Find all complex solutions for the following equation by hand.
[tex]\[
\frac{3}{x+5} - \frac{9}{x-5} = \frac{10x}{25-x^2}
\][/tex]

Select the correct choice below and, if necessary, fill in the answer box to complete your answer.

A. The solution set is [tex]\(\boxed{\phantom{3}}\)[/tex] (Type an integer or a simplified fraction.)
B. The solution set is the set of real numbers.
C. The solution set is an empty set.



Answer :

To solve the equation given by:
[tex]$ \frac{3}{x+5} - \frac{9}{x-5} = \frac{10x}{25 - x^2} $[/tex]

First, observe the denominator on the right-hand side. We recognize a difference of squares:
[tex]$ 25 - x^2 = (5 - x)(5 + x) $[/tex]

Thus, the equation becomes:
[tex]$ \frac{3}{x+5} - \frac{9}{x-5} = \frac{10x}{(5-x)(5+x)} $[/tex]

Next, we find a common denominator for the fractions on the left-hand side:
[tex]$ (x+5)(x-5) = x^2 - 25 $[/tex]

Rewrite each fraction with this common denominator:
[tex]$ \frac{3(x-5)}{(x+5)(x-5)} - \frac{9(x+5)}{(x+5)(x-5)} $[/tex]

Now combine the fractions:
[tex]$ \frac{3(x-5) - 9(x+5)}{x^2 - 25} = \frac{10x}{x^2 - 25} $[/tex]

Distribute and combine like terms in the numerator:
[tex]$ 3(x-5) - 9(x+5) = 3x - 15 - 9x - 45 = -6x - 60 $[/tex]

So the equation becomes:
[tex]$ \frac{-6x - 60}{x^2 - 25} = \frac{10x}{x^2 - 25} $[/tex]

Since the denominators are equal, we equate the numerators:
[tex]$ -6x - 60 = 10x $[/tex]

Solving for [tex]\( x \)[/tex]:
[tex]$ -6x - 10x = 60 \\ -16x = 60 \\ x = -\frac{60}{16} = -\frac{15}{4} $[/tex]

We have found a potential solution. However, we need to check that it does not make any denominator zero. Plugging [tex]\( x = -\frac{15}{4} \)[/tex] into the denominators:

For [tex]\( x+5 \)[/tex]:
[tex]$ -\frac{15}{4} + 5 = -\frac{15}{4} + \frac{20}{4} = \frac{5}{4} \neq 0, $[/tex]

For [tex]\( x-5 \)[/tex]:
[tex]$ -\frac{15}{4} - 5 = -\frac{15}{4} - \frac{20}{4} = -\frac{35}{4} \neq 0, $[/tex]

For [tex]\( 25 - x^2 \)[/tex]:
[tex]$ 25 - \left( -\frac{15}{4} \right)^2 = 25 - \frac{225}{16} = \frac{400}{16} - \frac{225}{16} = \frac{175}{16} \neq 0. $[/tex]

Thus, [tex]\( x = -\frac{15}{4} \)[/tex] does not make any of the denominators zero and is a valid solution.

Therefore, the solution set is:
A. The solution set is [tex]\( \left\{ - \frac{15}{4} \right\} \)[/tex]