sides of the equation by the least common denominator (LCD), which is $$(b-5)(b+5)$$

$$(b-5)(b+5) \left( \frac{b}{b-5} - \frac{2}{b+5} \right) = (b-5)(b+5) \left( \frac{50}{(b-5)(b+5)} \right)$$




Distribute and simplify each term

$$b(b+5) - 2(b-5) = 50$$

Expand the terms

$$b^2 + 5b - 2b + 10 = 50$$


Combine like terms

$$b^2 + 3b + 10 = 50$$


Move all terms to one side to set the equation to zero

$$b^2 + 3b + 10 - 50 = 0$$


$$b^2 + 3b - 40 = 0$$


Factor the quadratic equation

$$(b+8)(b-5) = 0$$

Set each factor equal to zero and solve for $$b$$

$$b+8 = 0 \quad \text{or} \quad b-5 = 0$$

$$b = -8 \quad \text{or} \quad b = 5$$

Check for extraneous solutions by plugging the values back into the original equation. Note that $$b=5$$
makes the denominators $$b-5$$
and $$b^{2-25}$$

equal to zero, which is undefined. Therefore, $$b=5$$
is an extraneous solution.
If $$b = 5$$
:

$$\frac{5}{5-5} - \frac{2}{5+5} = \frac{50}{5^2-25}$$





$$\frac{5}{0} - \frac{2}{10} = \frac{50}{0}$$




Since division by zero is undefined, $$b = 5$$
is an extraneous solution.

If $$b = -8$$
:

$$\frac{-8}{-8-5} - \frac{2}{-8+5} = \frac{50}{(-8)^2-25}$$





$$\frac{-8}{-13} - \frac{2}{-3} = \frac{50}{64-25}$$




$$\frac{8}{13} + \frac{2}{3} = \frac{50}{39}$$




$$\frac{24}{39} + \frac{26}{39} = \frac{50}{39}$$




$$\frac{50}{39} = \frac{50}{39}$$



The solution $$b = -8$$
is valid.

Answer 1

Your work and factoring are correct. The denominator on the right factors as b^2 - 25 = (b-5)(b+5), and multiplying through by the LCD (b-5)(b+5) gives

b(b+5) - 2(b-5) = 50
⇒ b^2 + 5b - 2b + 10 = 50
⇒ b^2 + 3b - 40 = 0
⇒ (b+8)(b-5) = 0.

So b = -8 or b = 5. Since b = 5 makes the original denominators zero, it is extraneous, leaving the only valid solution b = -8.

(There is a small typo in your write-up — "b^{2-25}" — but it does not affect the solution.)