Copyright © 2009 Pearson Education, Inc.

CHAPTER 9: Systems of Equations

and Matrices9.1 Systems of Equations in Two Variables

9.2 Systems of Equations in Three Variables

9.3 Matrices and Systems of Equations

9.4 Matrix Operations

9.5 Inverses of Matrices

9.6 Determinants and Cramer’s Rule

9.7 Systems of Inequalities and Linear Programming

9.8 Partial Fractions

Copyright © 2009 Pearson Education, Inc.

8.1Systems of Equations in

Two Variables

Solve a system of two linear equations in two variables by graphing.

Solve a system of two linear equations in two variables using the substitution and the elimination methods.

Use systems of two linear equations to solve applied problems.

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Systems of Equations

A system of equations is composed of two or more equations considered simultaneously.

Example: 5x y = 5

4x y = 3

This is a system of two linear equations in two variables. The solution set of this system consists of all ordered pairs that make both equations true. The ordered pair (2, 5) is a solution of this system.

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Solving Systems of Equations Graphically

When we graph a system of linear equations, each point at which the graphs intersect is a solution of both equations and therefore a solution of the system of equations.

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Solving Systems of Equations Graphically

Let’s solve the previous system graphically.

5x y = 5

4x y = 3

Solution: We see that the graph

intersects at the single point (2, 5), so this is the solution of the system of equations.

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Systems of Equations

If a system of equations has at least one solution, it is consistent. If the system has no solutions, it is inconsistent.

If a system of two linear equations in two variables has an infinite number of solutions, the equations are dependent. Otherwise, they are independent.

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Illustration of Graphs

Graphs of linear equations may be related to each other in one of three ways.

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Substitution Method

The substitution method is a technique that gives accurate results when solving systems of equations. It is most often used when a variable is alone on one side of an equation or when it is easy to solve for a variable. One equation is used to express one variable in terms of the other, then it is substituted in the other equation.

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ExampleUse substitution to solve the system

5x y = 5,

4x y = 3.

Solution

Solve the first equation for y: y = 5x 5

Then we substitute 5x 5 for y in the second equation to give an equation in one variable.

4x (5x 5) = 3

4x 5x + 5 = 3

x = 2

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Solution continued

Now we use back-substitution and substitute 2 for x in either original equation.

4x y = 3

4(2) y = 3

8 y = 3

y = 5

We find the solution to the system of equations to be (2, 5), once again.

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Elimination Method

Using the elimination method, we eliminate one variable by adding the two equations. If the coefficients of a variable are opposites, that variable can be eliminated by simply adding the original equations. If the coefficients are not opposites, it is necessary to multiply one or both equations by suitable constants, before we add.

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Example

Solve the system using the elimination method.

6x + 2y = 4

10x + 7y = 8Solution

If we multiply the first equation by 5 and the second equation by 3, we will be able to eliminate the x variable.

30x + 10y = 20 Substituting: 6x + 2y = 4

30x 21y = 24 6x + 2(4) = 4

11y = 44 6x 8 = 4

y = 4 6x = 12

The solution is (2, 4). x = 2

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Another ExampleSolve the system.

x 3y = 9 (1)

2x 6y = 3 (2)

Solution: 2x + 6y = 18 Mult. (1) by 2

2x 6y = 3

0 = 21

There are no values of x and y in which 0 = 21. So this system has no solution. The graphs of the equations are of parallel lines.

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Another Example

Solve the system. 9x + 6y = 48 (1) 3x + 2y = 16 (2)

Solution: 9x + 6y = 48 9x 6y = 48 Mult. (2) by 3

0 = 0 When we obtain the equation 0 = 0, we know the equations are dependent. There are infinitely many solutions. The graphs of the equations are identical.

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Application

Ethan and Ian are twins. They have decided to save all of the money they earn, at their part-time jobs, to buy a car to share at college. One week, Ethan worked 8 hours and Ian worked 14 hours. Together they saved $256. The next week, Ethan worked 12 hours and Ian worked 16 hours and they earned $324. How much does each twin make per hour?

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SolutionLetting E represent Ethan and I represent Ian, the following system can be obtained.

8E + 14I = 256 First week 12E + 16I = 324 Second week

Mult by 12 96E + 168I = 3072Mult by 8 96E 128I = 2592

40I = 480 I = 12

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Solution

Solve for E. 8E + 14(12) = 256 8E = 88 E = 11

Ian makes $12 per hour while Ethan makes $11 per hour.