If you need to refresh your skills on graphing linear inequalities,
see Linear Inequality Graphs.

Systems of linear inequalities are composed of two separate linear inequalities.
It is possible, however, that a system may be composed of
one linear inequality and one linear equation.

 System of Two Linear Inequalities

Graphing a system of linear inequalities involves graphing the two separate linear inequalities on the sames axes, and noting the sections of the graphs that overlap one another. Be sure you clearly label the overlapping section as the solution, and/or describe your solution.

 Graph this system of linear inequalities and label the solution area S: y < 2x - 6    and    y > -3x + 4 1. Replace the inequality symbols with equal signs and graph the straight lines. y = 2x - 6 and y = -2x + 3 This problem deals with strict inequalities, so the straight lines will be drawn dashed. 2. Determine which side of each line will be shaded to represent its inequality. The test point (0,0) is being used in this example. y < 2x - 6 0 < 2(0) - 6 0 < -6 False (shade the other side of the line) y > -3x + 4 0 > -3(0) + 4 0 > 4 False (shade the other side of the line) Find the overlapping section and label it S.

 Graph this system of linear inequalities and label the solution area S: 1. Replace the inequality symbols with equal signs and graph the straight lines. y = ½x + 2 and y = -x + 1 This problem deals with a "less than or equal to" inequality and one strict inequality. One line is drawn solid, and one dashed. 2. Determine which side of each line will be shaded to represent its inequality. The test point (0,0) is being used in this example. y ½x + 2 0 ½(0) + 2 0 2 True (shade the same side of the line) y > -x + 1 0 > -(0) + 1 0 > 1 False (shade the other side of the line) Find the overlapping section and label it S.

 Graph this system of linear inequalities and label the solution area S: 1. Replace the inequality symbols with equal signs and graph the straight lines. Notice that these two lines have the same slope and are parallel. y = -2x + 2 and y = -2x + 6 This problem deals with two "less than or equal to" inequalities, so both lines are drawn solid. 2. Determine which side of each line will be shaded to represent its inequality. The test point (0,0) is being used in this example. y -2x - 2 0 -2(0) - 2 0 -2 True (shade the same side of the line) y -2x + 6 0 -2(0) + 6 0 6 True (shade the same side of the line) Find the overlapping section and label it S. The area between the two parallel lines is the solution.

Let's revisit Example 3, but reverse the inequalities.
 Graph this system of linear inequalities and label the solution area S: 1. We have the same parallel lines we had in Example 3. y = -2x + 2 and y = -2x + 6 This problem deals with two "less than or equal to" inequalities, so both lines are drawn solid. 2. Determine which side of each line will be shaded to represent its inequality. The test point (0,0) is being used in this example. y -2x - 2 0 -2(0) - 2 0 -2 False (shade the other side of the line) y -2x + 6 0 -2(0) + 6 0 6 False (shade the other side of the line) The shaded areas do NOT overlap. There is NO solution.

 System of 1 Inequality and 1 Linear Equation

 Graph this system of a linear inequality and a linear equation: y - 2x < 4    and    y = -x + 1 1. Replace the inequality symbol with an equal sign and graph the straight line. y = 2x + 4 This problem deals with a strict inequality, so the line is drawn dashed. 2. Determine which side of the line will be shaded to represent the inequality. The test point (0,0) is being used in this example. y < 2x + 4 0 < 2(0) + 4 0 < 4 True (shade the same side of the line) Graph the straight line: y = -x + 1 The overlap of the graphs is the portion of the straight line graph which passes over the shaded region of the inequality. The line first intersects the inequality's shaded area at the point (-1,2), which is "open" due to the dashed line of the inequality. The solution can then be described as the line y = -x + 1 on the domain (-1, ∞).