To solve the system of equations simultaneously:

1. \( 3y + x = 3 \)
2. \( xy = 5 \)

### Step 1: Solve one equation for one variable.
From equation (1), solve for \( x \) in terms of \( y \):

\[
x = 3 - 3y
\]

### Step 2: Substitute into the second equation.
Now substitute this expression for \( x \) into equation (2):

\[
(3 - 3y)y = 5
\]

### Step 3: Simplify the equation.
Distribute \( y \) on the left-hand side:

\[
3y - 3y^2 = 5
\]

Rearrange the equation to form a quadratic equation:

\[
3y^2 - 3y + 5 = 0
\]

### Step 4: Solve the quadratic equation.
The quadratic equation is \( 3y^2 - 3y + 5 = 0 \). To solve this, use the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 3 \), \( b = -3 \), and \( c = 5 \). Substitute these into the quadratic formula:

\[
y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(3)(5)}}{2(3)}
\]

Simplify:

\[
y = \frac{3 \pm \sqrt{9 - 60}}{6}
\]
\[
y = \frac{3 \pm \sqrt{-51}}{6}
\]

Since the discriminant is negative (\( -51 \)), the solutions for \( y \) are complex:

\[
y = \frac{3 \pm i\sqrt{51}}{6}
\]
\[
y = \frac{1}{2} \pm \frac{i\sqrt{51}}{6}
\]

### Step 5: Find \( x \).
Now, substitute these values of \( y \) back into \( x = 3 - 3y \) to find the corresponding \( x \) values. Since the solutions for \( y \) are complex, the solutions for \( x \) will also be complex.

For \( y = \frac{1}{2} + \frac{i\sqrt{51}}{6} \):

\[
x = 3 - 3\left(\frac{1}{2} + \frac{i\sqrt{51}}{6}\right)
\]
\[
x = 3 - \frac{3}{2} - \frac{i\sqrt{51}}{2}
\]
\[
x = \frac{3}{2} - \frac{i\sqrt{51}}{2}
\]

For \( y = \frac{1}{2} - \frac{i\sqrt{51}}{6} \):

\[
x = 3 - 3\left(\frac{1}{2} - \frac{i\sqrt{51}}{6}\right)
\]
\[
x = \frac{3}{2} + \frac{i\sqrt{51}}{2}
\]

### Final solutions:
The solutions are:

\[
x = \frac{3}{2} \pm \frac{i\sqrt{51}}{2}, \quad y = \frac{1}{2} \pm \frac{i\sqrt{51}}{6}
\]

These are complex solutions.

Would you like more details on any step or have any questions?

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Here are 5 questions to expand this topic:
1. How do you handle systems of equations when both equations are quadratic?
2. What are real-world applications of complex numbers in solving equations?
3. How can the quadratic formula be derived?
4. What does the discriminant tell you about the nature of solutions?
5. How can simultaneous equations be solved graphically?

**Tip:** The discriminant \( b^2 - 4ac \) reveals whether the solutions are real or complex, and whether they are distinct or repeated.

solve 3y + x = 3 and xy = 5 simultaneously

Explore how to solve the simultaneous equations 3y + x = 3 and xy = 5, including the use of the quadratic formula and handling complex solutions. This step-by-step solution is designed for students in Grades 10-12, covering key algebraic concepts and theorems.

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