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 + 2} - \frac{1}{x - 2} = \frac{3x}{x^2 - 4}\\)

a. write the value or values of the variable that make a denominator zero
\\(x = \\) -2,2 (use a comma to separate answers as needed.)
b. what is the solution of the equation? select the correct choice below and, if necessary, fill in the answer box to complete your choice
\\(\bigcirc\\) a. the solution set is \\(\\{\\}\\) (use a comma to separate answers as needed.)
\\(\bigcirc\\) b. the solution set is \\(\\{x | x\\) is a real number\\}
\\(\bigcirc\\) c. the solution set is \\(\varnothing\\)

Explanation:

Part a

Step1: Identify denominators

The denominators are \(x + 2\), \(x - 2\), and \(x^2 - 4\). Note that \(x^2 - 4=(x + 2)(x - 2)\) (difference of squares: \(a^2 - b^2=(a + b)(a - b)\) with \(a = x\), \(b = 2\)).

Step2: Find values making denominators zero

• For \(x + 2 = 0\), solve: \(x=-2\).
• For \(x - 2 = 0\), solve: \(x = 2\).
• For \(x^2 - 4 = 0\), since \(x^2 - 4=(x + 2)(x - 2)\), the solutions are also \(x=-2\) and \(x = 2\) (from the zero - product property: if \(ab = 0\), then \(a = 0\) or \(b = 0\)).

Step1: Eliminate denominators

The least common denominator (LCD) of \(x + 2\), \(x - 2\), and \(x^2 - 4=(x + 2)(x - 2)\) is \((x + 2)(x - 2)\). Multiply each term in the equation \(\frac{2}{x + 2}-\frac{1}{x - 2}=\frac{3x}{x^2 - 4}\) by \((x + 2)(x - 2)\) (note \(x
eq - 2,2\) from part a):

$$
LATEXBLOCK0
$$

Step2: Simplify the left - hand side

Expand the left - hand side:

$$
LATEXBLOCK1
$$

Step3: Solve for \(x\)

Subtract \(x\) from both sides:

$$
LATEXBLOCK2
$$

Divide both sides by 2: \(x=-3\)

Step4: Check the solution

We need to check if \(x = - 3\) is not a restricted value (i.e., \(x
eq - 2,2\)). Since \(-3
eq - 2\) and \(-3
eq 2\), \(x=-3\) is a valid solution.

Answer:

\(-2,2\)

Part b