Question

which of the following is the solution to the equation below?
\\(\frac{1}{x - 4} + \frac{2}{x - 2} = \frac{2x - 3}{x^2 - 6x + 8}\\)
\\(\bigcirc\\) \\(x = 3\\)
\\(\bigcirc\\) \\(x = 7\\)
\\(\bigcirc\\) \\(x = 9\\)
\\(\bigcirc\\) \\(x = 13\\)

Explanation:

Step1: Factor the denominator

First, factor the quadratic denominator \(x^2 - 6x + 8\). We need two numbers that multiply to \(8\) and add to \(-6\), which are \(-4\) and \(-2\). So, \(x^2 - 6x + 8=(x - 4)(x - 2)\).
The equation becomes \(\frac{1}{x - 4}+\frac{2}{x - 2}=\frac{2x - 3}{(x - 4)(x - 2)}\).

Step2: Multiply through by the LCD

The least common denominator (LCD) of the fractions is \((x - 4)(x - 2)\). Multiply each term by \((x - 4)(x - 2)\) to eliminate the denominators:

$$
LATEXBLOCK0
$$

Step3: Simplify and solve for \(x\)

Expand the left - hand side: \(x - 2+2x-8 = 2x - 3\).
Combine like terms: \(3x-10 = 2x - 3\).
Subtract \(2x\) from both sides: \(3x-2x - 10=2x-2x - 3\), which gives \(x - 10=-3\).
Add \(10\) to both sides: \(x-10 + 10=-3 + 10\), so \(x = 7\).

We need to check for extraneous solutions. The original equation has restrictions \(x
eq4\) and \(x
eq2\). When \(x = 7\), the denominators \(x - 4=7 - 4 = 3
eq0\) and \(x - 2=7 - 2 = 5
eq0\), so \(x = 7\) is a valid solution.

Answer:

\(x = 7\) (corresponding to the option \(x = 7\))