If x is a real solution to the equation \frac{3x}{x - 1} + \frac{2x}{x - 2} = \frac{x + 2}… — SAT

x

is a real solution to the equation

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

what is the value of

5x - 4

-5

The correct answer is

-5

. Solving the equation by multiplying both sides by the least common denominator

(x-1)(x-2)

yields the quadratic equation

5x^2 - 9x - 2 = 0

. Factoring this equation gives the potential solutions

x = -1/5

and

x = 2

. However, substituting

x = 2

into the original equation results in division by zero, meaning

x = 2

is extraneous and must be discarded. The only valid solution is

x = -1/5

. Substituting this value into the expression

5x - 4

yields

5(-1/5) - 4 = -5

Identify the least common denominator (LCD) and multiply both sides of the equation to clear the denominators.

The LCD is

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

. Multiplying both sides by the LCD gives:

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

Multiplying by the LCD simplifies the rational equation into a polynomial equation.

Expand the terms and collect like terms to form a quadratic equation.

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

5x^2 - 8x = x + 2

5x^2 - 9x - 2 = 0

This puts the equation in the standard quadratic form

ax^2 + bx + c = 0

Factor the quadratic equation to find potential solutions.

The quadratic factors as:

(5x + 1)(x - 2) = 0

This yields potential solutions

x = -\frac{1}{5}

and

x = 2

Factoring allows us to isolate the candidate values of

x

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

For

x = 2

, the denominators

x - 2

and

x^2 - 3x + 2

become zero, which is undefined. Therefore,

x = 2

is an extraneous solution. For

x = -\frac{1}{5}

, all denominators are non-zero, so it is the only valid solution.

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

Substitute the valid solution into the target expression

5x - 4

5\left(-\frac{1}{5}\right) - 4 = -1 - 4 = -5

Evaluating the expression with the only valid solution gives the final answer.

Solving rational equations by clearing denominators and checking for extraneous solutions.

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