Question
Solve the rational equation for x: 3/x + x-3/x = 1/x^2
A
x=0x=0x=0
B
x=3x=3x=3
C
x=1x=1x=1
D
x=±1x=\pm1x=±1
A
x=0x=0x=0
B
x=3x=3x=3
C
x=1x=1x=1
D
x=±1x=\pm1x=±1
Answers
GPT-4o mini
To solve the equation
\[
\frac{3}{x} + \frac{x - 3}{x} = \frac{1}{x^2},
\]
first, we can combine the left side:
\[
\frac{3 + (x - 3)}{x} = \frac{1}{x^2}.
\]
Simplifying the left side,
\[
\frac{x}{x} = 1.
\]
Thus, the equation becomes:
\[
1 = \frac{1}{x^2}.
\]
Cross multiplying gives:
\[
x^2 = 1.
\]
Taking the square root of both sides, we get:
\[
x = 1 \quad \text{or} \quad x = -1.
\]
Now, we should verify that these solutions do not make any denominator zero. For \(x = 1\) and \(x = -1\), the denominators \(x\) and \(x^2\) are both non-zero.
Thus, the solutions to the equation are:
\[
x = 1 \quad \text{and} \quad x = -1.
\]
Therefore, the answer is:
\[
\text{D: } x = \pm 1.
\]
\[
\frac{3}{x} + \frac{x - 3}{x} = \frac{1}{x^2},
\]
first, we can combine the left side:
\[
\frac{3 + (x - 3)}{x} = \frac{1}{x^2}.
\]
Simplifying the left side,
\[
\frac{x}{x} = 1.
\]
Thus, the equation becomes:
\[
1 = \frac{1}{x^2}.
\]
Cross multiplying gives:
\[
x^2 = 1.
\]
Taking the square root of both sides, we get:
\[
x = 1 \quad \text{or} \quad x = -1.
\]
Now, we should verify that these solutions do not make any denominator zero. For \(x = 1\) and \(x = -1\), the denominators \(x\) and \(x^2\) are both non-zero.
Thus, the solutions to the equation are:
\[
x = 1 \quad \text{and} \quad x = -1.
\]
Therefore, the answer is:
\[
\text{D: } x = \pm 1.
\]