Question

lesson 2.5 review - solving rational equations
3
\\(\frac{x}{x - 2} - \frac{2}{x + 4} = \frac{12}{x^2 + 2x - 8}\\)
lcd = (x + 4)(x - 2)
① factor denominators
② find lcd
③ multiply to create lcd
④ solve using numerator
⑤ check for extraneous solutions!
\\(\frac{x}{(x - 2)} - \frac{2}{(x + 4)} = \frac{12}{(x + 4)(x - 2)}
\frac{x}{(x - 2)} \frac{(x + 4)}{(x + 4)} - \frac{2}{(x + 4)} \frac{(x - 2)}{(x - 2)} = \frac{12}{(x + 4)(x - 2)}
\\(x(x + 4) - 2(x - 2) = 12\\)
\\(x^2 + 4x - 2x + 4 = 12\\)
\\(x^2 + 2x - 8 = 0\\)
\\((x + 4)(x - 2) = 0\\)
\\(x + 4 = 0\\) \\(x = -4\\)
\\(x - 2 = 0\\) \\(x = 2\\)
multiple - choice question
you dont circle any answers here yet because we must
check if any of the xs we just got are 0.
check if any of the xs we just got are negative.
check if any of the xs we just got are extraneous
check if any of the xs we just got are irrational.

Explanation:

When solving rational equations, after finding potential solutions, we need to check for extraneous solutions. Extraneous solutions are values that make the original equation's denominator zero (undefined) or don't satisfy the equation. Checking if solutions are extraneous is a key step. The other options (checking if x is 0, negative, or irrational) are not relevant to verifying solutions of rational equations.

Answer:

C. check if any of the x’s we just got are extraneous (assuming the option with this text is labeled C; if the option identifiers are different, adjust accordingly, e.g., if it's the third option, "The third option: check if any of the x’s we just got are extraneous")