Question

solve the equation.
\\(\frac{3}{x - 5} - \frac{1}{x - 4} = -\frac{1}{x^2 - 9x + 20}\\)

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

a. \\(x = \square\\)
(type an integer or a fraction. use a comma to separate answers as needed.)

b. there is no solution.

Explanation:

Step1: Factor the denominator

Factor \(x^2 - 9x + 20\). We know that \(x^2 - 9x + 20=(x - 4)(x - 5)\) since \((x - 4)(x - 5)=x^2-5x-4x + 20=x^2-9x + 20\). So the equation becomes \(\frac{3}{x - 5}-\frac{1}{x - 4}=-\frac{1}{(x - 4)(x - 5)}\).

Step2: Find the common denominator

The common denominator of the left - hand side and the right - hand side is \((x - 4)(x - 5)\), where \(x
eq4\) and \(x
eq5\) (because if \(x = 4\) or \(x = 5\), the original fractions are undefined). Multiply each term by \((x - 4)(x - 5)\):

\(3(x - 4)-(x - 5)=- 1\)

Step3: Expand and simplify the left - hand side

Expand \(3(x - 4)-(x - 5)\):
\(3x-12 - x + 5=-1\)
Combine like terms: \((3x-x)+(-12 + 5)=-1\), so \(2x-7=-1\)

Step4: Solve for x

Add 7 to both sides of the equation: \(2x-7 + 7=-1 + 7\), which gives \(2x=6\)
Divide both sides by 2: \(x = 3\)

We need to check if \(x = 3\) makes the original denominators non - zero. When \(x = 3\), \(x-5=3 - 5=-2
eq0\) and \(x - 4=3-4=-1
eq0\), so \(x = 3\) is a valid solution.

Answer:

A. \(x = 3\)