Certainly! Let's find the solutions to the given system of equations step-by-step.
1. Solve the equation \( x = \frac{3}{x} \):
- First, multiply both sides by \( x \) to clear the fraction:
[tex]\[
x \cdot x = 3
\][/tex]
[tex]\[
x^2 = 3
\][/tex]
- Next, take the square root of both sides to solve for \( x \):
[tex]\[
x = \sqrt{3} \quad \text{or} \quad x = -\sqrt{3}
\][/tex]
So, the solutions for this equation are:
[tex]\[
x = \sqrt{3}, \quad x = -\sqrt{3}
\][/tex]
2. Solve the equation \( x = \frac{9}{x} \):
- Similar to the previous step, multiply both sides by \( x \) to clear the fraction:
[tex]\[
x \cdot x = 9
\][/tex]
[tex]\[
x^2 = 9
\][/tex]
- Take the square root of both sides to solve for \( x \):
[tex]\[
x = 3 \quad \text{or} \quad x = -3
\][/tex]
So, the solutions for this equation are:
[tex]\[
x = 3, \quad x = -3
\][/tex]
3. Solve the equation \( 0 = \frac{3}{x} + x \):
- Start by moving all terms to one side to set the equation to zero:
[tex]\[
\frac{3}{x} + x = 0
\][/tex]
- Multiply every term by \( x \) to clear the fraction:
[tex]\[
3 + x^2 = 0
\][/tex]
- Rearrange to isolate the quadratic term:
[tex]\[
x^2 = -3
\][/tex]
Since \( x^2 = -3 \) involves the square root of a negative number, it does not have any real solutions. Thus, for real numbers, there are no solutions to this equation:
[tex]\[
\text{No real solutions}
\][/tex]
4. Solve the equation \( 0 = \frac{9}{x} + x \):
- Similarly, start by moving all terms to one side:
[tex]\[
\frac{9}{x} + x = 0
\][/tex]
- Multiply every term by \( x \) to clear the fraction:
[tex]\[
9 + x^2 = 0
\][/tex]
- Rearrange to isolate the quadratic term:
[tex]\[
x^2 = -9
\][/tex]
Again, since \( x^2 = -9 \) involves the square root of a negative number, it does not have any real solutions. Thus, for real numbers, there are no solutions to this equation:
[tex]\[
\text{No real solutions}
\][/tex]
To summarize, the solutions to the system of equations are:
- For \( x = \frac{3}{x} \), the solutions are \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).
- For \( x = \frac{9}{x} \), the solutions are \( x = 3 \) and \( x = -3 \).
- Both \( 0 = \frac{3}{x} + x \) and \( 0 = \frac{9}{x} + x \) have no real solutions.
So, our final answers are:
[tex]\[
([1.7320508075688772, -1.7320508075688772], [3, -3], [], [])
\][/tex]
