If \frac{x}{x - 2} - \frac{8}{x + 2} = \frac{8}{x^2 - 4}, what is the value of x? — SAT

If $\frac{x}{x - 2} - \frac{8}{x + 2} = \frac{8}{x^2 - 4}$ , what is the value of $x$ ?

Answer: 4

To solve the rational equation, multiply both sides by the least common denominator,

x^2 - 4

. This results in the equation

x(x + 2) - 8(x - 2) = 8

. Expanding and combining like terms yields the quadratic equation

x^2 - 6x + 8 = 0

. Factoring this equation gives

(x - 2)(x - 4) = 0

, which yields the potential solutions

x = 2

and

x = 4

. However,

x = 2

must be excluded because it makes the denominators of the original equation equal to zero. Therefore, the only valid solution is

x = 4

Determine the least common denominator (LCD) for the terms in the equation.

The LCD is

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

Multiplying the entire equation by the LCD clears all rational expressions.

Multiply the entire equation by the LCD to eliminate the denominators.

x(x + 2) - 8(x - 2) = 8

This simplifies the rational equation into a polynomial equation.

Expand, simplify, and solve the resulting quadratic equation.

x^2 - 6x + 8 = 0 \Rightarrow (x - 2)(x - 4) = 0 \Rightarrow x = 2 \text{ or } x = 4

Factoring the quadratic equation gives the potential solutions.

Substitute each potential solution into the original equation to check for extraneous solutions.

Substituting

x = 2

leads to division by zero, making the expression undefined. Substituting

x = 4

yields a valid identity:

\frac{4}{2} - \frac{8}{6} = \frac{2}{3}

. Therefore,

x = 4

is the only valid solution.

Values that make any denominator in the original equation equal to zero must be excluded.

Solving rational equations and checking for extraneous solutions.

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