Question

solve: \\(\frac{1}{x-3} + \frac{3}{x+5} = \frac{12}{x^2+2x-15}\\)\
\\(\bigcirc\\) x=3\
\\(\bigcirc\\) x=4\
\\(\bigcirc\\) x=5\
\\(\bigcirc\\) x=15

Explanation:

Step1: Factor the denominator on the right

First, factor \(x^{2}+2x - 15\). We need two numbers that multiply to \(-15\) and add to \(2\). Those numbers are \(5\) and \(-3\), so \(x^{2}+2x - 15=(x + 5)(x - 3)\). The equation becomes \(\frac{1}{x - 3}+\frac{3}{x + 5}=\frac{12}{(x + 5)(x - 3)}\).

Step2: Find the common denominator

The common denominator for the left - hand side is \((x + 5)(x - 3)\). Multiply each fraction by the appropriate form of \(1\) to get the common denominator:
\(\frac{1\times(x + 5)}{(x - 3)(x + 5)}+\frac{3\times(x - 3)}{(x + 5)(x - 3)}=\frac{12}{(x + 5)(x - 3)}\)

Step3: Simplify the left - hand side

Simplify the numerators:
\(\frac{x + 5+3(x - 3)}{(x + 5)(x - 3)}=\frac{12}{(x + 5)(x - 3)}\)
Expand \(3(x - 3)=3x-9\), then \(x + 5+3x-9 = 4x-4\). So the equation is \(\frac{4x - 4}{(x + 5)(x - 3)}=\frac{12}{(x + 5)(x - 3)}\)

Step4: Solve for \(x\)

Since the denominators are the same (and \(x
eq - 5\) and \(x
eq3\) to avoid division by zero), we can set the numerators equal:
\(4x-4 = 12\)
Add \(4\) to both sides: \(4x=12 + 4=16\)
Divide both sides by \(4\): \(x = 4\)

We should check if \(x = 4\) makes the original denominators non - zero. For \(x-3=4 - 3 = 1
eq0\) and \(x + 5=4+5 = 9
eq0\), so \(x = 4\) is a valid solution.

Answer:

B. \(x = 4\) (assuming the options are labeled as A. \(x = 3\), B. \(x = 4\), C. \(x = 5\), D. \(x = 15\))