Question

perform the indicated operation and simplify the result. leave your answer in factored form.
\\(\frac{x}{x + 3} + \frac{8x - 1}{x - 3}\\)
\\(\frac{x}{x + 3} + \frac{8x - 1}{x - 3} = \square\\)

Explanation:

Step1: Find a common denominator

The denominators are \(x + 3\) and \(x - 3\), so the common denominator is \((x + 3)(x - 3)\).
Rewrite each fraction with the common denominator:
\(\frac{x}{x + 3}=\frac{x(x - 3)}{(x + 3)(x - 3)}\)
\(\frac{8x - 1}{x - 3}=\frac{(8x - 1)(x + 3)}{(x + 3)(x - 3)}\)

Step2: Add the fractions

Now that the fractions have the same denominator, we can add the numerators:

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Step3: Factor the numerator

Factor \(9x^{2}+20x - 3\). We need two numbers that multiply to \(9\times(-3)=-27\) and add to \(20\). The numbers are \(27\) and \(-1\).

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Step4: Simplify the fraction

Substitute the factored numerator back into the fraction:
\(\frac{(9x - 1)(x + 3)}{(x + 3)(x - 3)}\)
Cancel out the common factor \((x + 3)\) (assuming \(x
eq - 3\)):
\(\frac{9x - 1}{x - 3}\)

Answer:

\(\frac{9x - 1}{x - 3}\)