Ulearngo

Solving a System of Linear Inequalities By Graphing

Solving 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.

Example: How to Solve a System of Linear inequalities

Solve the system by graphing.

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

Solution

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.”

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.

Example

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

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.

Example

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

Solution

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.

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

Example

Solve the system by graphing. \(\begin{array}{c}4x+3y\ge 12\hfill \\ y<-\frac{4}{3}x+1\hfill \end{array}\)

Solution

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). 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.

Example

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

Solution

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 \(x-2y<-4\) 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 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 \(x-2y<-4\).

This lesson is part of:

Systems of Linear Equations I

View Full Tutorial

Track Your Learning Progress

Sign in to unlock unlimited practice exams, tutorial practice quizzes, personalized weak area practice, AI study assistance with Lexi, and detailed performance analytics.