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Mathematics LibreTexts

5.6: Graphing Systems of Linear Inequalities

  • Page ID
    18960
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    Learning Objectives

    By the end of this section, you will be able to:

    • 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

    Note

    Before you get started, take this readiness quiz.

    1. Graph x>2 on a number line.
      If you missed this problem, review [link].
    2. Solve the inequality 2a<5a+12.
      If you missed this problem, review [link].
    3. Determine whether the ordered pair \((3,\frac{1}{2})\) is a solution to the system \(\left\{\begin{array}{l}{x+2 y=4} \\ {y=6 x}\end{array}\right.\).
      If you missed this problem, review [link]

    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.

    SYSTEM OF LINEAR INEQUALITIES

    Two or more linear inequalities grouped together form a system of linear inequalities.

    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.

    \[\left\{\begin{array}{l}{x+4 y \geq 10} \\ {3 x-2 y<12}\end{array}\right.\]

    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)(x,y) that make both inequalities true.

    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.

    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.

    Exercise \(\PageIndex{1}\)

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

    1. (−2, 4)
    2. (3,1)
    Answer

    1. Is the ordered pair (−2, 4) a solution?
    This figure says, “We substitute x = -2 and y = 4 into both inequalities. The first inequality, x + 4 y is greater than or equal to 10 becomes -2 plus 4 times 4 is greater than or less than 10 or 14 is great than or less than 10 which is true. The second inequality, 3x – 2y is less than 12 becomes 3 times -2 – 2 times 4 is less than 12 or  -14 is less than 12 which is true.

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

    2. Is the ordered pair (3,1) a solution?
    This figure says, “We substitute x  3 and y = 1 into both inequalities.” The first inequality, x + 4y  is greater than or equal to 10 becomes 3 + 4 times 1 is greater than or equal to 10 or y is greater than or equal to 10 which is false. The second inequality, 3x -2y is less than 12 becomes 3 times 3 – two times 1 is less than 12 or 7 is less than 12 which is true.

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

    Exercise \(\PageIndex{2}\)

    Determine whether the ordered pair is a solution to the system.\(\left\{\begin{array}{l}{x-5 y>10} \\ {2 x+3 y>-2}\end{array}\right.\)

    1. (3,−1)
    2. (6,−3)
    Answer
    1. no
    2. yes

    Exercise \(\PageIndex{3}\)

    Determine whether the ordered pair is a solution to the system.\(\left\{\begin{array}{l}{y>4 x-2} \\ {4 x-y<20}\end{array}\right.\)

    1. (2,1)
    2. (4,−1)
    Answer
    1. no
    2. no

    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.

    Exercise \(\PageIndex{4}\): How to Solve a System of Linear inequalities

    Solve the system by graphing. \(\left\{\begin{array}{l}{y \geq 2 x-1} \\ {y<x+1}\end{array}\right.\)

    Answer

    This is a table with three columns and several rows. The first row says, “Step 1: Graph the first inequality. We will graph y is greater than or equal to 2x – 1.” There are two equations givens, y is greater than or equal to 2x – 1 and y is less than x + 1. The table then reads, “Graph the boundary line. We graph the line y = 2x – 1. It is a solid line because the inequality sign is greater than or equal to. Shade in the side of the boundary line where the inequality is true. We choose (0, 0) as a test point. It is a solution to y is greater than or equal to 2x – 1, so we shad in the left side of the boundary line.” There is a figure of a line graphed on an x y coordinate plane. The area to the left of the line is shaded.The second row then says, “Step 2: On the same grid, graph the second inequality. We will graph y is less than x + 1 on the same grid. Grph the boundary line. We graph the lin y = x + 1. It is a dashed line because the inequality sign is less than. There is a graph which shows two lines graphed on an x y coordinate plane. The area to the left of one line is shaded. The area to the right of the second line is shaded. There is a small area where the shaded areas overlap. The table then says, “Shade in the side of that boundary line where the inequality is true. Again we use (0, 0) as a test point. It is a solution so we shade in that side of the line y = x + 1.The third row then says, “Step 3: The solution is the region where the shading overlaps. The poing where the boundary lines intersect is not a solution because it is not a solution to y is less than x + 1. The solution is all points in the purple shaded region.”The fourth row then says, “Step 4: Check by choosing a test point. We’ll use (-1, -1) as a test point. Is (-1, -1) a solution to y is greater than or equal to 2x – 1? -1 is greater than or equal to 2 times -1 – 1 or -1 is greater than or equal to -3 true.”

    Exercise \(\PageIndex{5}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{y<3 x+2} \\ {y>-x-1}\end{array}\right.\)

    Answer

    This figure shows a graph on an x y-coordinate plane of y is less than 3x +2 and y is greater than –x – 1. The area to the right of each line is shaded slightly different colors with the overlapping area also shaded a slightly different color. Both lines are dotted.

    Exercise \(\PageIndex{6}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{y<-\frac{1}{2} x+3} \\ {y<3 x-4}\end{array}\right.\)

    Answer

    This figure shows a graph on an x y-coordinate plane of y is less than –(1/2)x + 3 and y is less than 3x – 4. The area to the right or below each line is shaded slightly different colors with the overlapping area also shaded a slightly different color. Both lines are dotted.

    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.

    Exercise \(\PageIndex{7}\)

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

    Answer
    Graph xy > 3, by graphing xy = 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<−15x+4 by graphing y=−15x+4
    using the slope m=−15 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.

    Exercise \(\PageIndex{8}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{x+y \leq 2} \\ {y \geq \frac{2}{3} x-1}\end{array}\right.\)

    Answer

    This figure shows a graph on an x y-coordinate plane of x + y is less than or equal to 2 and y is greater than or equal to (2/3)x – 1. The area to the left of each line is shaded different colors with the overlapping area also shaded a different color.

    Exercise \(\PageIndex{9}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{3 x-2 y \leq 6} \\ {y>-\frac{1}{4} x+5}\end{array}\right.\)

    Answer

    This figure shows a graph on an x y-coordinate plane 3 of 3x – 2y is less than or equal to 6 and y is greater than –(1/4)x + 5. The area to the left or above each line is shaded slightly different colors with the overlapping area also shaded a slightly different color. One line is dotted.

    Exercise \(\PageIndex{10}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{x-2 y<5} \\ {y>-4}\end{array}\right.\)

    Answer



    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). It makes the inequality true. So, shade the side
    that contains (0, 0) red.
    .



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

    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 is the darker-shaded region.

    Exercise \(\PageIndex{11}\)

    Solve the system by graphing.\(\left\{\begin{array}{l}{y \geq 3 x-2} \\ {y<-1}\end{array}\right.\)

    Answer

    This figure shows a graph on an x y-coordinate plane of y is greater than or equal to 3x - 2 and y is less than -1. The area to the left or below each line is shaded different colors with the overlapping area also shaded a different color. One line is dotted.

    Exercise \(\PageIndex{12}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{x>-4} \\ {x-2 y \leq-4}\end{array}\right.\)

    Answer

    This figure shows a graph on an x y-coordinate plane of x is greater than negative 4 and x – 2y is less than or equal to negative 4. The area to the right or below each line is shaded slightly different colors with the overlapping area also shaded a slightly different color. One line is dotted.

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

    Exercise \(\PageIndex{13}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{4 x+3 y \geq 12} \\ {y<-\frac{4}{3} x+1}\end{array}\right.\)

    Answer
    Graph \(4x+3y\geq 12\), 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). It makes the inequality false. So,
    shade the side that does not contain (0, 0) red.
    .
    Graph \(y<−\frac{4}{3}x+1\) by graphing \(y=−\frac{4}{3}x+1\) using the
    slope \(m = \frac{4}{3}\) and the y-intercept b = 1. The boundary line will be dashed.

    Test (0, 0). It makes the inequality true. So,
    shade the side that contains (0, 0) blue.
    .
    There is no point in both shaded regions, so the system has no solution. This system has no solution.

    Exercise \(\PageIndex{14}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{3 x-2 y \leq 12} \\ {y \geq \frac{3}{2} x+1}\end{array}\right.\)

    Answer

    no solution

    This figure shows a graph on an x y-coordinate plane of 3x – 2y is less than or equal 12 and y is greater than or equal to (3/2)x + 1. The area to the left or right of each line is shaded different colors. There is not overlapping area.

    Exercise \(\PageIndex{15}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{x+3 y>8} \\ {y<-\frac{1}{3} x-2}\end{array}\right.\)

    Answer

    no solution

    This figure shows a graph on an x y-coordinate plane of x + 3y is greater than 8 and y is less than –(1/3)x – 2. The area to the above or below each line is shaded slightly different colors. There is no overlapping area. Both lines are dotted.

    Exercise \(\PageIndex{16}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{y>\frac{1}{2} x-4} \\ {x-2 y<-4}\end{array}\right.\)

    Answer
    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). It makes the inequality true. So,
    shade the side that contains (0, 0) red.
    .
    Graph x−2y<−4x−2y<−4 by graphing x−2y=−4x−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 inequalities.
    .

    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.

    Exercise \(\PageIndex{17}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{y \geq 3 x+1} \\ {-3 x+y \geq-4}\end{array}\right.\)

    Answer

    \(y \geq 3 x+1\)

    This figure shows a graph on an x y-coordinate plane of y is greater than or equal to 3x + 1 and -3x + y is greater than or equal to -4. The area to the left of each line is shaded with the overlapping area shaded a slightly different color.

    Exercise \(\PageIndex{18}\)

    Solve the system by graphing. \(\left\{\begin{array}{l}{y \leq-\frac{1}{4} x+2} \\ {x+4 y \leq 4}\end{array}\right.\)

    Answer

    \(x+4 y \leq 4\)

    This figure shows a graph on an x y-coordinate plane of y is less than or equal to –(1/4)x + 2 and x + 4y is less than or equal to 4. The area to the below each line is shaded with the overlapping area shaded a slightly different color.

    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.

    Exercise \(\PageIndex{19}\)

    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.

    1. Write a system of inequalities to model this situation.
    2. Graph the system.
    3. Could she display 15 small and 5 large photos?
    4. Could she display 3 large and 22 small photos?
    Answer

    1.Let x= the number of small photos.

    y= the number of large photos
    To find the system of inequalities, translate the information.
    \(\begin{array}{c}{\text { She wants to have at least } 25 \text { photos. }} \\ {\text { The number of small plus the number of large should be at least } 25 .} \\ {x+y \geq 25} \\ {\$ 4 \text { for each small and } \$ 10 \text { for each large must be no more than } \$ 200} \\ {4 x+10 y \leq 200}\end{array}\)
    We have our system of inequalities. \(\left\{\begin{array}{l}{x+y \geq 25} \\ {4 x+10 y \leq 200}\end{array}\right.\)

    2.

    To graph \(x+y\geq 25\), 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 \(4x+10y\leq 200\), 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.

    3. 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.

    4. 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.

    Exercise \(\PageIndex{20}\)

    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.

    1. Write a system of inequalities to model this situation.
    2. Graph the system.
    3. Could 4 microwaves and 2 printers be carried on this trailer?
    4. Could 7 microwaves and 3 printers be carried on this trailer?
    Answer
    1. \(\left\{\begin{array}{l}{30 m+20 p \leq 160} \\ {2 m+3 p \leq 15}\end{array}\right.\)
    2.  

    This figure shows a graph on an x y-coordinate plane of 30m + 20p is less than or equal 160 and 2m + 3p is less than or equal to 15. The area to the left of each line is shaded with the overlapping area shaded a slightly different color.

    3. yes

    4. no

    Exercise \(\PageIndex{21}\)

    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.

    1. Write a system of inequalities to model this situation.
    2. Graph the system.
    3. Could Mary purchase 100 pencils and 100 answer sheets?
    4. Could Mary purchase 150 pencils and 150 answer sheets?
    Answer
    1. \(\left\{\begin{array}{l}{a \geq p+5} \\ {a+2 p \leq 400}\end{array}\right.\)
    2.  

    This figure shows a graph on an x y-coordinate plane of a is greater than or equal to p + 5 and a + 2p is less than or equal to 400. The area to the left of each line is shaded different colors with the overlapping area also shaded a different color.

    3. no

    4. no

    Exercise \(\PageIndex{22}\)

    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.

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

    Let h= the number of hamburgers.
    c= the number of cookies
    To find the system of inequalities, 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.

    \[240 h+160 c \geq 800\]

    The amount spent on hamburgers at $1.40 each, plus the amount spent on cookies at $0.50 each must be no more than $5.00.

    \[1.40 h+0.50 c \leq 5\]

    \(\text { We have our system of inequalities. } \quad \left\{\begin{array}{l}{240 h+160 c \geq 800} \\ {1.40 h+0.50 c \leq 5}\end{array}\right.\)

    2.

    To graph \(240h+160c\geq 800\) graph 240h+160c=800 as a solid line.
    Choose (0, 0) as a test point. it does not make the inequality true.
    So, shade (red) the side that does not include the point (0, 0).


    To graph \(1.40 h+0.50 c \leq 5\), graph 1.40h+0.50c=5 as a solid line.
    Choose (0,0) as a test point. It makes the inequality true. So, shade
    (blue) the side that includes the point.
    .

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

    3. To determine if 3 hamburgers and 2 cookies would meet Omar’s criteria, we see if the point (3, 1) is in the solution region. It is. He might choose to eat 3 hamburgers and 2 cookies.
    4. 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. He might choose to eat 2 hamburgers and 4 cookies.

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

    Exercise \(\PageIndex{23}\)

    Tension 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.

    1. Write a system of inequalities that models this situation.
    2. Graph the system.
    3. Can he buy 8 donuts and 4 energy drinks?
    4. Can he buy 1 donut and 3 energy drinks?
    Answer
    1. \(\left\{\begin{array}{l}{0.75 d+2 e \leq 25} \\ {360 d+110 e \geq 1000}\end{array}\right.\)
    2.  

    This figure shows a graph on an x y-coordinate plane of 0.75d + 2e is less than or equal to 25 and 360d + 110e is greater than or equal to 1000. The area to the left or right of each line is shaded slightly different colors with the overlapping area also shaded a slightly different color.

    3. yes

    4. no

    Exercise \(\PageIndex{24}\)

    Philip’s doctor tells him he should add at least 1000 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.

    1. Write a system of inequalities that models this situation.
    2. Graph the system.
    3. Can he buy 3 protein bars and 5 bottles of juice?
    4. Can he buy 5 protein bars and 3 bottles of juice?
    Answer
    1. \(\left\{\begin{array}{l}{140 p+125 j \geq 1000} \\ {1.80 p+1.25 j \leq 12}\end{array}\right.\)
    2.  

    This figure shows a graph on an x y-coordinate plane of 140p + 125j is greater than or equal to 1000 and 1.80p + 1.25j is less than or equal to 12. The area to the left or right of each line is shaded slightly different colors with the overlapping area also shaded a slightly different color.

    3. yes

    4. no

    Note

    Access these online resources for additional instruction and practice with graphing systems of linear inequalities.

    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.

     

    Glossary

    system of linear inequalities
    Two or more linear inequalities grouped together form a system of linear inequalities.