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]
