hawkes learning systems college algebra

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Copyright © 2011 Hawkes Learning Systems. All rights reserved.

Hawkes Learning SystemsCollege Algebra

Section 8.1: Solving Systems by Substitution and Elimination




Objectives

o Definition and classification of linear systems of equations.

o Solving systems of equations using substitution.o Solving systems of equations using elimination.o Larger systems of equations.o Applications of systems of equations.




Linear Systems of EquationsMany problems of a mathematical nature are most naturally described by two or more equations in two or more variables. A collection of linear equations is called a linear system of equations, or simultaneous linear equations. The three graphs below pictorially identify the three possible varieties of systems of two linear equations in two variables.




Linear Systems of Equations

Varieties of Linear SystemsThe chart below illustrates the only possible solutions to any linear system of equations. If a system of linear equations has

No Solutions

One Solution

Infinite Solutions

The system is

calledInconsistent Consistent Dependent




Solving Systems by Substitution

The solution method of substitution works by solving one equation and substituting the result into the remaining equations, a method that may require several repetitions. This method can be time consuming for large systems where there are many equations and variables, but for smaller systems, this method works well.





For example, we’ll demonstrate solving the system

by substitution.We’ll show two (of the four possible) ways of solving this system of equations by substitution.

2 15

x yx y

2 1x y 5x y Possibility 1: Possibility 2:

2 1y x

52 1x x

3 6x 2x 2 2 1y

2 1x y

3

Choose an equation.

Solve for a variable.

Insert the solution into the other equation.

Solve for the other variable.

Solution is (2,3).

5x y

2 5 1yy

9 3y3y

5 3x

5x y

2





If we observe the graph of the two linear equations in the system we solved on the previous slide, we can see that the solution we found, (2,3), does indeed satisfy the system.




Example 1: Solving Systems by Substitution

Solve the system by substitution.3 2 54 3x yx y

4 3x y

4 3y x

3 532 4xx

3 8 6 5x x 11 11x

1x 14 1 3y

Choose an equation. (You could have chosen either!)

Solve for a variable. We choose y.Insert the solution found for y into the other equation.Solve for the other variable. In this case, x.

Thus, the solution is (1,1).




Example 2: Solving Systems by Substitution

Solve the system by substitution.2 4

3 6 12x yx y

2 4x y

2 4x y

3 2 4 6 12y y

6 12 6 12y y

0 0

This is a true and trivial statement which implies that there are an infinite number of solutions. In other words, for any x-value, there exists a y-value that satisfies the system.




Solving Systems by Elimination

The method of elimination is based on the goal of eliminating one variable by adding two equations together (some texts call this the addition method). The elimination method works because of the following truth:

If and then .Proof: We know that .

Since , it must be true that . Similarly, so, . Thus, the above statement, “If and then ,” is true.

A B C D A C B D A C A C

A B A BC C C D A BC D

A B C DA C B D




Solving Systems by Elimination

For example, we’ll demonstrate solving the following system by the method of elimination.

First, we’ll try adding the equations to see if any variables are eliminated.

Adding the equations as they are doesn’t help us, so let’s try manipulating one (or both) of the equations by multiplying both sides of that equation by a constant. If we multiply the first equation by ̶ 3 and add the equations again, the x variable will be eliminated.

23 6 3x yx y

2y 6 3 3x y

x

4 5 1x y

3 3 3 6x y 3 6 3x y

9 9y

Continued on the next slide…

Not helpful!

3 6 3x y




Solving Systems by EliminationWe found

Now we can plug in y into either of the equations in the system.

Thus, the solution to the system is .

9 9y .1y

1 2x 1x

1, 1




Example 3: Solving Systems by Elimination

Solve the system by the method of elimination.

To eliminate a variable, you’ll need to multiply each equation by a unique constant. Let’s eliminate y. To do so, notice that we’ll have to multiply the first equation by 3 and the second equation by 2.

5 2 62 3 10x yx y

15 6 18x y 2 4 6 0x y

19 38x

5 2 6x y 2 3 10x y 32

2x 5 2 2 6y

2y Thus, the solution is . 2, 2




Example 4: Solving Systems by Elimination

Solve the system by the method of elimination.6

4 4 5x yx y

4 4 24x y + 4 4 5x y

0 29

6y 4 4 5x y

4

This equation is false, so the solution to this system is (there are no solutions).

x




Larger Systems of Equations

Algebraically, larger systems of equations can be dealt with in a similar manner to the systems we have been looking at. However, graphically, larger systems are very different. The table below highlights the differences between equations in three variables and equations in two variables.

An equation inIf a solution exists, it is called an

The graph contains The graph is

2 variables ordered pair(x,y) 2 axes 2 dimensional

3 variables ordered triple(x,y,z) 3 axes 3 dimensional




Larger Systems of Equations

Below is a graphical representation of the ordered triple, . 2,1,3 -axisz

-axisy

-axisx

2,1,3

1 2

3 origin 0,0,0




Example 5: Larger Systems of Equations

Solve the system

Select two equations and eliminate one variable. We’ll eliminate y from equations (I) and (II).

3

9 7 1

I

II

2 4 III5 17

x y z

x y z

x y z

I II

3z

9 7 1 x y z

y x

10x 8 2z





Select two other equations and eliminate the same variable, y. We chose equations (I) and (III).

We have now found two linear equations in two variables. Now we will solve those equations by the elimination method.

Example 5: Larger Systems of Equations(cont.)

I III

3z 2 4 5 17x y z

y x 5 2 4 5 17x y z 7x 9 2z

5 5 5 15x y z

10 8 2x z 7 9 2x z

7 10

70 56 14x z 70 90 20x z

34 34z 1z





Example 5: Larger Systems of Equations (cont.)

7 9 2x z 7 9 21x

7 9 2x 7 7x

We know that z = –1. We’ll back substitute into one of the linear equations in two variables. We chose the second equation found.

Back substitute the values found into an original equation; we chose equation (I).

3x y z 1 31y 0 3y

3y Solution: 1,3, 1

1x




Example 6: Applications of Systems of Equations

Joey has $128 in his pocket in $10, $5, and $1 bills. There are 19 bills in all and he has four more $5 bills than $1 bills. How many bills of each kind does Joey have?

x = the number of $10 billsy = the number of $5 billsz = the number of $1 bills

Come up with a system of equations from the information given.

19 Ix y z

There are 19 bills.

The total value is $128.

There are four more $5 than $1 bills.

10 5 I12 I8x y z

4 III y z Continued on the next slide…



Copyright © 2011 Hawkes Learning Systems. All rights reserved.Example 6: Applications of Systems of

Equations (cont.)Now we can solve the system of equations.

10 10 10 19 0x y z I II

19y z

9 62z

10 5 128x y z

10 x

128z 5y 10x

5y


Note: One of the equations in the system, y = z + 4 is already an equation in two variables, y and z.

5 9 624 zz 5 20 9 62z z

14 42z 3z



Copyright © 2011 Hawkes Learning Systems. All rights reserved.Example 6: Applications of Systems of

Equations (cont.) 3 4y 7y

3 97 1x

10 19x

9x

Solution: 9,7,3

Thus, Joey has nine $10 bills, seven $5 bills and three $1 bills.

hawkes learning systems college algebra

Documents

larger systems of equations

applications of systems

large systems

smaller systems

linear system of equations

system of linear equations

possible varieties of

simultaneous linear