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. \( x = \pm 1 \)
(note only one answer can be selected)

Answer 1

To solve the rational equation

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

we first simplify the left-hand side. We can combine the two fractions on the left:

\[
\frac{3 + (x - 3)}{x} = \frac{x}{x} = 1.
\]

Now, we rewrite our equation:

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

Next, we can multiply both sides by \( x^2 \) (assuming \( x \neq 0 \) since we have a division by \( x \)) to eliminate the fraction:

\[
x^2 = 1.
\]

Now, we solve for \( x \):

\[
x^2 - 1 = 0,
\]

factoring gives:

\[
(x - 1)(x + 1) = 0.
\]

Thus, the solutions are:

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

Now we check whether these solutions are valid in the original equation. Since we multiplied by \( x^2 \) and assumed \( x \neq 0 \), both solutions \( x = 1 \) and \( x = -1 \) are valid because they are not equal to zero.

Now we also check if the solutions correspond to the given options. The options are:

A. \( x = 0 \)

B. \( x = 3 \)

C. \( x = 1 \)

D. \( x = \pm 1 \)

Since we found solutions \( x = 1 \) and \( x = -1 \), the correct response from the list is D. \( x = \pm 1 \).

Therefore, the answer is:

\[
\boxed{D}
\]