1 Intersection of sets (Section P.1 , page 8)
2 Graphing a line (Section 2.3 , page 204)
3 Solving systems of equations (Sections 5.1 , 5.2, and 5.4)
The definition of horsepower was originated by the inventor James Watt (1736–1819). To demonstrate the power (work capability) of the steam engine that he invented and patented in 1783, he rigged his favorite horse with a rope and some pulleys and showed that his horse could raise a 3300-pound weight 10 feet in the air in one minute. Watt called this amount of work 1 horsepower (hp). So he defined
Because horses were one of the main energy sources during that time, the public easily understood and accepted the horsepower measure. Watt labeled his steam engines in terms of equivalent horsepower. A 10 hp engine could do the work of 10 horses, or could lift 5500 pounds up 1 foot in one second. In the usual definition of power=(force)(distance)time,
1 Graph a linear inequality in two variables.
The statements x+y>4, 2x+3y<7, y≥x,
Graph the inequality 2x+y>6.
First, graph the equation 2x+y=6.
Graph the inequality 3x+y<6.
In Figure 5.12(b), the graph of the equation 2x+y=6
In general, the graph of a linear inequality’s corresponding equation (found by changing the inequality symbol to the equality sign) is a line that separates the plane into two regions, each called a half plane. The line itself is called the boundary of each region. All of the points in one region satisfy the inequality, and none of the points in the other region are solutions. You can therefore test one point not on the boundary, called a test point, to determine which region represents the solution set.
Graph the linear inequality y≤−2x+4.
Sketch the graph of each inequality.
x≥2
y<3
x+y<4
Steps | Inequality | ||
---|---|---|---|
1. Change the inequality to an equality. | a. x≥2 x=2 |
b. y<3 y=3 |
c. x+y<4 x+y=4 |
2. Graph the equation in Step 1 with a dashed line (< or >) |
|||
3. Select a test point. | Test (0, 0) in x≥2; 0≥2 |
Test (0, 0) in y<3; 0<3 |
Test (0, 0) in x+y<4; |
4. Shade the solution set. |
Graph the inequality x+y>9.
2 Graph systems of linear inequalities in two variables.
An ordered pair (a, b) is a solution of a system of inequalities involving two variables if it is a solution of each of its inequalities. For example, (0, 1) is a solution of the system
because both 2(0)+1≤1
Graph the solution set of the system of inequalities.
Graph each inequality separately in the same coordinate plane.
Inequality 1 | Inequality 2 |
---|---|
2x+3y>6 |
y−x≤0 |
|
|
The graph of the solution set of the system of inequalities (1) and (2) (the region where the shadings overlap) is shown shaded in purple in Figure 5.14(c).
Graph the solution set of the system of inequalities.
The point P(65, 65)
In this case, however, the point P is not in the solution set of the given system because the ordered pair (65, 65)
Sketch the graph and label the vertices of the solution set of the system of linear inequalities.
First, on the same coordinate plane, sketch the graphs of the three linear equations that correspond to the three inequalities. Because either ≤
Notice that (0, 0) satisfies each of the inequalities but none of the equations. You can use (0, 0) as the test point for each inequality.
Make the following conclusions:
Because (0, 0) lies below the line 3x+2y=11,
Because (0, 0) is above the line x−y=2,
Finally, because (0, 0) lies below the line 7x−2y=−1,
In working with linear inequalities, you must use the test point to graph the solution set. Do not assume that just because the symbol ≤
The region common to the regions in (i), (ii), and (iii) (including the boundaries) is the solution set of the given system of inequalities. The solution set is the shaded region in Figure 5.16, including the sides of the triangle.
Figure 5.16 also shows the vertices of the solution set. These vertices are obtained by solving each pair of equations in the system. Because all vertices are solutions of the given system, they are shown as (solid) points.
To find the vertex (3, 1), solve the system
Solve equation (2) for x to obtain x=2+y.
Back-substitute y=1
Solve the system of equations {x−y=27x−2y=−1
Solve the system of equations {3x+2y=117x−2y=−1
Sketch the graph (and label the corner points) of the solution set of the system of inequalities.
3 Graph a nonlinear inequality in two variables.
We follow the same steps that we used in solving a linear inequality.
Graph y≤x2−2.
4 Graph systems of nonlinear inequalities in two variables.
The procedure for solving a nonlinear system of inequalities is identical to the procedure for solving a linear system of inequalities.
Graph the solution set of the system of inequalities.
Graph each inequality separately in the same coordinate plane. Because (0, 0) is not a solution of any of the corresponding equations, use (0, 0) as a test point for each inequality.
Inequality (1) | Inequality (2) | Inequality (3) |
---|---|---|
y≤4−x2 |
y≥32x−3 |
y≥−6x−3 |
Step 1 y=4−x2 |
y=32x−3 |
y=−6x−3 |
Step 2 The graph of y=4−x2 |
The graph of y=32x−3 |
The graph of y=−6x−3 |
Step 3 Testing (0, 0) in y≤4−x2 |
Testing (0, 0) in y≥32x−3 |
Testing (0, 0) in y≥−6x−3 |
Step 4 Shade the region as in Figure 5.17(a). | Shade the region as in Figure 5.17(b). | Shade the region as in Figure 5.17(c). |
The region common to all three graphs is the graph of the solution set of the given system of inequalities. See Figure 5.17(d).
Use the substitution method to locate the points of intersection A, B, and C of the corresponding equations. See Figure 5.17(d). Solve all pairs of corresponding equations.
The point A(2, 0) is the solution of the system
The point B(0, −3)
Finally, the point C(−1, 3)
Graph the solution set of the system of inequalities.
In the graph of x−3y>1,
If a test point from one of the two regions determined by an inequality’s corresponding equation satisfies the inequality, then in that region satisfy the inequality.
In a system of inequalities containing both 2x+y>5
In a nonlinear system of equations or inequalities, equation or inequality must be nonlinear.
True or False. There are always infinitely many solutions of an inequality ax+by<c,
True or False. If x2+2y<5
True or False. None of the points on the graph of the line 3x+5=7
True or False. The intersection point of the graphs of the equations of the system {2x+4y≤12x−7y<14
In Exercises 9–24, graph each inequality.
x≥0
y≥0
x>−1
y>2
x≥2
y≤3
y−x<0
y+2x≤0
x+2y<6
3x+2y<12
2x+3y≥12
2x+5y≥10
x−2y≤4
3x−4y≤12
3x+5y<15
5x+7y<35
In Exercises 25–30, determine which ordered pairs are solutions of each system of inequalities.
{x+y<22x+y≥6
(0, 0)
(−4, −1)
(3, 0)
(0, 3)
{x−y<23x+4y≥12
(0, 0)
(165, 35)
(4, 0)
(12, 12)
{y<x+2x+y≤4
(0, 0)
(1, 0)
(0, 1)
(1, 1)
{y≤2−xx+y≤1
(0, 0)
(0, 1)
(1, 2)
(−1, −1)
{3x−4y≤12x+y≤45x−2y≥6
(0, 0)
(2, 0)
(3, 1)
(2, 2)
{2+y≤3x−2y≤35x+2y≥3
(0, 0)
(1, 0)
(1, 2)
(2, 1)
In Exercises 31–46, graph the solution set of each system of linear inequalities and label the vertices (if any) of the solution set.
{x+y≤1x−y≤−1
{x+y≥1y−x≥1
{3x+5y≤152x+2y≤8
{−2x+y≤23x+2y≤4
{2x+3y≤64x+6y≥24
{x+2y≥62x+4y≤4
{3x+y≤84x+2y≥4
{x+2y≤122x+4y≥8
{x≥0y≥0x+y≤1
{x≥0y≥0x+y>2
{x≥0y≥02x+3y≥6
{x≥0y≥03x+4y≤12
{x+y≤1x−y≥12x+y≤1
{x+y≤1x−y≥12x+y≥1
{x+y≤1x−y<2−x+y≤3x+y≥−4
{x+y≤−12x−y>−2−2x+y≥−32x+2y≥−4
In Exercises 47–52, use the following figure to indicate the regions (A–G) that correspond to the graph of the given system of inequalities.
{x+y≤23x−2y≤1y−4x≤7
{x+y≥23x−2y≤1y−4x≤7
{x+y≥23x−2y≥1y−4x≤7
{x+y≥23x−2y≤1y−4x≥7
{x+y≤23x−2y≤1y−4x≥7
{x+y≤23x−2y≥1y−4x≥7
In Exercises 53–66, use the procedure for graphing a nonlinear inequality in two variables to graph each inequality.
y≤x2+2
y<x2−3
y>x2−1
y≥x2+1
y≤(x−1)2+3
y>−(x+1)2+4
x2+y2>4
x2+y2≤9
(x−1)2+(y−2)2≤9
(x+2)2+(y−1)2>4
y<2x
y≥ex
y<log x
y≥ln x
For Exercises 67–72, use the following figure to indicate the region or regions that correspond to the graph of each system of inequalities.
{y≥x2−5x2+y2≤25x≥0
{y≥x2−5x2+y2≥25x≥0
{y≤x2−5x2+y2≤25x≥0
{y≤x2−5x2+y2≥25y≥0
{y≤x2−5x2+y2≤25y≥0
{y≥x2−5x2+y2≤25y≥0
In Exercises 73–80, graph the region determined by each system of inequalities and label all points of intersection.
{x+y≤6y≥x2
{x+y≥6y≥x2
{y≥2x+1y≤−x2+2
{y≤2x+1y≤−x2+2
{y≥xx2+y2≤1
{y≤xx2+y2≤1
{x2+y2≤25x2+y≤5
{x2+y2≤25x2+y≥5
Grocer’s shelf space. Your local grocery store has shelf space for, at most, 40 cases of Coke and Sprite. The grocer wants at least ten cases of Coke and five cases of Sprite. Graph the possible stock arrangements and find the vertices.
DVD players. An electronic store sells two types, A and B, of DVD players. Each player of type A and type B costs $60 and $80, respectively. The store wants at least 20 players in stock and does not want its investment in the players to exceed $6000. Graph the possible inventory of the players and find the vertices.
Home builder. A builder has 43 units of material and 25 units of labor available to build one- and/or two-story houses within one year. Assume that a two-story house requires seven units of material and four units of labor and a one-story house requires five units of material and three units of labor. Graph the possible inventory of houses and find the vertices.
Financial planning. Sabrina has $100,000 to invest in stocks and bonds. The annual rate of return for stocks and bonds is 4% and 7%, respectively. She wants at least $5600 income from her investments. She also wants to invest at least $60,000 in bonds. Graph her possible investment portfolio and find the vertices.
Engine manufacturing. A company manufactures 3 hp and 5 hp engines. The time (in hours) required to assemble, test, and package each type of engine is shown in the table below.
Process | |||
---|---|---|---|
Engine | Assemble | Test | Package |
3 hp | 3 | 2 | 0.5 |
5 hp | 4.5 | 1 | 0.75 |
The maximum times (in hours) available for assembling, testing, and packaging are 360 hours, 200 hours, and 60 hours, respectively. Graph the possible processing activities and find the vertices.
Pet food. Vege Pet Food manufactures two types of dog food: Vegies and Yummies. Each bag of Vegies contains 3 pounds of vegetables and 2 pounds of cereal; each bag of Yummies contains 1.5 pounds of vegetables and 3.5 pounds of cereal. The total amount of vegetables and cereal available per week are 20,000 pounds and 27,000 pounds, respectively. Graph the possible number of bags of each type of food and find the vertices.
In Exercises 87–94, write a system of linear inequalities that has the given graph.
In Exercises 95–104, graph each inequality.
xy≥1
xy<3
y<2x
y≤2x−1
y≥3x−2
y>−3x
y≥ln x
y<−ln x
y>ln(x−1)
y≤−ln(x+1)
In Exercises 105 and 106, graph the solution set of each inequality.
1≤x2+y2≤9
x2≤y≤4x2
In Exercises 107 and 108, graph the solution set of each inequality.
|x|+|y|<1
|x+2y|<4
In Exercises 109–112, find the point of intersection of the two lines.
x+2y=40
3x+y=30
x−2y=2
3x+2y=12
In Exercises 113 and 114, find the vertices of the triangle whose sides lie along the given lines.
y−x=1, 2y+x=8,
y+x=3, y−3x=−5,
Consider the following system of linear inequalities:
Graph and shade the solution set of the system of inequalities.
List the coordinates of the corner points of the region in part (a).