An equation is given as follows: \frac{x}{x - 3} - \frac{4}{x + 2} = \frac{20}{x^2 - x - 6… — SAT

An equation is given as follows:

\frac{x}{x - 3} - \frac{4}{x + 2} = \frac{20}{x^2 - x - 6}

What is the only value of $x$ for which this equation is true?

Answer: 4

To solve the equation, factor the denominator on the right side:

x^2 - x - 6 = (x - 3)(x + 2)

. The least common denominator is

(x - 3)(x + 2)

. Multiplying both sides by

(x - 3)(x + 2)

clears the fractions, resulting in

x(x + 2) - 4(x - 3) = 20

. Expanding the terms gives

x^2 + 2x - 4x + 12 = 20

. Simplifying and writing this in standard form yields

x^2 - 2x - 8 = 0

. Factoring the quadratic gives

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

, which gives the candidate solutions

x = 4

and

x = -2

. Substituting

x = -2

back into the original equation results in a denominator of zero, so

x = -2

is an extraneous solution. The only valid solution is

x = 4

Factor the quadratic trinomial in the denominator of the right side of the equation.

x^2 - x - 6 = (x - 3)(x + 2)

Identifying the factors of the quadratic trinomial helps find the least common denominator of the rational equation.

Multiply the entire equation by the least common denominator

(x - 3)(x + 2)

to clear the fractions.

x(x + 2) - 4(x - 3) = 20

, with the constraints that

x \neq 3

and

x \neq -2

This simplifies the rational equation into a standard polynomial equation.

Distribute and combine like terms to write the equation in standard quadratic form.

x^2 - 2x - 8 = 0

Rewriting the equation in the form

ax^2 + bx + c = 0

is necessary to solve it by factoring.

Factor the quadratic equation.

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

Factoring allows us to apply the zero product property to find candidate solutions.

Check the candidate solutions

x = 4

and

x = -2

against the original equation to identify extraneous solutions.

x = 4

is the only valid solution because

x = -2

makes the denominators in the original equation equal to zero.

Any solution that makes a denominator in the original rational expression equal to zero is extraneous and must be excluded.

Solving rational equations by finding a common denominator, clearing fractions, and checking for extraneous solutions.

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