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 + 4)}{(x - 2)(x + 4)} - \frac{2(x - 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\\)
check \\(x = -4\\): \\(x \
eq -4\\)
check \\(x = 2\\)
multiple - choice question
do the substitution of \\(x = 2\\) into the lcd.
what can you conclude from the result?

• \\(x = 2\\) is also extraneous.
• \\(x = 2\\) is not extraneous.

rewatch

Explanation:

Step1: Substitute \( x = 2 \) into LCD

The LCD is \( (x + 4)(x - 2) \). Substitute \( x = 2 \):
\( (2 + 4)(2 - 2) = 6 \times 0 = 0 \).

Step2: Analyze the result

In a rational equation, a value that makes the denominator (or LCD) zero is an extraneous solution because it makes the original equation undefined. Since substituting \( x = 2 \) into the LCD gives \( 0 \), \( x = 2 \) makes the denominator zero.

Answer:

\( x = 2 \) is also extraneous.