Systems of Equationsand Inequalities

Copyright © Cengage Learning. All rights reserved.

5

Copyright © Cengage Learning. All rights reserved.

5.1 Systems of Linear Equations in Two Variables

3

Objectives

■ Systems of Linear Equations and Their Solutions

■ Substitution Method

■ Elimination Method

■ The Number of Solutions of a Linear System in Two Variables

■ Modeling with Linear Systems

4

Systems of Linear Equations and Their Solutions

5

Systems of Linear Equations and Their Solutions

A system of equations is a set of equations that involve the same variables. A system of linear equations is a system of equations in which each equation is linear. A solution of a system is an assignment of values for the variables that makes each equation in the system true.

To solve a system means to find all solutions of the system. Here is an example of a system of linear equations in two variables:

Equation 1

Equation 2

6

Systems of Linear Equations and Their Solutions

The solution of a system of equations is a point (see Figure below). Since the solution (3, 1) satisfies each equation, the point (3, 1) lies on each line. So it is the point of intersection of the two lines.

Figure 1

7

Substitution Method

8

Substitution MethodTo solve a system using the substitution method, we start with one equation in the system and solve for one variable in terms of the other variable.

9

Example 1 – Substitution MethodFind all solutions of the system.

Equation 1

Equation 2

10

Example 1 – SolutionFind all solutions of the system.

Solution:Solve for one variable. We solve for y in the first equation.

y = 1 – 2x

Equation 1

Equation 2

Solve for y in Equation 1

11

Example 1 – SolutionSubstitute. Now we substitute for y in the second equation and solve for x.

3x + 4(1 – 2x) = 14

3x + 4 – 8x = 14

–5x + 4 = 14

–5x = 10

x = –2

cont’d

Substitute y = 1 – 2x into Equation 2

Expand

Simplify

Subtract 4

Solve for x

12

Example 1 – SolutionBack-substitute. Next we back-substitute x = –2 into the equation y = 1 – 2x.

y = 1 – 2(–2) = 5

Thus x = –2 and y = 5, so the solution is the ordered pair (–2, 5). Figure 2 shows that the graphs of the two equations intersect at the point (–2, 5).

Back-substitute

Figure 2

cont’d

13

Example 1 – SolutionCheck your answer:x = –2, y = 5:

cont’d

14

Elimination Method

15

Elimination MethodTo solve a system using the elimination method, we try to combine the equations using sums or differences so as to eliminate one of the variables.

16

Example 2 – Elimination MethodFind all solutions of the system.

Equation 1

Equation 2

17

Example 2 – SolutionFind all solutions of the system.

Solution:Since the coefficients of the y-terms are negatives of each other, we can add the equations to eliminate y.

Equation 1

Equation 2

18

Example 2 – Solution

Now we back-substitute x = 4 into one of the original equations and solve for y.

Let’s choose the second equation because it looks simpler.

x – 2y = 2

System

Add

Solve for x

Equation 2

cont’d

19

Example 2 – Solution4 – 2y = 2

–2y = –2

y = 1

The solution is (4, 1). Figure 3 shows that the graphs of the equations in the system intersect at the point (4, 1).

Back-substitute x = 4 into Equation 2

Subtract 4

Solve for y

Figure 3

cont’d

20

The Number of Solutions of a LinearSystem in Two Variables

21

The Number of Solutions of a Linear System in Two Variables

22

The Number of Solutions of a Linear System in Two Variables

A system that has no solution is said to be inconsistent. A system with infinitely many solutions is called dependent.

(a) Lines intersect at asingle point. The system has one solution.

(b) Lines are parallel anddo not intersect. Thesystem has no solution.

(c) Lines coincide—equationsare for the same line. The system has infinitely manysolutions.

Figure 5

23

Example 4 Solve the system.

Equation 1

Equation 2

24

Example 4 – SolutionSolve the system.

Solution:We eliminate y from the equations and solve for x.

Equation 1

Equation 2

2 × Equation 1

Add

Solve for x

25

Example 4 – SolutionNow we back-substitute into the first equation and solve for y:

6(2) – 2y = 0

–2y = – 12

y = 6

Back-substitute x = 2

Subtract 12

Solve for y

cont’d

26

Example 4 – SolutionThe solution of the system is the ordered pair (2, 6), that is,

x = 2 y = 6

The graph in Figure 6 shows that the lines in the system intersect at the point (2, 6).

Check your answer:x = 2, y = 6

Figure 6

cont’d

27

Example 5Solve the system.

Equation 1Equation 2

28

Example 5 - SolutionSolve the system.

Solution:This time we try to find a suitable combination of the two equations to eliminate the variable y. Multiplying the first equation by 3 and the second equation by 2 gives

Equation 1Equation 2

3 × Equation 1

2 × Equation 2

Add

29

Example 5 – SolutionAdding the two equations eliminates both x and y in this case, and we end up with 0 = 29, which is obviously false. No matter what values we assign to x and y, we cannot make this statement true, so the system has no solution.

Figure 7 shows that the lines in the system are parallel so do not intersect. The system is inconsistent.

Figure 7

cont’d

30

Example 6Solve the system.

Equation 1Equation 2

31

Example 6 – SolutionSolve the system.

Solution:We multiply the first equation by 4 and the second equation by 3 to prepare for subtracting the equations to eliminate x. The new equations are

Equation 1Equation 2

4 × Equation 1

3 × Equation 2

32

Example 6 – SolutionWe see that the two equations in the original system are simply different ways of expressing the equation of one single line.

The coordinates of any point on this line give a solution of the system. Writing the equation in slope-intercept form, we have y = x – 2.

So if we let t represent any real number, we can write the solution as

x = t

y = t – 2

cont’d

33

Example 6 – SolutionWe can also write the solution in ordered-pair form as

(t, t – 2)

where t is any real number.

The system has infinitely many solutions (see Figure 8).

Figure 8

cont’d

34

Modeling with Linear Systems

35

Example 7 – A Distance-Speed-Time Problem

A woman rows a boat upstream from one point on a river to another point 4 mi away in 1 hours.

The return trip, traveling with the current, takes only 45 min. How fast does she row relative to the water, and at what speed is the current flowing?

36

Example 7 – Solution

(x – y) = 4 and (x + y) = 4

(The times have been converted to hours, since we are expressing the speeds in miles per hour.)

37

Example 7 – SolutionSolve the system. We multiply the equations by 2 and 4, respectively, to clear the denominators.

3(4) – 3y = 8–3y = 8 – 12

y =

The woman rows at 4 mi/h, and the current flows at mi/h.

2 × Equation 1

4 × Equation 2

Add

Solve for x

Back-substitute x = 4

Solve for y

cont’d

38

Example 7 – SolutionCheck your answer:Speed upstream is Speed downstream is

and this should equal and this should equal

rowing speed – current flow rowing speed + current flow

= 4 mi/h – mi/h = mi/h = 4 mi/h + mi/h = mi/h

cont’d

39Copyright © Cengage Learning. All rights reserved.

5.2 Systems of Linear Equations in Several Variables

40

Example 1 – Solving a Triangular System Using Back-Substitution

Solve the following system using back-substitution:

x – 2y – z = 1 y + 2z = 5

z = 3

Equation 1

Equation 2

Equation 3

41

Example 1 – SolutionSolve the following system using back-substitution:

x – 2y – z = 1 y + 2z = 5

z = 3

Solution:From the last equation we know that z = 3.

We back-substitute this into the second equation and solve for y.

y + 2(3) = 5

Equation 1

Equation 2

Equation 3

Back-substitute z = 3 into Equation 2

42

Example 1 – Solution

y = –1

Then we back-substitute y = –1 and z = 3 into the first equation and solve for x.

x – 2(–1) – (3) = 1

x = 2

The solution of the system is x = 2, y = –1, z = 3. We can also write the solution as the ordered triple (2, –1, 3).

cont’d

Solve for y

Back-substitute y = –1 and z = 3 into Equation 1

Solve for x

43

Example 2 – Solving a System of Three Equations in Three Variables

Solve the following system.

x – 2y + 3z = 1 x + 2y – z = 13

3x + 2y – 5z = 3

Equation 1

Equation 2

Equation 3

44

Example 2 – SolutionFirst add equation 2 & (-1) times equation 1 to eliminate the x term:

x + 2y - z = 13-x + 2y - 3z = -1

4y – 4z = 12

Now we add equation 3 & (-3) times equation 1 to eliminate the x term again:

(–1) × Equation 1

cont’d

45

Example 2 – SolutionNow you have a system of two equations with two variables that you can solve:

8y – 14z = 04y – 4z = 12

Multiply the second equation by (-2) & add them:

Now we have z = 4, and we can back-substitute to find y:4y - 4(4) = 12, so y = 7.

Then back-substitute into one of the original equations to find z: x – 2(7) + 3(4) = 1, so x = 3.

So the solutions is (3, 7, 4).

46

Example 2 – SolutionCheck Your Answer:x = 3, y = 7, z = 4:

(3) – 2(7) + 3(4) = 1

(3) + 2(7) – (4) = 13

3(3) + 2(7) – 5(4) = 3

cont’d

47

Example - #21Solve the following system.

x + y + z = 4 x + 3y + 3z = 10

2x + y – z = 3

Equation 1

Equation 2

Equation 3

48

Example - #21Solve the following system.

x + y + z = 4 x + 3y + 3z = 10

2x + y – z = 3

Equation 1

Equation 2

Equation 3

49

50

Example - #25Solve the following system.

2x + 4y – z = 2 x + 2y – 3z = – 4

3x – y + z = 1

Equation 1

Equation 2

Equation 3

51

Example - #25Solve the following system.

2x + 4y – z = 2 x + 2y – 3z = – 4

3x – y + z = 1

Equation 1

Equation 2

Equation 3

52