📌 What is a Linear-Quadratic System?

A linear-quadratic system consists of one linear equation and one quadratic equation. Common combinations include:

• A straight line and a parabola
• A straight line and a circle
• A straight line and an ellipse

🎯 Number of Solutions

A linear-quadratic system can have:

• 0 solutions - The graphs don't intersect
• 1 solution - The line is tangent to the curve
• 2 solutions - The line intersects the curve at two points

Steps:

1. Identify the graph of each equation (line, parabola, circle)
2. Graph the first equation on a coordinate plane
3. Graph the second equation on the same plane
4. Locate the points of intersection
5. Verify solutions by substituting into both equations

📝 Example:

Solve the system:

\( y = x^2 - 4 \) (parabola)
\( y = 2x - 1 \) (line)

Graph both equations and find where they intersect. The intersection points are the solutions.

Steps for Parabolas:

1. Solve the linear equation for one variable (usually \(y\))
2. Substitute this expression into the quadratic equation
3. Simplify to get a quadratic equation in one variable
4. Solve using factoring, quadratic formula, or completing the square
5. Find corresponding \(y\)-values by substituting \(x\)-values into the linear equation
6. Write solutions as ordered pairs \((x, y)\)

📝 Example:

Solve the system:

\( y = x^2 - 6x + 3 \)
\( y = -2x + 3 \)

Step 1: The linear equation is already solved for \(y\)

Step 2: Substitute: \( -2x + 3 = x^2 - 6x + 3 \)

Step 3: Simplify: \( 0 = x^2 - 4x \)

Step 4: Factor: \( 0 = x(x - 4) \), so \( x = 0 \) or \( x = 4 \)

Step 5: Find \(y\) values:

• When \( x = 0 \): \( y = -2(0) + 3 = 3 \)
• When \( x = 4 \): \( y = -2(4) + 3 = -5 \)

Solutions: \((0, 3)\) and \((4, -5)\)

Key Formulas:

Standard form of a circle:

\( (x - h)^2 + (y - k)^2 = r^2 \)

Where \((h, k)\) is the center and \(r\) is the radius

General form of a circle:

\( x^2 + y^2 + Dx + Ey + F = 0 \)

Steps:

1. Solve the linear equation for \(y\) (or \(x\))
2. Substitute into the circle equation
3. Expand and simplify to get a quadratic equation
4. Solve for the variable
5. Substitute back to find the other coordinate
6. Check your solutions in both equations

📝 Example:

Solve the system:

\( x^2 + y^2 = 9 \) (circle with center \((0,0)\), radius \(3\))
\( y = x - 3 \) (line)

Step 1: Linear equation is solved for \(y\)

Step 2: Substitute: \( x^2 + (x - 3)^2 = 9 \)

Step 3: Expand: \( x^2 + x^2 - 6x + 9 = 9 \)

Step 4: Simplify: \( 2x^2 - 6x = 0 \)

Step 5: Factor: \( 2x(x - 3) = 0 \), so \( x = 0 \) or \( x = 3 \)

Step 6: Find \(y\) values:

• When \( x = 0 \): \( y = 0 - 3 = -3 \)
• When \( x = 3 \): \( y = 3 - 3 = 0 \)

Solutions: \((0, -3)\) and \((3, 0)\)

When to Use Elimination:

Elimination works best when both variables are squared in both equations, such as:

• Two circles
• A circle and a hyperbola
• A circle and an ellipse

Steps:

1. Write both equations in standard form
2. Multiply one or both equations to make coefficients of one variable opposites
3. Add the equations to eliminate one variable
4. Solve for the remaining variable
5. Substitute each solution into one original equation
6. Solve for the other variable
7. Write all solutions as ordered pairs

📝 Example:

Solve the system:

\( x^2 + y^2 = 7 \) (circle)
\( x^2 - y^2 = 1 \) (hyperbola)

Step 1: Both equations are in standard form

Step 2: Notice \(y^2\) has opposite coefficients already

Step 3: Add equations: \( 2x^2 = 8 \)

Step 4: Solve: \( x^2 = 4 \), so \( x = \pm 2 \)

Step 5: Substitute \(x = 2\) and \(x = -2\) into first equation:

\( 4 + y^2 = 7 \)
\( y^2 = 3 \)
\( y = \pm \sqrt{3} \)

Solutions: \((2, \sqrt{3})\), \((2, -\sqrt{3})\), \((-2, \sqrt{3})\), \((-2, -\sqrt{3})\)

⚡ Quick Tips

• Substitution is easiest when one equation is linear
• Elimination works best when both variables are squared
• Always check your solutions in both original equations
• Use the discriminant \(b^2 - 4ac\) to determine if real solutions exist
• If discriminant \(< 0\), there are no real solutions
• Graphing helps visualize the number of solutions

📚 Important Formulas