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# Linear and quadratic systems | Lesson

A guide to linear and quadratic systems on the digital SAT

## What are linear and quadratic systems?

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'll be 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:
• Solving systems of linear equations
• Graphs of linear systems and inequalities
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 are graphed in the xy-plane. The line is represented by y=x, and the parabola is represented by y=x^2-6. The line and the parabola intersect at the points (-2,-2) and (3,3).
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.
A parabola and three horizontal lines are graphed in the xy-plane. The horizontal line that intersects the parabola twice is labeled "two solutions". The horizontal line that intersects the parabola once is labeled "one solution". The horizontal line that does not intersect the parabola is labeled "no solution".
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
A line and a parabola are graphed in the xy-plane. The line has a negative slope and passes through the points (0, 4) and (4, 0). The parabola opens downward, has a vertex located at (2, 4), and has x-intercepts located at (0, 0) and (4, 0). The line and the parabola share the points (1, 3) and (4, 0).
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 line and a parabola are graphed in the xy-plane. The line has a positive slope and passes through the points (2,3) and (8,9). The parabola opens downward and passes through the points (2,3), (6,11), and (8,9).
A system of equations is graphed in the x, y-plane above. Which of the following are solutions to the system?
I. left parenthesis, 2, comma, 3, right parenthesis
I, I. left parenthesis, 6, comma, 11, right parenthesis
I, I, I. left parenthesis, 8, comma, 9, right parenthesis

Practice: determine the number of solutions from a graph
A line and a parabola are graphed in the xy-plane. The line has a negative slope and passes through the points (0,5) and (2,2). The parabola opens upward and passes through the points (0,6), (2,2), (4,0), (5, -0.25), and (6,0).
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?
\begin{aligned} y &= 4x-2 \\\\ y &= x^2-4x+5 \end{aligned}