An equation is shown below. \frac{x}{x-3} - \frac{2}{x+1} = \frac{8}{x^2-2x-3} What is the… — SAT

An equation is shown below.

\frac{x}{x-3} - \frac{2}{x+1} = \frac{8}{x^2-2x-3}

What is the real solution to the equation above?

Answer: 2

The only real solution is 2.

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

(x-3)(x+1) = x^2-2x-3

. This yields

x(x+1) - 2(x-3) = 8

. Simplifying this equation gives

x^2 - x - 2 = 0

. Factoring the quadratic expression gives

(x-2)(x+1) = 0

, which yields potential solutions of

x = 2

and

x = -1

. However, substituting

x = -1

into the original equation results in division by zero, making it an extraneous solution. Substituting

x = 2

into the original equation is valid, so the only real solution is 2.

Factor the quadratic denominator on the right side of the equation.

x^2 - 2x - 3 = (x - 3)(x + 1)

This helps identify the least common denominator of the rational terms.

Multiply all terms of the equation by the least common denominator,

(x - 3)(x + 1)

, to eliminate the denominators.

x(x + 1) - 2(x - 3) = 8

Multiplying by the LCD clears the rational expressions, converting the equation into a polynomial equation, under the restriction that

x \neq 3

and

x \neq -1

Expand and simplify the resulting equation.

x^2 + x - 2x + 6 = 8 \Rightarrow x^2 - x - 2 = 0

This puts the equation into standard quadratic form:

ax^2 + bx + c = 0

Factor the quadratic equation.

(x - 2)(x + 1) = 0

Factoring allows us to find the potential solutions by setting each factor equal to zero.

Find the roots of the equation.

x = 2 \text{ or } x = -1

These are the values of

x

that satisfy the factored quadratic equation.

Check the potential solutions in the original equation to identify any extraneous solutions.

Substituting

x = -1

results in division by zero in the terms

\frac{2}{x+1}

and

\frac{8}{x^2-2x-3}

, so

x = -1

is extraneous. Substituting

x = 2

yields a valid statement:

-\frac{8}{3} = -\frac{8}{3}

Solutions that make any denominator in the original rational equation equal to zero are extraneous and must be excluded.

Solving rational equations and identifying extraneous solutions.

Question