Systems of Inequalities Help (page 3)
Introduction to Systems of Inequalities
The solution (if any) for a system of inequalities is usually a region in the plane. The solution to a polynomial inequality (the only kind in this book) is the region above or below the curve. We will begin with linear inequalities.
Graphing the Solution Region for Linear Inequalities
When sketching the graph for an inequality, we will use a solid graph for “≤” and “≥” inequalities, and a dashed graph for “<” and “>” inequalities. We can decide which side of the graph to shade by choosing any point not on the graph itself. We will put this point into the inequality. If it makes the inequality true, we will shade the side that has that point. If it makes the inequality false, we will shade the other side. Every point in the shaded region is a solution to the inequality.
- 2 x + 3 y ≤ 6
We will sketch the line 2 x + 3 y = 6, using a solid line because the inequality is “≤.”
We will always use the origin, (0, 0) in our inequalities unless the graph goes through the origin. Does x = 0 and y = 0 make 2 x + 3 y ≤ 6 true? 2(0) + 3(0) ≤ 6 is a true statement, so we will shade the side that has the origin.
- x − 2 y > 4
We will sketch the line x − 2 y = 4 using a dashed line because the inequality is “>.”
Now we need to decide which side of the line to shade. When we put (0, 0) in x − 2 y > 4, we get the false statement 0 − 2(0) > 4. We need to shade the side of the line that does not have the origin.
- y < 3 x
We use a dashed line to sketch the line y = 3 x . Because the line goes through (0, 0), we cannot use it to determine which side of the line to shade. This is because any point on the line makes the equality true. We want to know where the inequality is true. The point (1, 0) is not on the line, so we can use it. 0 < 3(1) is true so we will shade the side of the line that has the point (1,0), which is the right side.
- x ≥ −3
The line x = −3 is a vertical line through x = −3. Because we want x ≥ −3 we will shade to the right of the line.
- y < 2
The line y = 2 is a horizontal line at y = 2. Because we want y < 2, we will shade below the line.
Graphing the Solution Region for Nonlinear Inequalities
Graphing the solution region for nonlinear inequalities is done the same way—graph the inequality, using a solid graph for “≤” and “≥” inequalities and a dashed graph for “<” and “>” inequalities, then checking a point to see which side of the graph to shade.
- y ≤ x 2 − x − 2
The equality is y = x 2 − x − 2 = ( x − 2)( x + 1). The graph for this equation is a parabola.
Because (0, 0) is not on the graph, we can use it to decide which side to shade; 0 ≤ 0 2 − 0 − 2 is false, so we shade below the graph, the side that does not contain (0, 0).
- y > ( x + 2)( x − 2)( x − 4)
When we check (0, 0) in the inequality, we get the false statement 0 > (0 + 2)(0 − 2)(0 − 4). We will shade above the graph, the region that does not contain (0, 0).
Graphing the Solution for Two or More Inequalities
The solution (if there is one) to a system of two or more inequalities is the region that is part of each solution for the individual inequalities. For example, if we have a system of two inequalities and shade the solution to one inequality in blue and the other in yellow, then the solution to the system would be the region in green.
Sketch the solution for each inequality. The solution to x − y < 3 is the region shaded vertically. The solution to x + 2 y > 1 is the region shaded horizontally.
The region that is in both solutions is above and between the lines.
The solution to y ≤ 4 − x 2 is the region shaded vertically. The solution to x − 7 y ≤ 4 is the region shaded horizontally. The region that is in both solutions is above the line and inside the parabola.
Because a solid graph indicates that points on the graph are also solutions, to be absolutely accurate, the correct solution uses dashed graphs for the part of the graphs that are not on the border of the shaded region.
We will not quibble with this technicality here.
The inequalities x ≥ 0 and y ≥ 0 mean that we only need the top right corner of the graph. These inequalities are common in word problems.
The solution to the system is the region in the top right corner of the graph below the line 2x + y = 5 .
Some systems of inequalities have no solution. In the following example, the regions do not overlap, so there are no ordered pairs (points) that make both inequalities true.
Graphing the Solution Region for Three or More Inequalities
It is easy to lose track of the solution for a system of three or more inequalities. There are a couple of things you can do to make it easier. First, make sure the graph is large enough, using graph paper if possible. Second, shade the solution for each inequality in a different way, with different colors or shaded with horizontal, vertical, and slanted lines. The solution (if there is one) would be shaded all different ways. You could also shade one region at a time, erasing the part of the previous region that is not part of the inequality.
First we will shade the solution for x + y ≤ 4.
The region for x ≥ 1 is the right of the line x = 1, so we will erase the region to the left of x = 1.
The solution to y ≤ x is the region below the line y = x , so we will erase the shading above the line y = x .
The shaded region in Figure 10.27 is the solution for the system.
We will begin with y > x 2 − 16.
The solution to x < 2 is the region to the left of the line x = 2. We will erase the shading to the right of x = 2.
The solution to y < −5 is the region below the line y = −5. We will erase the shading above the line y = − 5.
The solution to − x + y < −8 is the region below the line − x + y = −8, so we will erase the shading above the line. The solution to the system is in Figure 10.31 .
Systems of Inequalities Practice Problems
Graph the solution.
- 2 x − 4 y < 4
- x > 1
- y ≤ −1
- y ≤ x 2 − 4
- y > x 3
- y < |x|
- y ≤ ( x − 3)( x + 1)( x + 3)
Practice problems for this concept can be found at: Systems of Equations and Inequalities Practice Test.
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