We have two equations here:

1. \( x + y = 8 \)
2. \( xy = 48 \)

We can solve this system of equations by using substitution or by solving it as a quadratic equation. I'll use substitution to get to the answer.

### Step-by-Step Solution:

1. From the first equation, solve for one variable in terms of the other:
   \[
   x + y = 8 \implies y = 8 - x
   \]

2. Substitute this expression for \( y \) into the second equation:
   \[
   x(8 - x) = 48
   \]

3. Expand and simplify the equation:
   \[
   8x - x^2 = 48
   \]
   \[
   -x^2 + 8x - 48 = 0
   \]
   Multiply through by \(-1\) to simplify:
   \[
   x^2 - 8x + 48 = 0
   \]

4. Now solve this quadratic equation using the quadratic formula:
   The quadratic formula is given by:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 1 \), \( b = -8 \), and \( c = 48 \).

   First, calculate the discriminant:
   \[
   \Delta = b^2 - 4ac = (-8)^2 - 4(1)(48) = 64 - 192 = -128
   \]

Since the discriminant is negative (\(\Delta = -128\)), the solutions will be complex numbers.

5. Now apply the quadratic formula:
   \[
   x = \frac{-(-8) \pm \sqrt{-128}}{2(1)} = \frac{8 \pm \sqrt{-128}}{2}
   \]
   \[
   x = \frac{8 \pm \sqrt{128}i}{2} = \frac{8 \pm 8\sqrt{2}i}{2}
   \]
   Simplifying:
   \[
   x = 4 \pm 4\sqrt{2}i
   \]

Thus, the two complex solutions for \( x \) are:
\[
x = 4 + 4\sqrt{2}i \quad \text{or} \quad x = 4 - 4\sqrt{2}i
\]

6. To find the corresponding values of \( y \), substitute these values of \( x \) back into the equation \( y = 8 - x \):
   - If \( x = 4 + 4\sqrt{2}i \), then \( y = 8 - (4 + 4\sqrt{2}i) = 4 - 4\sqrt{2}i \)
   - If \( x = 4 - 4\sqrt{2}i \), then \( y = 8 - (4 - 4\sqrt{2}i) = 4 + 4\sqrt{2}i \)

### Final Solutions:
- \( x = 4 + 4\sqrt{2}i \) and \( y = 4 - 4\sqrt{2}i \)
- \( x = 4 - 4\sqrt{2}i \) and \( y = 4 + 4\sqrt{2}i \)

Would you like more details, or do you have any questions?

### 5 Related Questions:
1. How do we handle quadratic equations with real solutions?
2. What happens if the discriminant is zero in the quadratic formula?
3. How are complex numbers used in real-world applications?
4. Can we graph the solutions of complex numbers?
5. How can we solve systems of nonlinear equations without using the quadratic formula?

### Tip:
When solving quadratic equations, always check the discriminant first. It tells you whether the solutions will be real, complex, or repeated.

Learn how to solve the system of equations x + y = 8 and xy = 48 using substitution and the quadratic formula. This problem involves algebraic concepts and complex numbers. The solution is suitable for students in Grades 9-11.

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