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# Linear and quadratic systems | Lesson

## What are linear and quadratic systems, and how frequently do they appear on the test?

Linear and quadratic systems are systems of equations with one linear equation and one quadratic equation.
\begin{aligned} y &=x+1&\purpleD{\text{linear equation}} \\ \\ y &= x^2-1 &\purpleD{\text{quadratic equation}} \end{aligned}
On the test, you're expected to find the solution(s) to systems like the one shown above either algebraically or graphically.
In this lesson, we'll:
1. Explore the graphs of linear and quadratic systems
2. Determine the number of solutions for linear and quadratic systems
3. Learn how to solve linear and quadratic systems algebraically
This lesson builds upon the following skills:
You can learn anything. Let's do this!

## How are linear and quadratic systems represented graphically?

### Quadratic systems: a line and a parabola

Quadratic systems: a line and a parabolaSee video transcript

### Identifying solutions to linear and quadratic systems from graphs

A linear and quadratic system can be represented by a line and a parabola in the x, y-plane. Each intersection of the line and the parabola represents a solution to the system.
For example, the system graphed below has two solutions: left parenthesis, minus, 2, comma, minus, 2, right parenthesis and left parenthesis, 3, comma, 3, right parenthesis.
A line and a parabola can intersect zero, one, or two times, which means a linear and quadratic system can have zero, one, or two solutions.
If a graph of the system is not provided, then the ability to quickly draw graphs based on equations is essential.

Example:
\begin{aligned} y &= -x^2+4 \\\\ y &= c \end{aligned}
If the system above has exactly one solution, what is the value of c ?

### Try it!

TRY: determine the number solutions graphically
The graph of a system of equations is shown above. Because the line and the parabola intersect
, the system has
.

## How do I solve linear and quadratic systems algebraically?

### Quadratic system with no solutions

Quadratic system with no solutionsSee video transcript

### Solving linear and quadratic systems algebraically

Our goal when solving a system of equations is to reduce two equations with two variables down to a single equation with one variable. Since each equation in the system has two variables, one way to reduce the number of variables in an equation is to substitute an expression for a variable.
Consider the following example:
\begin{aligned} y &=x+1 \\ \\ y &= x^2-1 \end{aligned}
In a system of equations, both equations are simultaneously true. In other words, since the first equation tells us that y is equal to x, plus, 1, the y in the second equation is also equal to x, plus, 1. Therefore, we can plug in x, plus, 1 as a substitute for y in the second equation:
\begin{aligned} y&=x^2-1\\\\ x+1&=x^2-1 \end{aligned}
From here, we can solve the quadratic equation for x, which gives us the x-values of the solutions to the system. Then, we can use the x-values and either equation in the system to calculate the y-values.
To solve a linear and quadratic system:
1. Isolate one of the two variables in one of the equations. In most cases, isolating y is easier.
2. Substitute the expression that is equal to the isolated variable from Step 1 into the other equation. This should result in a quadratic equation with only one variable.
3. Solve the resulting quadratic equation to find the x-value(s) of the solution(s).
4. Substitute the x-value(s) into either equation to calculate the corresponding y-values.

Example:
\begin{aligned} y &= x\\\\ y &= x^2-6 \end{aligned}
What are the solutions to the system above?

### Try it!

try: follow the steps for substitution
\begin{aligned} x+y&=5 \\\\ y&=x^2+x+5 \end{aligned}
In the system of equations above, isolating y in the first equation gives us y, equals, 5, minus, x. This means we can replace the y in the second equation with
to reduce the equation to a quadratic equation with a single variable.
After rearranging some terms, we can solve the equation x, squared, plus, 2, x, equals, 0 for x. The values of x are 0 and
.
Finally, since y, equals, 5, minus, x, we can substitute the values of x into the equation to calculate y. The values of y are 5 and
.

Practice: identify solutions from a graph
A system of equations is graphed in the x, y-plane above. Which of the following are solutions to the system?
start text, I, end text. left parenthesis, 2, comma, 3, right parenthesis
start text, I, I, end text. left parenthesis, 6, comma, 11, right parenthesis
start text, I, I, I, end text. left parenthesis, 8, comma, 9, right parenthesis

Practice: determine the number of solutions from a graph
A system of equations is graphed in the x, y-plane above. How many solutions does the system have?

Practice: solve a linear and quadratic system
\begin{aligned} 3x-y &= 3 \\\\ y &= x^2+6x-7 \end{aligned}
Which of the following is a solution to the system of equations above?

Practice: solve a linear and quadratic system
\begin{aligned} y &= 4x-2 \\\\ y &= x^2-4x+5 \end{aligned}
If a and b are x-values of the solutions to the system above, what is the value of a, plus, b ?

## Want to join the conversation?

• Can you put more links to find more questions like this? • How to find that a system of quadratic equations have zero, one or two solution. Moreover, can a system of quadratic equation has infinite solutions ? • To solve a system of quadratic equations, you still want to cancel one of the variables so that you have a one-variable system. Your strategies for solving the system are the same as for linear systems: either solving one equation for a variable and then substituting that into the other equation or multiplying an equation by a constant so that you can eliminate one variable if you add them together.
When finding the number of solutions, try thinking in terms of the graph. If the two lines aren't touching at all, then you have zero solutions. If they have one or two intersection points, then you have that many solutions. And yes, a system of quadratic equations can have infinitely many solutions if the equations refer to the same line.

As an example, let's take solving for the solutions of the system: (1) y = 3x^2 and (2) y = -(x-3)^2 + 7
Both of these equations already come nicely solved for y, so the easiest strategy to use is substitution and then to combine like terms until you get a form you're comfortable with. From there you can solve the quadratic equation like you normally would, getting you the x-coordinate(s) of the solution(s):
3x^2 = -(x-3)^2 + 7
3x^2 + (x-3)^2 - 7 = 0
3x^2 + x^2 - 6x + 9 - 7 = 0
4x^2 - 6x + 2 = 0
2x^2 - 3x + 1 = 0
(2x - 1)(x - 1) = 0
x = 1/2, 1
You can't really know the number of solutions beforehand unless you're good at visualizing the graphs and intersection points, or you solve the system to the point where you can calculate the determinant to see how many solutions you'll end up getting from the quadratic formula.
• Suppose we have a system of linear and quadratic equations with rational coefficients and we want to find out whether this system has a real solution or not. How would we figure that out?
(1 vote) • First, we can try visualizing what we're trying to find. The solutions of a system of equations are the intersection points between those equations plotted on a graph. A line and a parabola couldn't ever have infinitely many solutions, because they're different shapes. A line could intersect the parabola at one point, at two, or not at all. Not intersecting at all would mean that the system has no real solutions.
To find out if a nonlinear system has a real solution or not, you just have to try solving it. Let's try the system of y = x^2 and y = x - 1 (which don't intersect) as an example:
y = x^2 and y = x - 1
x^2 = x - 1
If we try completing the square:
x^2 - x + 1/4 = -1 + 1/4
(x - 1/2)^2 = -3/4
x - 1/2 = sqrt(-3/4)
As you can see, when we went to solve it we end up trying to take the square root of a negative number, which provides no real solutions. Whenever you get to this point, whether you use completing the square or the quadratic formula or whatever, you can say that the system has no real solutions.  