Question

1. solve the rational equation $\frac{1}{x^{2}-16} = \frac{1}{x-4}$ and test the solution(s) by substitution. identify the extraneous solution(s).$x = -3LXB0x = -3$ and $x = 4$there are no extraneous solutions.

Explanation:

Step1: Factor denominator left side

$x^2-16=(x-4)(x+4)$

Step2: Rewrite the equation

$\frac{1}{(x-4)(x+4)}=\frac{1}{x-4}$

Step3: Eliminate denominators

Multiply both sides by $(x-4)(x+4)$:
$1 = x+4$

Step4: Solve for x

$x=1-4=-3$

Step5: Check domain restrictions

Original equation: $x^2-16
eq0 \implies x
eq4,-4$; $x-4
eq0 \implies x
eq4$

Step6: Test $x=-3$

Substitute into original equation:
$\frac{1}{(-3)^2-16}=\frac{1}{9-16}=-\frac{1}{7}$
$\frac{1}{-3-4}=-\frac{1}{7}$. Valid solution.

Step7: Test $x=4$

Substitute into original equation:
Left side: $\frac{1}{4^2-16}=\frac{1}{0}$ (undefined); Right side: $\frac{1}{4-4}=\frac{1}{0}$ (undefined). $x=4$ is extraneous.

Answer:

B. $x=4$