Question

Solve the rational equation for \( x \):



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



A. \( x = 0 \)

B. \( x = 3 \)

C. \( x = 1 \)

D. \( x = \pm 1 \)

Answer 1

To solve the rational equation

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

we start by simplifying the left-hand side. The left side can be combined as follows:

\[
\frac{3}{x} + \frac{x - 3}{x} = \frac{3 + (x - 3)}{x} = \frac{x}{x} = 1 \quad (\text{for } x \neq 0).
\]

Now, the equation simplifies to

\[
1 = \frac{1}{x^2}.
\]

Next, we cross-multiply to eliminate the fraction:

\[
x^2 = 1.
\]

Taking the square root of both sides, we get:

\[
x = 1 \quad \text{or} \quad x = -1.
\]

Next, we need to check if these values are valid in the context of the original equation. They must not cause a division by zero.

1. **For** \( x = 1 \):

Substituting back into the original equation:

\[
\frac{3}{1} + \frac{1 - 3}{1} = \frac{1}{1^2} \implies 3 - 2 = 1 \implies 1 = 1 \quad (\text{true}).
\]

2. **For** \( x = -1 \):

Substituting back into the original equation:

\[
\frac{3}{-1} + \frac{-1 - 3}{-1} = \frac{1}{(-1)^2} \implies -3 + 4 = 1 \implies 1 = 1 \quad (\text{true}).
\]

Therefore, both \( x = 1 \) and \( x = -1 \) are valid solutions.

Now, let's examine the solution options:

A. \( x = 0 \) (Not valid, as it causes division by zero)

B. \( x = 3 \) (Let's check: \(\frac{3}{3} + \frac{3 - 3}{3} = \frac{1}{3^2} \implies 1 + 0 \neq \frac{1}{9}\))

C. \( x = 1 \) (Valid solution)

D. \( x = \pm 1 \) (Valid solutions)

The valid solutions are:

- \( x = 1 \)
- \( x = -1 \)

Thus, the answer is:

\[
\text{C. } x = 1 \quad \text{and} \quad \text{D. } x = \pm 1.
\]