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5.6 Graphing systems of linear inequalities  (Page 2/10)

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Solve the system by graphing. { y < 3 x + 2 y > − x − 1

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.

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Solve the system by graphing. { y < − 1 2 x + 3 y < 3 x − 4

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.

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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.

Solve the system by graphing. { x − y > 3 y < − 1 5 x + 4

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 < − 1 5 x + 4 by graphing y = − 1 5 x + 4
using the slope m = − 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.

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Solve the system by graphing. { x + y ≤ 2 y ≥ 2 3 x − 1

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.

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Solve the system by graphing. { 3 x − 2 y ≤ 6 y > − 1 4 x + 5

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.

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Solve the system by graphing. { x − 2 y < 5 y > −4

Solution

Graph x − 2 y < 5 , by graphing x − 2 y = 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.

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Solve the system by graphing. { y ≥ 3 x − 2 y < −1

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.

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Solve the system by graphing. { x > −4 x − 2 y ≤ − 4

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.

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

Solve the system by graphing. { 4 x + 3 y ≥ 12 y < − 4 3 x + 1

Solution

Graph 4 x + 3 y ≥ 12 , by graphing 4 x + 3 y = 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 < − 4 3 x + 1 by graphing y = − 4 3 x + 1 using the
slope m = 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.

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Solve the system by graphing. { 3 x − 2 y ≤ 12 y ≥ 3 2 x + 1

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.

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Solve the system by graphing. { x + 3 y > 8 y < − 1 3 x − 2

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.

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Solve the system by graphing. { y > 1 2 x − 4 x − 2 y < −4

Solution

Graph y > 1 2 x − 4 by graphing y = 1 2 x − 4
using the slope m = 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 − 2 y < − 4 by graphing x − 2 y = − 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 − 2 y < −4 .

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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