# Solving Nonlinear Systems with a Quadratic & a Linear Equation

Instructor: Laura Pennington

Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.

In this lesson, we'll learn how to solve non-linear systems with a quadratic and linear equation both algebraically and graphically. We'll also look at the different possible solutions for this type of system.

## Non-Linear Systems of Equations

Have you ever been to a carnival? I don't know about you, but I love carnivals! Imagine you go to a carnival, and there is a booth there that gives a prize if you can solve a number puzzle.

You really want that prize, so you decide to solve the puzzle: find two numbers such that the square of the first number plus the second number gives you 9, and the first number minus the second number gives you 3. You think about something you learned in math class and realize that you have all the information you need to solve the puzzle! Yay!

What you remembered from math class is that this number puzzle corresponds to a non-linear system of equations. A non-linear system of equations is a collection of two or more equations with the same variables and where at least one of the equations is not a linear equation. You quickly grab a piece of paper and a pencil and get to work!

To set up this system, let's let x equal the first number and y equal the second number. We know that the square of the first number, x 2, plus the second number, y, must equal 9. Let's set up our first equation.

x 2 + y = 9

We also know that the first number, x, minus the second number, y, must equal 3, as shown in our second equation.

x - y = 3

When we put these two equations together, we have the following system of equations.

x 2 + y = 9

x - y = 3

Here we have a non-linear system of equations, where one of the equations is a quadratic equation, or an equation with the highest exponent being 2, and the other is a linear equation, or an equation with the highest exponent being 1.

Now that we've set up the problem, you are well on your way to that prize! Now, let's just look at how to solve it!

## Solving a System Algebraically

When we have a non-linear system of equations with a quadratic and a linear equation, often substitution is the best way to go about solving it. To do this, we follow these steps.

1. Solve for one of the variables in the linear equation.
2. Replace the variable in the quadratic equation with the solution from step one.
3. You now have an equation in one variable. Solve for that variable.
4. Replace the variable in the linear equation with the solution from step three. Solve for the other variable. You will have to do this for each of the values found. Your answers are your pairs of solutions.

Let's go ahead and take our system through these steps so we can get you that prize!

The first step is to solve for one of the variables in the linear equation. Either variable will work, but since we will be plugging it into the quadratic equation, it would be easier to solve for y. Then, when we plug it in, we don't have to square an expression, so let's go ahead and solve for y.

We get y = x - 3. The next step is to plug this expression in for y in our quadratic equation.

We now have the quadratic equation x 2 + x - 3 = 9. The third step is to solve for x in this equation.

We get that x = -4 or x = 3. The last step is to plug these values into the linear equation in our system and solve for y.

We get that when x = -4, y = -7, and when x = 3, y = 0. Both of these pairs of numbers are solutions to the problem. Quick! Go tell the booth attendant you have solved the puzzle so you can get your well-deserved prize!

## Solving Graphically

We can also solve nonlinear systems of equations graphically. To do so, we follow these steps.

1. Graph both equations on the same graph.
2. Identify their intersection points. These intersection points are your solutions.

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