Question

lesson 2.5 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)\\)
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)} \cdot \frac{(x + 4)}{(x + 4)} - \frac{2}{(x + 4)} \cdot \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\\)
check \\(x = -4\\)
check \\(x = 2\\)
multiple - choice question
does \\(x = -4\\) make the lcd, \\((x + 4)(x - 2)\\), equal to 0?
\\(\square\\) no it does not. so \\(x = -4\\) is not extraneous.
\\(\square\\) yes it does. so \\(x = -4\\) is extraneous.
rewatch
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Explanation:

Step1: Substitute x = -4 into LCD

The LCD is \((x + 4)(x - 2)\). Substitute \(x=-4\) into the LCD: \((-4 + 4)(-4 - 2)\).

Step2: Simplify the expression

Simplify \((-4 + 4)\) to get \(0\), and \((-4 - 2)\) to get \(-6\). Then the product is \(0\times(-6)=0\). Since the LCD becomes \(0\) when \(x = - 4\), \(x=-4\) is extraneous.

Answer:

Yes it does. So x = -4 is extraneous.