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### Course: Digital SAT Math>Unit 2

Lesson 8: Graphs of linear systems and inequalities: foundations

# Graphs of linear systems and inequalities | Lesson

A guide to graphs of linear systems and inequalities on the digital SAT

## What are graphs of linear systems and inequalities problems?

Graphs of linear systems and inequality problems deal with graphs of the following in the $xy$-plane:
In this lesson, we'll learn to:
1. Identify solutions to linear systems and inequalities
2. Match graphs with systems of linear equations and inequalities
This lesson builds upon an understanding of the following skills:
• Solving linear equations and linear inequalities
• Graphs of linear equations and functions
• Solving systems of linear equations
You can learn anything. Let's do this!

## How are systems of linear equations related to their graphs?

### Systems of equations with graphing

Systems of equations with graphingSee video transcript

### Intersection points as solutions

In a system of linear equations, each equation is represented by a line in the $xy$-plane. The intersection of these lines represents the solution to the system.
For example, the graphs of $y=x$ and $y=-x$ intersect at $\left(0,0\right)$, which is the solution to the system of equations.

### Try it!

Try: identify the solution of a system of equations by its graph
In the graph above, the line with the positive slope has a slope of $1$ and a $y$-intercept of
. The line with the negative slope has a slope of
and a $y$-intercept of $1$.
The two lines intersect at $\left($
,
), which is the solution $\left(x,y\right)$ to the following system of equations $y=x-2$ and $y=-2x+1$.

## How do I graph linear inequalities?

### Intro to graphing two-variable inequalities

Intro to graphing two-variable inequalitiesSee video transcript

### Representing solutions with shaded areas

When we learned to solve linear inequalities, we learned that multiple values can satisfy an inequality. How do we graph the infinite number of $\left(x,y\right)$ values that can satisfy a linear inequality?
We do it by shading the $xy$-plane! A linear inequality is defined by a line in the $xy$-plane. The line divides the plane into two halves, and which half of the plane we shade depends on the inequality sign.
It's the easier to determine which half of the plane to shade when the inequality is in slope-intercept form. When we have linear inequalities in slope-intercept form:
• If $y$ is greater than $mx+b$, shade above the line.
• If $y$ is less than $mx+b$, shade below the line.
But what about points on the line? In slope-intercept form, inequalities with...
• Greater than ($>$) or less than ($<$) signs do not include points on the line in the solution set. We use a dashed line to show that the points on the line are not included.
• Greater than or equal to ($\ge$) or less than or equal to ($\le$) signs do include points on the line in the solution set. We use a solid line to show that points on the line are included.
For example, for $y>x$, we draw $y=x$ as a dashed line and shade above the line:

### Try it!

Try: verify solutions to a linear inequality
The inequality $y>-\frac{1}{3}x-2$ is graphed in the $xy$-plane above.
The point $\left(1,1\right)$ is
a solution to the inequality.
The point $\left(-3,-4\right)$ is
a solution to the inequality.
The point $\left(0,-2\right)$ is on the line $y=-\frac{1}{3}x-2$. It
a solution to the inequality because the $>$ sign indicates that points on the line
part of the solution set.

## How do I graph systems of linear inequalities?

### Intro to graphing systems of inequalities

Intro to graphing systems of inequalitiesSee video transcript

### How do I identify the region representing a system of linear inequalities?

Two intersecting lines will always divide the $xy$-plane into four regions. Points in each of the four regions represent solutions to a different system of linear inequalities.
In the graph below, the equations of the lines defining the four regions are:
$\begin{array}{rl}y& =3x-2\\ \\ y& =-\frac{1}{3}x+\frac{4}{3}\end{array}$
Replacing the equal signs with inequality signs lets us specify one of the four regions.
When we have linear inequalities in slope-intercept form:
• If $y$ is greater than $mx+b$, shade above the line.
• If $y$ is less than $mx+b$, shade below the line.
For example, let's look at the following system:
$\begin{array}{rl}y& \le 3x-2\\ \\ y& \ge -\frac{1}{3}x+\frac{4}{3}\end{array}$
This means the solutions to the system of linear inequalities are represented by the region below the line $y=3x-2$ and above the line $y=\frac{1}{3}x+\frac{4}{3}$:

### Try it!

TRY: DETERMINE THE INEQUALITY SIGNS FROM A GRAPH
The the graph above represents a system of linear inequalities.
Because the region
the line $y=-\frac{1}{3}x+\frac{4}{3}$ is shaded, $y$ is
or equal to $-\frac{1}{3}x+\frac{4}{3}$.
Because the region
the line $y=3x-2$ is shaded, $y$ is
or equal to $3x-2$.

practice: find the intersection of two lines
The graph of a line in the $xy$-plane has a slope of $3$ and passes through the origin. The graph of a second line has a slope of $6$ and passes through the point $\left(-1,-9\right)$. If the two lines intersect at the point $\left(a,b\right)$, what is the value of $a$ ?

practice: match a linear inequality to its graph
If the shaded region in the graph above represents the solution set to an inequality, which of the following could be the inequality?

Practice: match a system of linear inequalities to its graph
$\begin{array}{rl}& x-y\ge 4\\ \\ & y\le -\frac{3}{4}x-\frac{1}{2}\end{array}$
In which of the following does the shaded region represent the solution set in the $xy$-plane to the system of inequalities above?

## Things to remember

When a system of linear equations is graphed, the solution appears where the lines intersect.
When we have linear inequalities in slope-intercept form:
• If $y$ is greater than $mx+b$, shade above the line.
• If $y$ is less than $mx+b$, shade below the line.
In slope-intercept form, inequalities with...
• Greater than ($>$) or less than ($<$) signs do not include points on the line in the solution set. We use a dashed line to show that the points on the line are not included.
• Greater than or equal to ($\ge$) or less than or equal to ($\le$) signs do include points on the line in the solution set. We use a solid line to show that points on the line are included.

## Want to join the conversation?

• is it just me or is this lesson hard to understand than the others
• to be honest with you......i am confused
• hey, they make a mistake on the last question.4 times -1 suppose equal to -4, not 4
• There is, in fact, a typo in the second line, but I'm pretty sure that doesn't lead to a mistake, as you're claiming (and I believe that the typo added extra confusion in your mind). They should either have just flipped the inequality sign and multiplied one side by (-1) or have left the flipping of the inequality from "greater than or equal to" to "less than or equal to" to be done in the following line. However, they have omitted the multiplication of negative one on the right-hand side from line three.
• why i can't understand this lesson , i have watched the videos 3 times
• Tip: if you're solving for y in an inequality and move it to the other side, make sure to pay attention that the inequality sign has "switched" for the shaded region.
• The second line has a slope of 6 and passes through the point (−1, −9). This means the line also passes through the point (-1 + 1, -9 + 6), or (0, - 3). Its equation is y = 6x - 3

Please someone explain this to me. Why does a line that has a slope of 6 and passes through (−1, −9) also pass through point 1 and 6?
• We're given a slope and a point on the line (-1,-9)
Now, we are aware that our slope is ∆y/∆x. That means, our slope is 6/1 which shows that for every 1 we move in x, we move up 6 in y. That also means if we change x by 1,we change y by 6.
Hence, (-1+1,-9+6)
The original points were (-1,-9)
We want to change x by 1 and y by 6 to get our y-intercept. (x+1,y+6)

I hope this helps
• I am struggling with most of these questions, I am not really good when it comes to graphs. Any suggestions pls
• practice is the only way
• What?
• mistake in the last practice, x−y≥4, after isolating y in will be y=<x-4