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To solve this system of equations:

\[
\begin{cases}
x + y = 4 \\
x^2 - 2xy - 3 = 0
\end{cases}
\]

we'll proceed algebraically first, and then sketch the graphs to find the points of intersection.

### Step 1: Substitute \( y = 4 - x \) into the second equation
From the first equation, we have:
\[
y = 4 - x
\]
Substitute \( y = 4 - x \) into the second equation:
\[
x^2 - 2x(4 - x) - 3 = 0
\]

### Step 2: Expand and simplify
Expanding \( -2x(4 - x) \):
\[
x^2 - 8x + 2x^2 - 3 = 0
\]
Combine like terms:
\[
3x^2 - 8x - 3 = 0
\]

### Step 3: Solve the quadratic equation
Now we solve \( 3x^2 - 8x - 3 = 0 \) using the quadratic formula:
\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3}
\]
\[
x = \frac{8 \pm \sqrt{64 + 36}}{6}
\]
\[
x = \frac{8 \pm \sqrt{100}}{6}
\]
\[
x = \frac{8 \pm 10}{6}
\]

This gives us two solutions for \( x \):
1. \( x = \frac{18}{6} = 3 \)
2. \( x = \frac{-2}{6} = -\frac{1}{3} \)

### Step 4: Find corresponding \( y \) values
For each \( x \)-value, substitute back into \( y = 4 - x \):

1. When \( x = 3 \):
   \[
   y = 4 - 3 = 1
   \]
   So, one solution is \( (3, 1) \).

2. When \( x = -\frac{1}{3} \):
   \[
   y = 4 - \left(-\frac{1}{3}\right) = 4 + \frac{1}{3} = \frac{12 + 1}{3} = \frac{13}{3}
   \]
   So, the other solution is \( \left(-\frac{1}{3}, \frac{13}{3}\right) \).

### Final Answer
The solutions to the system are:
\[
(3, 1) \quad \text{and} \quad \left(-\frac{1}{3}, \frac{13}{3}\right)
\]

### Verification by Graphing
To confirm, you would graph both equations:
1. The line \( x + y = 4 \).
2. The hyperbola \( x^2 - 2xy - 3 = 0 \).

The points of intersection on the graph will correspond to the solutions we found: \( (3, 1) \) and \( \left(-\frac{1}{3}, \frac{13}{3}\right) \).

Would you like a graphical sketch for verification or further explanation on any of the steps?

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### Related Questions
1. How would you solve a system of equations if both were quadratic?
2. What is the significance of finding points of intersection in a system of equations?
3. Can a linear equation intersect a hyperbola at more than two points?
4. How can we check the solutions of a nonlinear system using a calculator?
5. What other methods can be used to solve nonlinear systems algebraically?

### Tip
When dealing with systems involving a linear and nonlinear equation, substituting from the linear equation often simplifies the solution process significantly.

Solve the following system of equations. Verify your answers algebraically and graphically.

x + y = 4
x^2 - 2xy - 3 = 0

Learn how to solve the system of equations with a linear and quadratic equation: x + y = 4 and x^2 - 2xy - 3 = 0. This problem involves algebraic and graphical solutions, covering key concepts in systems of equations and quadratic equations. Suitable for high school students in Grades 10-12.

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