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{2}{x + 3} - \frac{1}{x - 3} = \frac{2x}{x^2 - 9}\\)\\(\quad\\)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: Identify denominators

Denominators: $x+3$, $x-3$, $x^2-9$

Step2: Factor $x^2-9$

$x^2-9=(x+3)(x-3)$

Step3: Solve for zero denominators

Set $x+3=0$ → $x=-3$
Set $x-3=0$ → $x=3$

Step4: Solve the rational equation

Multiply all terms by $(x+3)(x-3)$:
$2(x-3) - 1(x+3) = 2x$

Step5: Expand and simplify

$2x-6 -x-3 = 2x$
$x-9=2x$

Step6: Isolate $x$

$x-2x=9$ → $-x=9$ → $x=-9$
Verify $x=-9$ does not make denominators zero.

Answer:

a. $x=-3, 3$
b. $x=-9$