Alright, let's solve the equation step by step.
Given the equation:
[tex]\[
\frac{9}{x-3} - \frac{27}{x^2 - 3x} = -3
\][/tex]
First, we need to simplify the second term [tex]\(\frac{27}{x^2 - 3x}\)[/tex]. Notice that the denominator [tex]\(x^2 - 3x\)[/tex] can be factored:
[tex]\[
x^2 - 3x = x(x - 3)
\][/tex]
So the equation becomes:
[tex]\[
\frac{9}{x-3} - \frac{27}{x(x-3)} = -3
\][/tex]
Next, we find a common denominator for the fractions on the left side of the equation. The common denominator will be [tex]\(x(x-3)\)[/tex]:
[tex]\[
\frac{9x}{x(x-3)} - \frac{27}{x(x-3)} = -3
\][/tex]
Combine the fractions:
[tex]\[
\frac{9x - 27}{x(x-3)} = -3
\][/tex]
To clear the fraction, multiply both sides of the equation by [tex]\(x(x-3)\)[/tex]:
[tex]\[
9x - 27 = -3x(x-3)
\][/tex]
Expand the right-hand side:
[tex]\[
9x - 27 = -3x^2 + 9x
\][/tex]
Subtract [tex]\(9x\)[/tex] from both sides:
[tex]\[
-27 = -3x^2
\][/tex]
Divide both sides by [tex]\(-3\)[/tex]:
[tex]\[
9 = x^2
\][/tex]
Take the square root of both sides:
[tex]\[
x = \pm 3
\][/tex]
Now we need to check if these solutions are valid in the context of the original equation. We substitute [tex]\(x = 3\)[/tex] and [tex]\(x = -3\)[/tex] back into the original equation:
For [tex]\(x = 3\)[/tex]:
[tex]\[
\frac{9}{3-3} - \frac{27}{3^2-3 \cdot 3} = \frac{9}{0} - \frac{27}{0}
\][/tex]
This results in division by zero, which is undefined. Therefore, [tex]\(x = 3\)[/tex] is not a valid solution.
For [tex]\(x = -3\)[/tex]:
[tex]\[
\frac{9}{-3-3} - \frac{27}{(-3)^2 - 3 \cdot (-3)} = \frac{9}{-6} - \frac{27}{9 + 9} = -\frac{3}{2} - \frac{27}{18} = -\frac{3}{2} - \frac{3}{2} = -3
\][/tex]
This satisfies the original equation. Therefore, the valid solution is:
[tex]\[
x = -3
\][/tex]
