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### Course: Digital SAT Math>Unit 8

Lesson 8: Linear and quadratic systems: medium

# 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.
$\begin{array}{rlr}y& =x+1& \text{linear equation}\\ \\ y& ={x}^{2}-1& \text{quadratic equation}\end{array}$
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
• Solving quadratic equations
• Quadratic graphs
You can learn anything. Let's do this!

## How are linear and quadratic systems represented graphically?

### Quadratic systems: a line and a parabola

Khan Academy video wrapper
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 $xy$-plane. Each intersection of the line and the parabola represents a solution to the system.
For example, the system graphed below has two solutions: $\left(-2,-2\right)$ and $\left(3,3\right)$.
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:
$\begin{array}{rl}y& =-{x}^{2}+4\\ \\ y& =c\end{array}$
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

Khan Academy video wrapper
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:
$\begin{array}{rl}y& =x+1\\ \\ y& ={x}^{2}-1\end{array}$
In a system of equations, both equations are simultaneously true. In other words, since the first equation tells us that $y$ is equal to $x+1$, the $y$ in the second equation is also equal to $x+1$. Therefore, we can plug in $x+1$ as a substitute for $y$ in the second equation:
$\begin{array}{rl}y& ={x}^{2}-1\\ \\ x+1& ={x}^{2}-1\end{array}$
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:
$\begin{array}{rl}y& =x\\ \\ y& ={x}^{2}-6\end{array}$
What are the solutions to the system above?

### Try it!

try: follow the steps for substitution
$\begin{array}{rl}x+y& =5\\ \\ y& ={x}^{2}+x+5\end{array}$
In the system of equations above, isolating $y$ in the first equation gives us $y=5-x$. 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 ${x}^{2}+2x=0$ for $x$. The values of $x$ are $0$ and
.
Finally, since $y=5-x$, we can substitute the values of $x$ into the equation to calculate $y$. The values of $y$ are $5$ and
.

## Your turn!

Practice: identify solutions from a graph
A system of equations is graphed in the $xy$-plane above. Which of the following are solutions to the system?
$\mathrm{I}$. $\left(2,3\right)$
$\mathrm{II}$. $\left(6,11\right)$
$\mathrm{III}$. $\left(8,9\right)$
Choose 1 answer:

Practice: determine the number of solutions from a graph
A system of equations is graphed in the $xy$-plane above. How many solutions does the system have?
Choose 1 answer:

Practice: solve a linear and quadratic system
$\begin{array}{rl}3x-y& =3\\ \\ y& ={x}^{2}+6x-7\end{array}$
Which of the following is a solution to the system of equations above?
Choose 1 answer:

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

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