Sure, let's solve the given equation step-by-step.
We need to solve the equation:
[tex]\[
1 - \frac{5}{x-5} = \frac{3x}{x^2 - 13x + 40}
\][/tex]
First, let's simplify the denominator on the right-hand side:
[tex]\[
x^2 - 13x + 40
\][/tex]
We notice that it can be factored:
[tex]\[
x^2 - 13x + 40 = (x-5)(x-8)
\][/tex]
So the equation becomes:
[tex]\[
1 - \frac{5}{x-5} = \frac{3x}{(x-5)(x-8)}
\][/tex]
Next, find a common denominator on the left side. The common denominator would be [tex]\((x-5)\)[/tex]:
[tex]\[
1 - \frac{5}{x-5} = \frac{x-5}{x-5} - \frac{5}{x-5} = \frac{(x-5) - 5}{x-5} = \frac{x-10}{x-5}
\][/tex]
Now the equation is:
[tex]\[
\frac{x-10}{x-5} = \frac{3x}{(x-5)(x-8)}
\][/tex]
We can eliminate the denominators by multiplying both sides of the equation by [tex]\((x-5)(x-8)\)[/tex]:
[tex]\[
(x-10)(x-8) = 3x
\][/tex]
Expand the left-hand side:
[tex]\[
(x-10)(x-8) = x^2 - 8x - 10x + 80 = x^2 - 18x + 80
\][/tex]
Now we have:
[tex]\[
x^2 - 18x + 80 = 3x
\][/tex]
Rearrange all terms to one side of the equation:
[tex]\[
x^2 - 18x + 80 - 3x = 0
\][/tex]
Simplify:
[tex]\[
x^2 - 21x + 80 = 0
\][/tex]
This is a quadratic equation. We can solve it using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -21\)[/tex], and [tex]\(c = 80\)[/tex]:
[tex]\[
x = \frac{-(-21) \pm \sqrt{(-21)^2 - 4 \cdot 1 \cdot 80}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{21 \pm \sqrt{441 - 320}}{2}
\][/tex]
[tex]\[
x = \frac{21 \pm \sqrt{121}}{2}
\][/tex]
[tex]\[
x = \frac{21 \pm 11}{2}
\][/tex]
This gives us two potential solutions:
[tex]\[
x = \frac{21 + 11}{2} = \frac{32}{2} = 16
\][/tex]
[tex]\[
x = \frac{21 - 11}{2} = \frac{10}{2} = 5
\][/tex]
We must check if these solutions are valid for the original equation. Recall the original equation:
[tex]\[
1 - \frac{5}{x-5} = \frac{3x}{x^2 - 13x + 40}
\][/tex]
For [tex]\(x=5\)[/tex], the left side has a division by zero ([tex]\(x-5 = 0\)[/tex]). Thus, [tex]\(x = 5\)[/tex] is not a valid solution.
Therefore, the only valid solution is:
[tex]\[
x = 16
\][/tex]
Thus, the solution to the equation [tex]\(\boxed{16}\)[/tex].
