If x is a real number that satisfies the equation \frac{x}{x-3} - \frac{6}{x^2-3x} = \frac… — SAT

If $x$ is a real number that satisfies the equation $\frac{x}{x-3} - \frac{6}{x^2-3x} = \frac{2}{x}$ , what is the value of $x$ ?

Answer: 2

The correct answer is 2. Multiplying the entire equation by the lowest common denominator,

x(x-3)

, yields

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

. Distributing the right side gives

x^2 - 6 = 2x - 6

. Adding 6 to both sides and subtracting

2x

results in

x^2 - 2x = 0

. Factoring this equation gives

x(x-2) = 0

, which has solutions

x = 0

and

x = 2

. However,

x = 0

is extraneous because it results in division by zero in the original equation. Thus, the only valid real solution is 2.

Identify the common denominator and multiply the equation by it.

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

To clear the fractions and simplify the rational equation.

Distribute and simplify the equation.

x^2 - 2x = 0

To set up the quadratic equation in standard form.

Factor and solve for

x

x = 0

x = 2

To find potential solutions to the equation.

Check for extraneous solutions by substituting the values back into the original denominators.

x = 2

x = 0

makes the denominators equal to zero and is therefore extraneous.

Solving rational equations and identifying extraneous solutions.

Question