Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
rational equations
which solution to the equation \\( \frac{3}{2g + 8} = \frac{g + 2}{g^2 - 16} \\) is extraneous?
options:

• \\( g = -4 \\) and \\( g = 16 \\)
• neither \\( g = -4 \\) nor \\( g = 16 \\)
• \\( g = -4 \\)
• \\( g = 16 \\)

Explanation:

Step1: Identify the equation

The equation is $\frac{3}{2g + 8}=\frac{g + 2}{g^2 - 16}$. First, factor the denominators: $2g + 8 = 2(g + 4)$ and $g^2 - 16=(g - 4)(g + 4)$. So the equation becomes $\frac{3}{2(g + 4)}=\frac{g + 2}{(g - 4)(g + 4)}$. The domain restrictions are $g
eq - 4$ and $g
eq4$ (since these make denominators zero).

Step2: Eliminate denominators

Multiply both sides by $2(g + 4)(g - 4)$ (the least common denominator) to get:
$3(g - 4)=2(g + 2)$

Step3: Solve the linear equation

Expand both sides: $3g-12 = 2g + 4$. Subtract $2g$ from both sides: $g-12 = 4$. Add 12 to both sides: $g = 16$.

Step4: Check for extraneous solutions

Check $g=-4$: It's in the domain restrictions (denominator zero), so it's extraneous. Check $g = 16$: Substitute into original equation. Left side: $\frac{3}{2(16)+8}=\frac{3}{40}$. Right side: $\frac{16 + 2}{16^2-16}=\frac{18}{240}=\frac{3}{40}$. So $g = 16$ is valid, $g=-4$ is extraneous.

Answer:

$g = 16$ (the option corresponding to $g = 16$)