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Graphing Systems of Linear Inequalities

Determine whether an ordered pair is a solution of a system of linear inequalities

The definition of a system of linear inequalities is very similar to the definition of a system of linear equations.

A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here.

\[\{\begin{array}{l}x+4y\ge 10 \\ 3x-2y<12\end{array}\]

To solve a system of linear inequalities, we will find values of the variables that are solutions to both inequalities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will find the region on the plane that contains all ordered pairs $(x,y)$ that make both inequalities true.

To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system.

Example

Try it.

Determine whether the ordered pair is a solution to the system $\{\begin{array}{l}x+4y\ge 10 \\ 3x-2y<12\end{array}.$

ⓐ $(-2,4)$ ⓑ $(3,1)$

Solution

ⓐ Is the ordered pair $(-2,4)$ a solution?

The ordered pair $(-2,4)$ made both inequalities true. Therefore $(-2,4)$ is a solution to this system.

ⓑ Is the ordered pair $(3,1)$ a solution?

The ordered pair $(3,1)$ made one inequality true, but the other one false. Therefore $(3,1)$ is not a solution to this system.

Solve a System of Linear Inequalities by Graphing

The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To find this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph.

How to Solve a System of Linear Inequalities by Graphing

Try it.

Solve the system by graphing: \(\{\begin{array}{l}y\ge 2x-1 \\ y

Solution

Example

Try it.

Solve the system by graphing: $\{\begin{array}{l}x-y>3 \\ y<-\frac{1}{5}x+4\end{array}.$

Solution

	$\{\begin{array}{l}x-y>3 \\ y<-\frac{1}{5}x+4\end{array}$
Graph x − y > 3, by graphing x − y = 3 and testing a point. The intercepts are x = 3 and y = −3 and the boundary line will be dashed. Test (0, 0) which makes the inequality false so shade (red) the side that does not contain (0, 0).

Graph $y<-\frac{1}{5}x+4$ by graphing $y=-\frac{1}{5}x+4$
using the slope $m=-\frac{1}{5}$ and y-intercept b = 4.
The boundary line will be dashed

Test (0, 0) which makes the inequality true, so
shade (blue) the side that contains (0, 0).

Choose a test point in the solution and verify that it is a solution to both inequalties.

The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Solve Applications of Systems of Inequalities

The first thing we’ll need to do to solve applications of systems of inequalities is to translate each condition into an inequality. Then we graph the system, as we did above, to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so we add inequalities to the system as additional requirements.

When we use variables other than x and y to define an unknown quantity, we must change the names of the axes of the graph as well.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Solutions of a System of Linear Inequalities: Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the x, y coordinate system that includes all the points whose ordered pairs make the inequalities true.
How to solve a system of linear inequalities by graphing.
1. Graph the first inequality.
  Graph the boundary line.
  Shade in the side of the boundary line where the inequality is true.
2. On the same grid, graph the second inequality.
  Graph the boundary line.
  Shade in the side of that boundary line where the inequality is true.
3. The solution is the region where the shading overlaps.
4. Check by choosing a test point.

Chapter Practice Test

In the following exercises, solve the following systems by graphing.

In the following exercises, solve each system of equations. Use either substitution or elimination.

Solve the system of equations using a matrix.

Solve using Cramer’s rule.

In the following exercises, translate to a system of equations and solve.

Determine Whether an Ordered Pair is a Solution of a System of Linear Inequalities

The definition of a system of linear inequalities is very similar to the definition of a system of linear equations.

A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown below.

\[\{\begin{array}{l}x+4y\ge 10 \\ 3x-2y<12\end{array}\]

Example

Try it.

Determine whether the ordered pair is a solution to the system. $\{\begin{array}{l}x+4y\ge 10 \\ 3x-2y<12\end{array}$

ⓐ (−2, 4) ⓑ (3,1)

Solution

ⓐ Is the ordered pair (−2, 4) a solution?

The ordered pair (−2, 4) made both inequalities true. Therefore (−2, 4) is a solution to this system.

ⓑ Is the ordered pair (3,1) a solution?

The ordered pair (3,1) made one inequality true, but the other one false. Therefore (3,1) is not a solution to this system.

Solve a System of Linear Inequalities by Graphing

How to Solve a System of Linear inequalities

Try it.

Solve the system by graphing.

\(\{\begin{array}{l}y\ge 2x-1 \\ y

Solution

Example

Try it.

Solve the system by graphing. $\{\begin{array}{l}x-y>3 \\ y<-\frac{1}{5}x+4\end{array}$

Solution

Graph x − y > 3, by graphing x − y = 3 and
testing a point.

The intercepts are x = 3 and y = −3 and the boundary
line will be dashed.

Test (0, 0). It makes the inequality false. So,
shade the side that does not contain (0, 0) red.

Graph $y<-\frac{1}{5}x+4$ by graphing $y=-\frac{1}{5}x+4$
using the slope $m=-\frac{1}{5}$ and y−intercept
b = 4. The boundary line will be dashed.

Test (0, 0). It makes the inequality true, so shade the side that contains (0, 0) blue.

Choose a test point in the solution and verify that it is a solution to both inequalities.

The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice which is the darker-shaded region.

Systems of linear inequalities where the boundary lines are parallel might have no solution. We’ll see this in .

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Solve Applications of Systems of Inequalities

The first thing we’ll need to do to solve applications of systems of inequalities is to translate each condition into an inequality. Then we graph the system as we did above to see the region that contains the solutions. Many situations will be realistic only if both variables are positive, so their graphs will only show Quadrant I.

Example

Try it.

Christy sells her photographs at a booth at a street fair. At the start of the day, she wants to have at least 25 photos to display at her booth. Each small photo she displays costs her $4 and each large photo costs her $10. She doesn’t want to spend more than $200 on photos to display.

ⓐ Write a system of inequalities to model this situation.

ⓑ Graph the system.

ⓓ Could she display 3 large and 22 small photos?

Solution

ⓐ Let $x=$ the number of small photos.
$\ y=$ the number of large photos
To find the system of inequalities, translate the information.
$\begin{array}{lllll} & & & & \text{She wants to have at least 25 photos.} \\ & & & & \text{The number of small plus the number of large should be at least 25.} \\ & & & & x+y\ge 25 \\ & & & & \text{\$4 for each small and \$10 for each large must be no more than \$200} \\ & & & & 4x+10y\le 200\end{array}$
We have our system of inequalities. $\{\begin{array}{l}x+y\ge 25 \\ 4x+10y\le 200\end{array}$

ⓑ

To graph $x+y\ge 25$, graph x + y = 25 as a solid line.
Choose (0, 0) as a test point. Since it does not make the inequality
true, shade the side that does not include the point (0, 0) red.

To graph $4x+10y\le 200$, graph 4x + 10y = 200 as a solid line.
Choose (0, 0) as a test point. Since it does not make the inequality
true, shade the side that includes the point (0, 0) blue.

The solution of the system is the region of the graph that is double shaded and so is shaded darker.

ⓒ To determine if 10 small and 20 large photos would work, we see if the point (10, 20) is in the solution region. It is not. Christy would not display 10 small and 20 large photos.
ⓓ To determine if 20 small and 10 large photos would work, we see if the point (20, 10) is in the solution region. It is. Christy could choose to display 20 small and 10 large photos.

Notice that we could also test the possible solutions by substituting the values into each inequality.

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Key Concepts

To Solve a System of Linear Inequalities by Graphing
1. Graph the first inequality.
  - Graph the boundary line.
  - Shade in the side of the boundary line where the inequality is true.
2. On the same grid, graph the second inequality.
  - Graph the boundary line.
  - Shade in the side of that boundary line where the inequality is true.
3. The solution is the region where the shading overlaps.
4. Check by choosing a test point.

Chapter 5 Review Exercises

Determine Whether an Ordered Pair is a Solution of a System of Equations.

In the following exercises, determine if the following points are solutions to the given system of equations.

Try it.

$\{\begin{array}{l}x+3y=-9 \\ 2x-4y=12\end{array}$

ⓐ $(-3,-2)$ ⓑ $(0,-3)$

Solution

ⓐ no ⓑ yes

Try it.

$\{\begin{array}{l}x+y=8 \\ y=x-4\end{array}$

ⓐ $(6,2)$ ⓑ $(9,-1)$

Solve a System of Linear Equations by Graphing

In the following exercises, solve the following systems of equations by graphing.

Try it.

$\{\begin{array}{l}3x+y=6 \\ x+3y=-6\end{array}$

Solution

$(3,-3)$

Try it.

$\{\begin{array}{l}y=x-2 \\ y=-2x-2\end{array}$

Try it.

$\{\begin{array}{l}2x-y=6 \\ y=4\end{array}$

Solution

$(5,4)$

Try it.

$\{\begin{array}{l}x+4y=-1 \\ x=3\end{array}$

Try it.

$\{\begin{array}{l}2x-y=5 \\ 4x-2y=10\end{array}$

Solution

coincident lines

Try it.

$\{\begin{array}{l}-x+2y=4 \\ y=\frac{1}{2}x-3\end{array}$

Determine the Number of Solutions of a Linear System

In the following exercises, without graphing determine the number of solutions and then classify the system of equations.

Try it.

$\{\begin{array}{l}y=\frac{2}{5}x+2 \\ -2x+5y=10\end{array}$

Solution

infinitely many solutions, consistent system, dependent equations

Try it.

$\{\begin{array}{l}3x+2y=6 \\ y=-3x+4\end{array}$

Try it.

$\{\begin{array}{l}5x-4y=0 \\ y=\frac{5}{4}x-5\end{array}$

Solution

no solutions, inconsistent system, independent equations

Try it.

$\{\begin{array}{l}y=-\frac{3}{4}x+1 \\ 6x+8y=8\end{array}$

Solve Applications of Systems of Equations by Graphing

Try it.

LaVelle is making a pitcher of caffe mocha. For each ounce of chocolate syrup, she uses five ounces of coffee. How many ounces of chocolate syrup and how many ounces of coffee does she need to make 48 ounces of caffe mocha?

Solution

LaVelle needs 8 ounces of chocolate syrup and 40 ounces of coffee.

Try it.

Eli is making a party mix that contains pretzels and chex. For each cup of pretzels, he uses three cups of chex. How many cups of pretzels and how many cups of chex does he need to make 12 cups of party mix?

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve the inequality $2a<5a+12.$
If you missed this problem, review .

Bonisa impendulo

\(-4
Determine whether the ordered pair $(3,\frac{1}{2})$ is a solution to the system $y>2x+3.$
If you missed this problem, review .

Bonisa impendulo

no
Determine whether the ordered pair is a solution to the system $\{\begin{array}{l}x+4y\ge 10 \\ 3x-2y<12\end{array}.$
ⓐ $(-2,4)$ ⓑ $(3,1)$

Bonisa impendulo

ⓐ Is the ordered pair $(-2,4)$ a solution?
The ordered pair $(-2,4)$ made both inequalities true. Therefore $(-2,4)$ is a solution to this system.
ⓑ Is the ordered pair $(3,1)$ a solution?
The ordered pair $(3,1)$ made one inequality true, but the other one false. Therefore $(3,1)$ is not a solution to this system.
Determine whether the ordered pair is a solution to the system: $\{\begin{array}{l}x-5y>10 \\ 2x+3y>-2\end{array}.$
ⓐ $(3,-1)$ ⓑ $(6,-3)$

Bonisa impendulo

ⓐ no ⓑ yes
Determine whether the ordered pair is a solution to the system: $\{\begin{array}{l}y>4x-2 \\ 4x-y<20\end{array}.$
ⓐ $(-2,1)$ ⓑ $(4,-1)$

Bonisa impendulo

ⓐ yes ⓑ no
Solve the system by graphing: \(\{\begin{array}{l}y\ge 2x-1 \\ y
Solve the system by graphing: $\{\begin{array}{l}y<3x+2 \\ y>\text{-}x-1\end{array}.$

Bonisa impendulo

The solution is the grey region.
Solve the system by graphing: $\{\begin{array}{l}y<-\frac{1}{2}x+3 \\ y<3x-4\end{array}.$

Bonisa impendulo

The solution is the grey region.

Solve the system by graphing: $\{\begin{array}{l}x-y>3 \\ y<-\frac{1}{5}x+4\end{array}.$

Bonisa impendulo

	$\{\begin{array}{l}x-y>3 \\ y<-\frac{1}{5}x+4\end{array}$
Graph x − y > 3, by graphing x − y = 3 and testing a point. The intercepts are x = 3 and y = −3 and the boundary line will be dashed. Test (0, 0) which makes the inequality false so shade (red) the side that does not contain (0, 0).

The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice—which appears as the darkest shaded region.

Solve the system by graphing: $\{\begin{array}{l}x+y\le 2 \\ y\ge \frac{2}{3}x-1\end{array}.$

Bonisa impendulo

The solution is the grey region.
Solve the system by graphing: $\{\begin{array}{l}3x-2y\le 6 \\ y>\text{-}\frac{1}{4}x+5\end{array}.$

Bonisa impendulo

The solution is the grey region.

Solve the system by graphing: $\{\begin{array}{l}x-2y<5 \\ y>-4\end{array}.$

Bonisa impendulo

	$\ \{\begin{array}{l}x-2y<5 \\ y>-4\end{array}$
Graph $x-2y<5,$ by graphing $x-2y=5$ and testing a point. The intercepts are x = 5 and y = −2.5 and the boundary line will be dashed. Test (0, 0) which makes the inequality true, so shade (red) the side that contains (0, 0).

Graph $y>-4,$ by graphing $y=-4$ and
recognizing that it is a horizontal line
through $y=-4.$ The boundary line will
be dashed.

Test (0, 0) which makes the inequality
true so shade (blue) the side that contains (0, 0).

The point $(0,0)$ is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed.

The solution is the area shaded twice—which appears as the darkest shaded region.

Solve the system by graphing: $\{\begin{array}{l}y\ge 3x-2 \\ y<-1\end{array}.$

Bonisa impendulo

The solution is the grey region.
Solve the system by graphing: $\{\begin{array}{l}x>-4 \\ x-2y\ge -4\end{array}.$

Bonisa impendulo

The solution is the grey region.

Solve the system by graphing: $\{\begin{array}{l}4x+3y\ge 12 \\ y<-\frac{4}{3}x+1\end{array}.$

Bonisa impendulo

	$\{\begin{array}{l}4x+3y\ge 12 \\ y<-\frac{4}{3}x+1\end{array}$
Graph $4x+3y\ge 12,$ by graphing $4x+3y=12$ and testing a point. The intercepts are x = 3 and y = 4 and the boundary line will be solid. Test (0, 0) which makes the inequality false, so shade (red) the side that does not contain (0, 0).
Graph $y<-\frac{4}{3}x+1$ by graphing $y=-\frac{4}{3}x+1$ using the slope $m=-\frac{4}{3}$ and y-intercept b = 1. The boundary line will be dashed. Test (0, 0) which makes the inequality true, so shade (blue) the side that contains (0, 0).

There is no point in both shaded regions, so the system has no solution.

Solve the system by graphing: $\{\begin{array}{l}3x-2y\ge 12 \\ y\ge \frac{3}{2}x+1\end{array}.$

Bonisa impendulo

No solution.
Solve the system by graphing: $\{\begin{array}{l}x+3y>8 \\ y<-\frac{1}{3}x-2\end{array}.$

Bonisa impendulo

No solution.

Solve the system by graphing: $\{\begin{array}{l}y>\frac{1}{2}x-4 \\ x-2y<-4\end{array}.$

Bonisa impendulo

	$\ \{\begin{array}{l}y>\frac{1}{2}x-4 \\ x-2y<-4\end{array}$
Graph $y>\frac{1}{2}x-4$ by graphing $y=\frac{1}{2}x-4$ using the slope $m=\frac{1}{2}$ and the intercept b = −4. The boundary line will be dashed. Test (0, 0) which makes the inequality true, so shade (red) the side that contains (0, 0).
Graph $x-2y<-4$ by graphing $x-2y=-4$ and testing a point. The intercepts are x = −4 and y = 2 and the boundary line will be dashed. Choose a test point in the solution and verify that it is a solution to both inequalties. Test (0, 0) which makes the inequality false, so shade (blue) the side that does not contain (0, 0).

No point on the boundary lines is included in the solution as both lines are dashed.

The solution is the region that is shaded twice which is also the solution to $x-2y<-4.$

Solve the system by graphing: $\{\begin{array}{l}y\ge 3x+1 \\ -3x+y\ge -4\end{array}.$

Bonisa impendulo

The solution is the grey region.
Solve the system by graphing: $\{\begin{array}{l}y\le -\frac{1}{4}x+2 \\ x+4y\le 4\end{array}.$

Bonisa impendulo

The solution is the grey region.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could she display 10 small and 20 large photos?
ⓓ Could she display 20 small and 10 large photos?

Bonisa impendulo

ⓐ

Let $x=\text{the number of small photos}.$

$\ y=\text{the number of large photos}$

To find the system of equations translate the information.

$\ \text{She wants to have at least 25 photos.}$

$\ \text{The number of small plus the number of large should be at least 25.}$

$x+y\ge 25$

$\ \text{\$4 for each small and \$10 for each large must be no more than \$200}$

$4x+10y\le 200$

$\ \text{The number of small photos must be greater than or equal to 0.}$

$x\ge 0$

$\ \text{The number of large photos must be greater than or equal to 0.}$

$y\ge 0$

We have our system of equations.

$\ \{\begin{array}{l}x+y\ge 25 \\ 4x+10y\le 200 \\ x\ge 0 \\ y\ge 0\end{array}\$

ⓑ
Since $x\ge 0$ and $y\ge 0$ (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one.

To graph $x+y\ge 25,$ graph $x+y=25$ as a solid line.
Choose (0, 0) as a test point. Since it does not make the inequality true, shade (red) the side that does not include the point (0, 0).

To graph $4x+10y\le 200,$ graph $4x+10y=200$ as a solid line.
Choose (0, 0) as a test point. Since it does make the inequality true, shade (blue) the side that include the point (0, 0).

The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly-shaded section are included in the solution as are the points on the x-axis from (25, 0) to (55, 0).

ⓒ To determine if 10 small and 20 large photos would work, we look at the graph to see if the point (10, 20) is in the solution region. We could also test the point to see if it is a solution of both equations.

It is not, Christy would not display 10 small and 20 large photos.

ⓓ To determine if 20 small and 10 large photos would work, we look at the graph to see if the point (20, 10) is in the solution region. We could also test the point to see if it is a solution of both equations.

It is, so Christy could choose to display 20 small and 10 large photos.

Notice that we could also test the possible solutions by substituting the values into each inequality.

A trailer can carry a maximum weight of 160 pounds and a maximum volume of 15 cubic feet. A microwave oven weighs 30 pounds and has 2 cubic feet of volume, while a printer weighs 20 pounds and has 3 cubic feet of space.
ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could 4 microwaves and 2 printers be carried on this trailer?
ⓓ Could 7 microwaves and 3 printers be carried on this trailer?

Bonisa impendulo

ⓐ $\{\begin{array}{l}30m+20p\le 160 \\ 2m+3p\le 15\end{array}$
ⓑ

ⓒ yes
ⓓ no
Mary needs to purchase supplies of answer sheets and pencils for a standardized test to be given to the juniors at her high school. The number of the answer sheets needed is at least 5 more than the number of pencils. The pencils cost $2 and the answer sheets cost $1. Mary’s budget for these supplies allows for a maximum cost of $400.
ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could Mary purchase 100 pencils and 100 answer sheets?
ⓓ Could Mary purchase 150 pencils and 150 answer sheets?

Bonisa impendulo

ⓐ $\{\begin{array}{l}a\ge p+5 \\ a+2p\le 400\end{array}$
ⓑ

ⓒ no
ⓓ no

Omar needs to eat at least 800 calories before going to his team practice. All he wants is hamburgers and cookies, and he doesn’t want to spend more than $5. At the hamburger restaurant near his college, each hamburger has 240 calories and costs $1.40. Each cookie has 160 calories and costs $0.50.

ⓐ Write a system of inequalities to model this situation.
ⓑ Graph the system.
ⓒ Could he eat 3 hamburgers and 1 cookie?
ⓓ Could he eat 2 hamburgers and 4 cookies?

Bonisa impendulo

ⓐ

Let $h=\text{the number of hamburgers}.$

$\ c=\text{the number of cookies}$

To find the system of equations translate the information.

The calories from hamburgers at 240 calories each, plus the calories from cookies at 160 calories each

must be more that 800.

$240h+160c\ge 800$

$\ \text{The amount spent on hamburgers at \$1.40 each, plus the amount spent on cookies at \$0.50 each}$

$\ \text{must be no more than \$5.00.}$

$1.40h+0.50c\le 5$

$\ \text{The number of hamburgers must be greater than or equal to 0.}$

$h\ge 0$

$\ \text{The number of cookies must be greater than or equal to 0.}$

$c\ge 0$

We have our system of equations. $\ \{\begin{array}{l}240h+160c\ge 800 \\ 1.40h+0.50c\le 5 \\ h\ge 0 \\ c\ge 0\end{array}$

ⓑ
Since $h>=0$ and $c>=0$ (both are greater than or equal to) all solutions will be in the first quadrant. As a result, our graph shows only quadrant one.

To graph $240h+160c\ge 800,$ graph $240h+160c=800$ as a solid line.

Choose (0, 0) as a test point. Since it does not make the inequality true, shade (red) the side that does not include the point (0, 0).

Graph $1.40h+0.50c\le 5.$ The boundary line is $1.40h+0.50c=5.$ We test (0, 0) and it makes the inequality true. We shade the side of the line that includes (0, 0).

The solution of the system is the region of the graph that is shaded the darkest. The boundary line sections that border the darkly shaded section are included in the solution as are the points on the x-axis from (5, 0) to (10, 0).

ⓓ To determine if 2 hamburgers and 4 cookies would meet Omar’s criteria, we see if the point (2, 4) is in the solution region. It is, Omar might choose to eat 2 hamburgers and 4 cookies.

We could also test the possible solutions by substituting the values into each inequality.

Tenison needs to eat at least an extra 1,000 calories a day to prepare for running a marathon. He has only $25 to spend on the extra food he needs and will spend it on $0.75 donuts which have 360 calories each and $2 energy drinks which have 110 calories.
ⓐ Write a system of inequalities that models this situation.
ⓑ Graph the system.
ⓒ Can he buy 8 donuts and 4 energy drinks and satisfy his caloric needs?
ⓓ Can he buy 1 donut and 3 energy drinks and satisfy his caloric needs?

Bonisa impendulo

ⓐ $\{\begin{array}{l}0.75d+2e\le 25 \\ 360d+110e\ge 1000\end{array}$
ⓑ

ⓒ yes
ⓓ no
Philip’s doctor tells him he should add at least 1,000 more calories per day to his usual diet. Philip wants to buy protein bars that cost $1.80 each and have 140 calories and juice that costs $1.25 per bottle and have 125 calories. He doesn’t want to spend more than $12.
ⓐ Write a system of inequalities that models this situation.
ⓑ Graph the system.
ⓒ Can he buy 3 protein bars and 5 bottles of juice?
ⓓ Can he buy 5 protein bars and 3 bottles of juice?

Bonisa impendulo

ⓐ $\{\begin{array}{l}140p+125j\ge 1000 \\ 1.80p+1.25j\le 12\end{array}$
ⓑ

ⓒ yes
ⓓ no
$\{\begin{array}{l}3x+y>5 \\ 2x-y\le 10\end{array}$
ⓐ $(3,-3)$ ⓑ $(7,1)$
$\{\begin{array}{l}4x-y<10 \\ -2x+2y>-8\end{array}$
ⓐ $(5,-2)$ ⓑ $(-1,3)$

Bonisa impendulo

ⓐ false ⓑ true
$\{\begin{array}{l}y>\frac{2}{3}x-5 \\ x+\frac{1}{2}y\le 4\end{array}$
ⓐ $\text{(6, -4)}$ ⓑ $\text{(3, 0)}$
$\{\begin{array}{l}y<\frac{3}{2}x+3 \\ \frac{3}{4}x-2y<5\end{array}$
ⓐ $(-4,-1)$ ⓑ $\text{(8, 3)}$

Bonisa impendulo

ⓐ false ⓑ true
$\{\begin{array}{l}7x+2y>14 \\ 5x-y\le 8\end{array}$
ⓐ $\text{(2, 3)}$ ⓑ $\text{(7, -1)}$
$\{\begin{array}{l}6x-5y<20 \\ -2x+7y>-8\end{array}$
ⓐ $\text{(1, -3)}$ ⓑ $\text{(-4, 4)}$

Bonisa impendulo

ⓐ false ⓑ true
$\{\begin{array}{l}y\le 3x+2 \\ y>x-1\end{array}$
$\{\begin{array}{l}y<-2x+2 \\ y\ge -x-1\end{array}$

Bonisa impendulo

The solution is the grey region.
$\{\begin{array}{l}y<2x-1 \\ y\le -\frac{1}{2}x+4\end{array}$
$\{\begin{array}{l}y\ge -\frac{2}{3}x+2 \\ y>2x-3\end{array}$

Bonisa impendulo

The solution is the grey region.
$\left\{\begin{array}{l}x-y>1 \\ y<-\frac{1}{4}x+3\end{array}\right$
\(\{\begin{array}{l}x+2y<4 \\ y

Bonisa impendulo

The solution is the grey region.
$\{\begin{array}{l}3x-y\ge 6 \\ y\ge -\frac{1}{2}x\end{array}$
$\{\begin{array}{l}2x+4y\ge 8 \\ y\le \frac{3}{4}x\end{array}$

Bonisa impendulo

The solution is the grey region.

Symbols used here

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Graphing Systems of Linear Inequalities

Determine whether an ordered pair is a solution of a system of linear inequalities
Solve a system of linear inequalities by graphing
Solve applications of systems of inequalities
Graph the first inequality.
Graph the boundary line.
Shade in the side of the boundary line where the inequality is true.
On the same grid, graph the second inequality.
Graph the boundary line.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Zama wena

Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0), OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Okuningi Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

	\(\{\begin{array}{l}x-y>3 \\ y<-\frac{1}{5}x+4\end{array}\)
Graph x − y > 3, by graphing x − y = 3 and testing a point. The intercepts are x = 3 and y = −3 and the boundary line will be dashed. Test (0, 0) which makes the inequality false so shade (red) the side that does not contain (0, 0).

	\(\ \{\begin{array}{l}x-2y<5 \\ y>-4\end{array}\)
Graph \(x-2y<5,\) by graphing \(x-2y=5\) and testing a point. The intercepts are x = 5 and y = −2.5 and the boundary line will be dashed. Test (0, 0) which makes the inequality true, so shade (red) the side that contains (0, 0).

	\(\{\begin{array}{l}4x+3y\ge 12 \\ y<-\frac{4}{3}x+1\end{array}\)
Graph \(4x+3y\ge 12,\) by graphing \(4x+3y=12\) and testing a point. The intercepts are x = 3 and y = 4 and the boundary line will be solid. Test (0, 0) which makes the inequality false, so shade (red) the side that does not contain (0, 0).
Graph \(y<-\frac{4}{3}x+1\) by graphing \(y=-\frac{4}{3}x+1\) using the slope \(m=-\frac{4}{3}\) and y-intercept b = 1. The boundary line will be dashed. Test (0, 0) which makes the inequality true, so shade (blue) the side that contains (0, 0).

	\(\ \{\begin{array}{l}y>\frac{1}{2}x-4 \\ x-2y<-4\end{array}\)
Graph \(y>\frac{1}{2}x-4\) by graphing \(y=\frac{1}{2}x-4\) using the slope \(m=\frac{1}{2}\) and the intercept b = −4. The boundary line will be dashed. Test (0, 0) which makes the inequality true, so shade (red) the side that contains (0, 0).
Graph \(x-2y<-4\) by graphing \(x-2y=-4\) and testing a point. The intercepts are x = −4 and y = 2 and the boundary line will be dashed. Choose a test point in the solution and verify that it is a solution to both inequalties. Test (0, 0) which makes the inequality false, so shade (blue) the side that does not contain (0, 0).