Question

solve the equation and check your solution.
\\(\frac{x^2}{x - 3} = \frac{9}{x - 3}\\)

select the correct choice below and, if necessary, fill in the answer box to complete your choice.
\\(\bigcirc\\) a. \\(x = \square\\) (simplify your answer. use a comma to separate answers as needed.)
\\(\bigcirc\\) b. the solution is all real numbers.
\\(\bigcirc\\) c. there is no solution.

Explanation:

Step1: Eliminate denominator (x≠3)

Multiply both sides by \( x - 3 \) (note \( x
eq 3 \) to avoid division by zero), getting \( x^2 = 9 \).

Step2: Solve quadratic equation

Factor \( x^2 - 9 = 0 \) as \( (x - 3)(x + 3) = 0 \), so \( x = 3 \) or \( x = -3 \).

Step3: Check for valid solutions

\( x = 3 \) makes original denominator zero (invalid). \( x = -3 \): substitute into original equation, \( \frac{(-3)^2}{-3 - 3} = \frac{9}{-6} \), \( \frac{9}{-3 - 3} = \frac{9}{-6} \), both sides equal. So only \( x = -3 \) is valid? Wait, no—wait, when we multiplied by \( x - 3 \), we assumed \( x
eq 3 \), but solving \( x^2 = 9 \) gives \( x = 3 \) or \( x = -3 \). But \( x = 3 \) is excluded, so check \( x = -3 \): works. Wait, but wait, original equation: \( \frac{x^2}{x - 3} = \frac{9}{x - 3} \). If we subtract \( \frac{9}{x - 3} \) from both sides, we get \( \frac{x^2 - 9}{x - 3} = 0 \), which simplifies to \( \frac{(x - 3)(x + 3)}{x - 3} = 0 \), then cancel \( x - 3 \) (x≠3) to get \( x + 3 = 0 \), so \( x = -3 \). So the solution is \( x = -3 \). Wait, but the options: A is x= (fill in), B all real, C no solution. So x = -3. Wait, but let's recheck. Wait, when x=3, denominator is zero, so undefined. So solving gives x=-3. So the correct choice is A with x=-3. Wait, but let's do the steps again.

Wait, step by step:

1. Given \( \frac{x^2}{x - 3} = \frac{9}{x - 3} \). The domain is all real numbers except \( x = 3 \) (since denominator can't be zero).

1. Multiply both sides by \( x - 3 \) (valid since \( x

eq 3 \)): \( x^2 = 9 \).

1. Solve \( x^2 = 9 \): \( x = 3 \) or \( x = -3 \).

1. But \( x = 3 \) is not in the domain, so we discard it. Check \( x = -3 \): substitute into original equation. Left side: \( \frac{(-3)^2}{-3 - 3} = \frac{9}{-6} = -\frac{3}{2} \). Right side: \( \frac{9}{-3 - 3} = \frac{9}{-6} = -\frac{3}{2} \). So \( x = -3 \) is a valid solution.

Wait, but earlier mistake: when we cancel \( x - 3 \), we have to remember that \( x
eq 3 \), so the equation simplifies to \( x + 3 = 0 \) (after canceling \( x - 3 \) in \( \frac{(x - 3)(x + 3)}{x - 3} \)), so \( x = -3 \). So the solution is \( x = -3 \). So option A, x = -3.

Wait, but wait, maybe I made a mistake. Let's check again. Original equation: \( \frac{x^2}{x - 3} = \frac{9}{x - 3} \). If x ≠ 3, we can multiply both sides by x - 3, getting x² = 9, so x = 3 or x = -3. But x = 3 is excluded, so only x = -3 is valid. So the solution is x = -3, so option A with x = -3.

Answer:

A. \( x = -3 \)