Question

for the following equation, a. write the value or values of the variable that make a denominator zero. these are the restrictions on the variable. b. keeping the restrictions in mind, solve the equation.\\(\frac{4}{x + 12}-\frac{3}{x - 12}=\frac{5x}{x^2 - 144}\\)\
a. write the value or values of the variable that make a denominator zero.\
x = \\(\square\\) (use a comma to separate answers as needed.)

Explanation:

Step1: Find zeros of first denominator

Set $x+12=0$, solve for $x$:
$x = -12$

Step2: Find zeros of second denominator

Set $x-12=0$, solve for $x$:
$x = 12$

Step3: Find zeros of third denominator

Factor $x^2-144=(x+12)(x-12)$. Set equal to 0:
$(x+12)(x-12)=0$, so $x=-12, 12$

Step4: Identify unique restrictions

Combine results, exclude duplicate values.

Step1: Factor right-hand denominator

$x^2-144=(x+12)(x-12)$

Step2: Multiply by common denominator

Multiply all terms by $(x+12)(x-12)$ to eliminate denominators:
$4(x-12) - 3(x+12) = 5x$

Step3: Expand and simplify left side

$4x - 48 - 3x - 36 = 5x$
$x - 84 = 5x$

Step4: Isolate variable terms

Subtract $x$ from both sides:
$-84 = 4x$

Step5: Solve for x

Divide both sides by 4:
$x = \frac{-84}{4} = -21$

Step6: Verify against restrictions

$-21$ is not $-12$ or $12$, so it is valid.

Answer:

$x=-12, 12$

---

Now solving part b (included as it is part of the given problem):