Question

lesson 25 review - solving rational equations
③ \\(\frac{x}{x - 2} - \frac{2}{x + 4} = \frac{12}{x^2 + 2x - 8}\\) \\(\text{lcd} = (x + 4)(x - 2)\\)
\\(\frac{x}{(x - 2)} - \frac{2}{(x + 4)} = \frac{12}{(x + 4)(x - 2)}
\frac{x}{(x - 2)}(x + 4) - \frac{2}{(x + 4)}(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\\)
① factor denominators
② find lcd
③ multiply to create lcd
④ solve using numerator
⑤ check for extraneous solutions!
check \\(x = -4\\): \\(x \
eq -4\\); check \\(x = 2\\): \\(x \
eq 2\\); \\(\text{no solution}\\)
multiple - choice question
powerful.
so in other words, if you check your answers into
the lcd, and end up having two extraneous
solutions, this actual means that
\\(\square\\) there is no solution to the original equation.
\\(\square\\) there are two solutions to the original equation.
\\(\square\\) there is only one solution to the original
equation.

Explanation:

Extraneous solutions are values that make the original equation's denominators zero (undefined) or don't satisfy the equation. If both potential solutions (from solving the transformed equation) are extraneous (fail the check, like \(x = -4\) and \(x = 2\) here, which make denominators zero), the original equation has no valid solutions. The first option states this, the second is wrong (extraneous means they aren't solutions), and the third is wrong (two extraneous means no solutions, not one).

Answer:

A. there is no solution to the original equation.