{{ item.displayTitle }}

No history yet!

Student

Teacher

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} We have been given a system of linear inequalities and asked to graph the solution sets. Let's look at graph of each individual inequality first and then combine the resulting solution sets.

The first thing we need to do in order to graph a linear inequality is draw the boundary line. In this case, our inequality is not in slope-intercept form so let's rewrite that first: $x+y>2⇒y>-x+2.$ The boundary line is drawn by considering the inequality as though it were an equation and then choosing the line to be dashed or solid based on the symbol. This time, our boundary line is $y=-x+2.$ It will be dashed because our symbol is $>$ and points lying on the line are not solutions.

Now we need to choose which side of the line to shade for our inequality. We can test a point in the original inequality to see if it's a solution. Let's use $(0,0).$ Since the point $(0,0)$ does not satisfy the inequality, we should shade the region that does not contain that point.Now we can go through the same process with the second inequality. We will start by graphing the boundary line. This time the line should remain solid because the symbol $≥$ indicates that points lying on the line are valid solutions.

Let's use $(0,0)$ again to test for our shaded region. Since $(0,0)$ satisfies the inequality, we should shade the region containing the point.Now we can combine the two solution sets for the individual inequalities to find the solution set for the system of inequalities.

Finally, let's cut away the unnecessary information so that we can see exactly where the final solution set lies. We only need the overlapping section of the graph.