Linear-Quadratic Systems: How to Solve

A linear-quadratic system is a line and a parabola, like $y = x^2 - 1$ and $y = x + 1$ . Its solutions are the points where the two graphs cross. To solve it, set the two expressions for $y$ equal, solve the quadratic, then find each $y$ . A system can have two, one or no solutions.

Before this lesson: You should know systems of equations, the substitution method and solving quadratic equations by factoring.

What Is a Linear-Quadratic System?

A system is two or more equations that share the same variables. In this kind of system, one equation is linear and one is quadratic. The graph of a quadratic function is a parabola, and the linear graph is a line.

A solution is a point that makes both equations true. On the graph, it is a point where the line and the parabola meet.

The parabola y = x² − 1 in blue and the line y = x + 1 in green on a grid from −3 to 4 across and −2 to 5 up. They cross at two points, marked in orange: (−1, 0) and (2, 3). These two points are the solutions of the system. — The line crosses the parabola at (−1, 0) and (2, 3), so the system has two solutions.

A line can cross a parabola in two places, touch it at one point, or miss it. So the system has two, one or no real solutions.

Number of solutions	What the graph shows
Two	The line crosses the parabola twice
One	The line just touches the parabola
None	The line misses the parabola

How to Solve a Linear-Quadratic System

Solve the linear equation for $y$ , if it isn’t already.
Set the two expressions for $y$ equal. This is substitution.
Move every term to one side, so the equation equals 0.
Solve the quadratic by factoring or the quadratic formula.
Put each $x$ into the linear equation to find $y$ .
Write each solution as a point.

Worked Examples

Example 1: Two solutions

Solve the system $y = x^2 - 1$ and $y = x + 1$ .

\begin{aligned} x^2 - 1 &= x + 1 \\ x^2 - x - 2 &= 0 \\ (x - 2)(x + 1) &= 0 \\ x = 2 \quad &\text{or} \quad x = -1 \end{aligned}

Now use $y = x + 1$ . When $x = 2$ , $y = 3$ . When $x = -1$ , $y = 0$ . The solutions are $(2, 3)$ and $(-1, 0)$ , the two points on the graph above.

Example 2: The x terms cancel

Solve $y = x^2 + 2x - 3$ and $y = 2x + 1$ .

\begin{aligned} x^2 + 2x - 3 &= 2x + 1 \\ x^2 - 4 &= 0 \\ x &= \pm 2 \end{aligned}

When $x = 2$ , $y = 2(2) + 1 = 5$ . When $x = -2$ , $y = 2(-2) + 1 = -3$ . The solutions are $(2, 5)$ and $(-2, -3)$ .

Example 3: One solution

Solve $y = x^2 + 3$ and $y = 2x + 2$ .

\begin{aligned} x^2 + 3 &= 2x + 2 \\ x^2 - 2x + 1 &= 0 \\ (x - 1)^2 &= 0 \\ x &= 1 \end{aligned}

When $x = 1$ , $y = 2(1) + 2 = 4$ . The only solution is $(1, 4)$ . The line just touches the parabola there.

Example 4: No solution

Solve $y = x^2$ and $y = x - 2$ .

Set them equal: $x^2 = x - 2$ , so $x^2 - x + 2 = 0$ . This doesn’t factor, so check the discriminant.

b^2 - 4ac = (-1)^2 - 4(1)(2) = 1 - 8 = -7

The discriminant is negative, so there are no real solutions. The line never meets the parabola.

Example 5: Solve the line for y first

Solve $y = x^2 - 4x + 5$ and $x + y = 5$ .

First, solve the line for $y$ : $y = 5 - x$ .

\begin{aligned} x^2 - 4x + 5 &= 5 - x \\ x^2 - 3x &= 0 \\ x(x - 3) &= 0 \\ x = 0 \quad &\text{or} \quad x = 3 \end{aligned}

When $x = 0$ , $y = 5$ . When $x = 3$ , $y = 2$ . The solutions are $(0, 5)$ and $(3, 2)$ .

Example 6: Word problem

A ball is kicked up. Its height is $h = -5t^2 + 20t$ meters after $t$ seconds. A drone starts 10 meters up and rises with height $h = 5t + 10$ . When are they at the same height?

\begin{aligned} -5t^2 + 20t &= 5t + 10 \\ -5t^2 + 15t - 10 &= 0 \\ t^2 - 3t + 2 &= 0 \\ (t - 1)(t - 2) &= 0 \end{aligned}

So $t = 1$ or $t = 2$ . At 1 second, both are at $5(1) + 10 = 15$ meters. At 2 seconds, both are at $5(2) + 10 = 20$ meters.

Common Mistakes

Mistake	Example of the error	Fix
Stopping at x	Answer: $x = 2$ or $x = -1$	Find $y$ for each $x$ . Answers are points: $(2, 3)$ , $(-1, 0)$ .
Sign errors when moving terms	$x^2 - 1 = x + 1$ becomes $x^2 + x - 2 = 0$	Subtract $x$ and 1 from both sides: $x^2 - x - 2 = 0$ .
Not solving the line for y	Setting $x^2 - 4x + 5$ equal to $x + y$	Rewrite $x + y = 5$ as $y = 5 - x$ first.
Expecting two answers every time	Giving up when $x^2 - x + 2 = 0$ won’t factor	Check the discriminant. The system may have one solution or none.
Mixing up the y-values	Pairing $x = 2$ with the other point’s $y$	Put each $x$ into the line. Keep each $x$ with its own $y$ .

Solving by Graphing

You can also graph both equations and read off where they cross. This shows how many solutions there are at a glance. For exact answers, check each point in both equations, or solve with algebra.

If the quadratic part doesn’t factor, use the quadratic formula. Then put each $x$ into the line.

Practice Problems

Solve $y = x^2$ and $y = 4$ .
Solve $y = x^2$ and $y = x + 6$ .
Solve $y = x^2 - 3x$ and $y = x - 4$ .
Solve $y = x^2 + 1$ and $y = -x$ .
Solve $y = x^2 - 2x - 3$ and $y = x + 1$ .
Solve $y = 2x^2 - 5$ and $y = 3x$ .
Solve $y = x^2 + 4x$ and $2x - y = 1$ .
How many solutions does the system $y = x^2 + 5$ and $y = 2x + 1$ have?
A company’s revenue is $R = -x^2 + 14x$ and its cost is $C = 2x + 20$ , both in thousands of dollars, where $x$ is hundreds of items sold. Find the break-even points, where $R = C$ .

Show answers

$x^2 = 4$ , so $x = \pm 2$ . Solutions: $(2, 4)$ and $(-2, 4)$ .
$x^2 - x - 6 = 0$ , so $(x - 3)(x + 2) = 0$ . Solutions: $(3, 9)$ and $(-2, 4)$ .
$x^2 - 4x + 4 = 0$ , so $(x - 2)^2 = 0$ and $x = 2$ . One solution: $(2, -2)$ .
$x^2 + x + 1 = 0$ . The discriminant is $1 - 4 = -3$ , so there are no real solutions.
$x^2 - 3x - 4 = 0$ , so $(x - 4)(x + 1) = 0$ . Solutions: $(4, 5)$ and $(-1, 0)$ .
$2x^2 - 3x - 5 = 0$ , so $(2x - 5)(x + 1) = 0$ . Solutions: $\left(\frac{5}{2}, \frac{15}{2}\right)$ and $(-1, -3)$ .
Solve the line for $y$ : $y = 2x - 1$ . Then $x^2 + 2x + 1 = 0$ , so $x = -1$ . One solution: $(-1, -3)$ .
$x^2 - 2x + 4 = 0$ . The discriminant is $4 - 16 = -12$ , so there are no solutions.
$-x^2 + 14x = 2x + 20$ , so $x^2 - 12x + 20 = 0$ and $(x - 2)(x - 10) = 0$ . The company breaks even at 200 items, where revenue and cost are both 24 thousand dollars, and at 1,000 items, where both are 40 thousand dollars.

FAQ

How do you solve a linear-quadratic system?

Set the two expressions for $y$ equal and solve the quadratic. Then put each $x$ into the linear equation to find its $y$ .

How many solutions can a linear-quadratic system have?

It can have two, one or no real solutions. A line can cross a parabola twice, touch it once or miss it.

Can you solve a linear-quadratic system by graphing?

Yes. Graph the line and the parabola, and find where they meet. Check each point in both equations.

Summary

A linear-quadratic system is a line and a parabola.
Its solutions are the points where the graphs meet.
Set the $y$ expressions equal, solve the quadratic, then find each $y$ .
There can be two, one or no real solutions. The discriminant tells you which.
Always write each answer as a point.

Previous lesson: The Discriminant

Next lesson: Linear, Quadratic and Exponential Models

Back to the Algebra 1 Index