Question

solve the rational equation:
\\(\frac{x}{x + 3} = \frac{18}{x^2 - 9} + 4\\)

select one:
\\(\circ\\) a. no solution.
\\(\circ\\) b. \\(x = -2\\)
\\(\circ\\) c. \\(x = 2\\)
\\(\circ\\) d. \\(x = 2, -3\\)

Explanation:

Step1: Factor denominator

Note that $x^2 - 9 = (x+3)(x-3)$.

Step2: Eliminate denominators

Multiply all terms by $(x+3)(x-3)$:
$$x(x-3) = 18 + 4(x+3)(x-3)$$

Step3: Expand all terms

$$x^2 - 3x = 18 + 4(x^2 - 9)$$
$$x^2 - 3x = 18 + 4x^2 - 36$$

Step4: Simplify to quadratic form

Combine like terms and rearrange:
$$0 = 3x^2 + 3x - 18$$
Divide by 3:
$$x^2 + x - 6 = 0$$

Step5: Factor quadratic

$$(x+3)(x-2) = 0$$
Solutions: $x=-3, x=2$

Step6: Check for extraneous solutions

$x=-3$ makes original denominators $x+3=0$ and $x^2-9=0$, so it is invalid.

Answer:

C. $x = 2$