Chapter 10 Linear-Quadratic Systems of Equations

What Are Linear-Quadratic Systems?

A linear-quadratic system is simply a system where one equation is linear (forms a straight line) and the other is quadratic (forms a parabola). You're trying to find the points where these two curves actually intersect.

That's it. Two equations, two different shapes, and you're hunting for where they meet. Mathematicians call these intersection points solutions.

Most systems you'll encounter look like this:

• A linear equation: y = mx + b
• A quadratic equation: y = ax² + bx + c

You can have 0, 1, or 2 real solutions. The parabola might miss the line entirely. It might touch it at exactly one point. Or it could cut through it twice. That's the geometry. Now let's solve them.

The Two Methods That Actually Work

Forget everything else you've heard. There are only two methods worth your time: substitution and graphing. Elimination doesn't help here because one equation has x² and the other doesn't.

The Substitution Method

Substitution works every time. Here's how it goes:

1. Solve the linear equation for y (or x)
2. Plug that expression into the quadratic equation
3. Solve the resulting quadratic
4. Back-substitute to find the matching coordinates

This method is algebra-heavy but reliable. No guessing, no graphing errors.

The Graphing Method

Graph both equations on the same coordinate plane. The intersection points are your solutions. That's the whole method.

The catch? You need to be accurate. Slightly off on your parabola? Your answers will be wrong. Graphing is great for understanding what's happening visually, but substitution gives you exact answers.

How to Solve: Step-by-Step Example

Let's work through this system:

y = 2x + 3
y = x² - 4x + 1

Step 1: Notice the linear equation is already solved for y. That's your substitution expression.

Step 2: Replace y in the quadratic with 2x + 3:

x² - 4x + 1 = 2x + 3

Step 3: Rearrange everything to one side:

x² - 4x - 2x + 1 - 3 = 0
x² - 6x - 2 = 0

Step 4: Solve using the quadratic formula:

x = (-(-6) ± √((-6)² - 4(1)(-2))) / 2(1)
x = (6 ± √(36 + 8)) / 2
x = (6 ± √44) / 2
x = (6 ± 2√11) / 2
x = 3 ± √11

So x₁ ≈ 6.32 and x₂ ≈ -0.32

Step 5: Find the matching y values by plugging back into y = 2x + 3:

For x ≈ 6.32: y ≈ 2(6.32) + 3 ≈ 15.64
For x ≈ -0.32: y ≈ 2(-0.32) + 3 ≈ 2.36

Solutions: (6.32, 15.64) and (-0.32, 2.36)

Verify both points satisfy the original equations. They should.

When the Discriminant Tells You Everything

After substituting, you'll always end up with a quadratic equation. The discriminant (b² - 4ac) tells you how many solutions exist before you finish solving:

• Positive → 2 real solutions (line cuts through the parabola twice)
• Zero → 1 real solution (line is tangent to the parabola)
• Negative → 0 real solutions (line and parabola never meet)

Use this to check your work. If you get 2 solutions but your discriminant is negative, you messed up somewhere.

Common Mistakes That Ruin Everything

• Forgetting to set equations equal — After substitution, you must have everything on one side. "y = y" doesn't solve anything.
• Messy algebra during distribution — Take your time expanding (2x+3)². Errors here cascade through the entire problem.
• Dropping the negative sign — When moving terms across the equals sign, signs flip. Don't forget this.
• Rounding too early — Keep exact values until the final answer. Rounding mid-calculation compounds errors.

Method Comparison

When You'll Actually Use This

Physics problems involving projectile motion. Business problems with revenue curves versus cost lines. Engineering problems with parabolic and linear relationships. The applications exist, but most students won't encounter them outside a classroom.

What's actually useful: these problems train you to handle systems with different equation types. That skill transfers to anything involving multiple constraints.

Quick Reference

When solving any linear-quadratic system:

1. Identify which equation is linear, which is quadratic
2. Solve the linear equation for y
3. Substitute into the quadratic
4. Solve the resulting quadratic equation
5. Back-substitute to find y-values
6. Check discriminant to verify solution count

That's the complete process. No magic, no shortcuts that work every time, no special cases beyond what substitution handles naturally.