Question

solve the following rational equation.\\(\frac{2x}{x^{2}-7x - 8}-\frac{5}{x - 8}=\frac{1}{x + 1}\\)\\(\\)\\(\\)select the correct choice below and, if necessary, fill in the answer box to comp\\(\\)\\(\\)\\(\circ\\) a. \\(x = \\) (simplify your answer. type an integer or a simplified fraction.)\\(\\)\\(\circ\\) b. the solution is all real numbers.\\(\\)\\(\circ\\) c. there is no solution.

Explanation:

Step1: Factor the denominator

Factor \(x^2 - 7x - 8\) as \((x - 8)(x + 1)\). So the equation becomes \(\frac{2x}{(x - 8)(x + 1)} - \frac{5}{x - 8} = \frac{1}{x + 1}\).

Step2: Find the common denominator

The common denominator of the three fractions is \((x - 8)(x + 1)\). Multiply each term by the common denominator to eliminate the fractions:
\[

$$\begin{align*} \frac{2x}{(x - 8)(x + 1)} \times (x - 8)(x + 1) - \frac{5}{x - 8} \times (x - 8)(x + 1) &= \frac{1}{x + 1} \times (x - 8)(x + 1)\\ 2x - 5(x + 1) &= x - 8 \end{align*}$$

\]

Step3: Simplify the equation

Expand and simplify the left side: \(2x - 5x - 5 = x - 8\), which becomes \(-3x - 5 = x - 8\).

Step4: Solve for x

Add \(3x\) to both sides: \(-5 = 4x - 8\). Then add 8 to both sides: \(3 = 4x\). Divide both sides by 4: \(x = \frac{3}{4}\).

Step5: Check for extraneous solutions

Check if \(x = \frac{3}{4}\) makes the original denominators zero. For \(x - 8\), \(\frac{3}{4}-8
eq0\). For \(x + 1\), \(\frac{3}{4}+1
eq0\). So it is a valid solution.

Answer:

A. \(x = \frac{3}{4}\)