4.4       Linear Inequalities in 2 Variables
Homework: p. 247; EOO 1-41

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DSPM 0800 Class Notes

Solutions of linear inequalities in 2 variables are ordered pairs (x, y) whose coordinates when inserted into the inequality in the place of x and  y will make a true statement.

Example:  Determine if given points are solutions of the following inequality.

                       

Is (-2, 2) a solution?    2 > 3(-2) – 2

                                    2 > -6 – 2

                                    2 > -8  TRUE

Is (1, 4) a solution?       4 > 3(1) – 2

                                    4 > 3 – 2

                                    4 > 1    TRUE

Is (5, 4) a solution?       4 > 3(5) – 2

                                    4 > 15 – 2

                                    4 > 13  FALSE


We see that more than one point will be a solution. In fact, there are infinite solutions. So we will graph the solution set on the rectangular coordinate system.

1. Put inequality in slope-intercept form to first graph the line.

            less than or greater than: the points on the line are not part of the solution set—dotted line

            less than or equal to, greater than or equal to: the points on line are solutions—solid line

2. Choose a point not on the line. Plug its coordinates into the inequality to see if it is a solution or not.

            Yes, the point is a solution—Shade on side of checkpoint

            No, the point is not a solution—Shade on the other side

3. You can use the origin (0, 0) as your checkpoint unless the line passes through the origin. If the line passes through the origin, choose a point that is obviously not on the line and plug its coordinates in to determine if it is a solution.

 

Example:  Graph the solution set for  y < -x + 5.

The inequality is already in slope-intercept form, so begin by using the y-intercept (0, 5) and the slope –1 to graph the line

           

Note: The above line should be dotted. The points on the line are not solutions because of no equal to.

We can use (0, 0) as our checkpoint since the line does not go through the origin.

            0 < 0 + 5

            0 < 5  TRUE

We will shade the region in which the origin lies.

           

 

To graph a system of inequalities, graph each inequality (line and shade). The solution set for the system is the region in which the shading overlaps. These are the points that make both inequalities true.

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