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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.

[tex]\frac{3}{x+2} - \frac{1}{x-2} = \frac{4x}{x^2-4}[/tex]

a. Write the value or values of the variable that make a denominator zero.
[tex]x = \square[/tex] (Use a comma to separate answers as needed.)

Sagot :

GinnyAnswer
Let's start by identifying the denominators in the given equation:

[tex]\[ \frac{3}{x+2}-\frac{1}{x-2}=\frac{4 x}{x^2-4} \][/tex]

We have three denominators in this equation:
1. \( x + 2 \)
2. \( x - 2 \)
3. \( x^2 - 4 \)

To find the values of \( x \) that make each of these denominators zero, we can set each denominator equal to zero and solve for \( x \).

1. For \( x + 2 = 0 \):
[tex]\[ x + 2 = 0 \][/tex]
[tex]\[ x = -2 \][/tex]

2. For \( x - 2 = 0 \):
[tex]\[ x - 2 = 0 \][/tex]
[tex]\[ x = 2 \][/tex]

3. For \( x^2 - 4 = 0 \):
[tex]\[ x^2 - 4 = 0 \][/tex]
[tex]\[ (x - 2)(x + 2) = 0 \][/tex]
[tex]\[ x = 2 \text{ or } x = -2 \][/tex]

Thus, the values of \( x \) that make the denominators zero are \( x = -2 \) and \( x = 2 \). These are the restrictions on the variable.

So, the values of \( x \) that make a denominator zero are:
[tex]\[ x = -2, 2 \][/tex]

### Part b. Solving the equation with these restrictions in mind:

The given equation is:
[tex]\[ \frac{3}{x+2} - \frac{1}{x-2} = \frac{4x}{x^2-4} \][/tex]

Notice that \( x^2 - 4 \) can be factored as \( (x+2)(x-2) \). Therefore, the right-hand side can be written as:
[tex]\[ \frac{4x}{(x+2)(x-2)} \][/tex]

Now, the equation becomes:
[tex]\[ \frac{3}{x+2} - \frac{1}{x-2} = \frac{4x}{(x+2)(x-2)} \][/tex]

To solve this equation, we perform algebraic steps and then check if the solutions violate any of the restrictions:

By finding a common denominator on the left side of the equation, we have:
[tex]\[ \frac{3(x-2) - 1(x+2)}{(x+2)(x-2)} = \frac{4x}{(x+2)(x-2)} \][/tex]

Simplifying the numerator on the left side:
[tex]\[ \frac{3x - 6 - x - 2}{(x+2)(x-2)} = \frac{4x}{(x+2)(x-2)} \][/tex]
[tex]\[ \frac{2x - 8}{(x+2)(x-2)} = \frac{4x}{(x+2)(x-2)} \][/tex]

Multiplying both sides by the common denominator \( (x+2)(x-2) \), we get:
[tex]\[ 2x - 8 = 4x \][/tex]

Solving this for \( x \):
[tex]\[ 2x - 4x = 8 \][/tex]
[tex]\[ -2x = 8 \][/tex]
[tex]\[ x = -4 \][/tex]

Finally, we need to check if \( x = -4 \) violates any of the restrictions. The restrictions are \( x \neq -2 \) and \( x \neq 2 \), and it is clear that \( x = -4 \) does not violate these restrictions.

Therefore, the solution to the equation, considering the restrictions, is:
[tex]\[ x = -4 \][/tex]
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