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Solving Linear–Quadratic
Systems

Find where lines meet curves by solving systems that pair a linear equation with a quadratic equation: algebraically and graphically.

Algebra 2

What Is a Linear–Quadratic System?

A linear–quadratic system is a set of two equations in two variables where one equation is linear (degree 1) and the other is quadratic (degree 2). The goal is to find every \((x, y)\) pair that satisfies both equations simultaneously.

Linear Equation
Graphs as a straight line. General form: \(y = mx + b\)
Quadratic Equation
Graphs as a parabola (or circle, ellipse). General form: \(y = ax^2 + bx + c\)

Seeing Solutions Graphically

When you graph a line and a parabola on the same axes, the number of intersection points tells you how many solutions the system has.

2 solutions 1 solution 0 solutions parabola line intersections

Why Linear–Quadratic Systems Matter

These systems show up whenever a constant-rate process (linear) interacts with a changing-rate process (quadratic). Real-world examples include:

Projectile Motion
When does a launched ball reach a certain height? The height is quadratic; the target height is a horizontal line.
Revenue vs. Cost
A company's revenue might be quadratic (e.g., based on demand) and its costs linear. The intersection points are the break-even points.
Geometry
Finding where a line intersects a circle (e.g., \(x^2 + y^2 = r^2\)) is exactly this type of system.

The Substitution Method

The most reliable algebraic technique is substitution: replace the variable isolated in the linear equation into the quadratic equation, then solve the resulting single-variable quadratic.

When the Quadratic Is a Circle

The standard-form equation of a circle centered at the origin is \(x^2 + y^2 = r^2\). This is still a quadratic in two variables, so the same substitution method applies.

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

Linear–Linear vs. Linear–Quadratic

Linear–Linear Linear–Quadratic
Equations Both degree 1 Degree 1 + Degree 2
Graph shapes Two straight lines Line + parabola/circle
Max solutions 1 (or 0, or ∞) 2 (or 1, or 0)
Solving step Always linear Produces a quadratic
Discriminant? Not needed Determines # of solutions

Step-by-Step Strategy

1
Isolate a variable in the linear equation (usually solve for \(y\)).
2
Substitute that expression into the quadratic equation so only one variable remains.
3
Simplify and collect terms into standard quadratic form: \(Ax^2 + Bx + C = 0\).
4
Solve the quadratic by factoring, completing the square, or using \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\).
5
Back-substitute each \(x\)-value into the linear equation to find the matching \(y\)-value.
6
Check every solution in both original equations.

Example 1: Two Intersections

Solve the system: \(y = x + 1\) and \(y = x^2 - 1\).

1
Substitute: \(x + 1 = x^2 - 1\)
2
Rearrange: \(0 = x^2 - x - 2\)
3
Factor: \(0 = (x - 2)(x + 1)\), so \(x = 2\) or \(x = -1\)
4
Back-substitute: \(x = 2 \Rightarrow y = 3\); \(x = -1 \Rightarrow y = 0\)

Example 1: Graph

x y 1 2 3 -1 -2 1 2 3 -1 (-1, 0) (2, 3) \(y = x^2 - 1\) \(y = x + 1\)

Example 2: One Intersection (Tangent)

Solve: \(y = 2x - 1\) and \(y = x^2\).

1
Substitute: \(2x - 1 = x^2\)
2
Rearrange: \(x^2 - 2x + 1 = 0\)
3
Factor: \((x-1)^2 = 0\), so \(x = 1\) (double root)
4
Back-substitute: \(y = 2(1) - 1 = 1\)

Example 3: Line Meets Circle

Solve: \(y = -3x\) and \(x^2 + y^2 = 3\).

1
Substitute \(y = -3x\): \(x^2 + (-3x)^2 = 3\)
2
Simplify: \(x^2 + 9x^2 = 3 \Rightarrow 10x^2 = 3\)
3
Solve: \(x^2 = \frac{3}{10}\), so \(x = \pm\sqrt{\frac{3}{10}}\)
4
Back-substitute: \(y = -3\sqrt{\frac{3}{10}}\) or \(y = 3\sqrt{\frac{3}{10}}\)

Checking for Solutions

Let's check a system using the discriminant: \(y = x + 5\) and \(y = x^2\).

1
Substitute: \(x + 5 = x^2 \Rightarrow x^2 - x - 5 = 0\)
2
Discriminant: \(\Delta = (-1)^2 - 4(1)(-5) = 1 + 20 = 21\)

Now, let's look at a true no-solution case: \(y = x - 2\) and \(y = x^2 + 1\). Substituting gives \(x^2 - x + 3 = 0\). The discriminant is \(\Delta = (-1)^2 - 4(1)(3) = 1 - 12 = -11\).

Common Mistakes to Avoid

Forgetting to back-substitute
Finding \(x\)-values but not computing the matching \(y\)-values. Solutions are points \((x, y)\), not just \(x\)-values.
Sign errors when rearranging
Moving terms across the equals sign and losing a negative. Double-check every sign when forming \(Ax^2 + Bx + C = 0\).
Dropping a solution
A quadratic can have two roots. Do not stop after finding the first \(x\): always check for the \(\pm\) from the quadratic formula or the second factor.

Summary of Cases

Discriminant \(\Delta\) # of Solutions Geometric Meaning
\(\Delta > 0\) 2 Line crosses curve twice
\(\Delta = 0\) 1 Line is tangent to curve
\(\Delta < 0\) 0 Line misses curve entirely

Question 1 of 5

Solve: \(y = x + 2\) and \(y = x^2\). Which points are solutions?

Question 2 of 5

How many solutions does the system \(y = 3x\) and \(y = x^2 + 4\) have?

Question 3 of 5

For the system \(y = 4x - 4\) and \(y = x^2\), the discriminant of the resulting quadratic is \(\Delta = 0\). What does this tell you?

Question 4 of 5

Find the points of intersection of \(y = -3x\) and \(x^2 + y^2 = 3\).

Question 5 of 5

What is the first step in solving the system \(2x + y = 5\) and \(y = x^2 - 3x + 1\) algebraically?