1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of...

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Systems of EquationsCHAPTER

9.1 Solving Systems of Linear Equations Graphically9.2 Solving Systems of Linear Equations by

Substitution9.3 Solving Systems of Linear Equations by

Elimination9.4 Solving Systems of Linear Equations in Three

Variables9.5 Solving Systems of Linear Equations Using

Matrices9.6 Solving Systems of Linear Equations Using

Cramer’s Rule9.7 Solving Systems of Linear Inequalities

99


Solving Systems of Linear Equations Graphically9.19.1

1. Determine if an ordered pair is a solution for a system of equations.

2. Solve a system of linear equations graphically.3. Classify systems of linear equations in two unknowns.

4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Objective 1

Determine if an ordered pair is a solution for a system of equations.


System of equations: A group of two or more equations.

Solution for a system of equations: An ordered set of numbers that makes all equations in a system true.

5 (Equation 1)

3 4 8 (Equation 2)

x y

x y


Checking a Solution to a System of EquationsTo verify or check a solution to a system of equations: 1. Replace each variable in each equation with its

corresponding value. 2. Verify that each equation is true.


ExampleDetermine whether each ordered pair is a solution to the system of equations.

a. (3, 2) b) (3, 4)Solutiona. (3, 2) x + y = 7 y = 3x – 2

3 + 2 = 7 2 = 3(3) – 2 1 = 7 2 = 11 False False

7 (Equation 1)

3 2 (Equation 2)

x y

y x


continued

b. (3, 4) x + y = 7 y = 3x – 2 3 + 4 = 7 4 = 3(2) – 2

7 = 7 4 = 4True True

Because (3, 2) does not satisfy both equations, it is not a solution for the system.Because (3, 4) satisfies both equations, it is a solution to the system of equations.


Objective 2

Solve a system of linear equations graphically.


A system of two linear equations in two variables can have one solution, no solution, or an infinite number of solutions.

2

5

4

x

y x

y

The graphs intersect at a single point.

3 2

3 1x

y x

y

2 8

2 4

4x y

x y

The equations have the same slope, the graphs are parallel.

The graphs are identical. There are an infinite number of solutions.


Solving Systems of Equations GraphicallyTo solve a system of linear equations graphically:1. Graph each equation.

a. If the lines intersect at a single point, then the coordinates of that point form the solution.b. If the lines are parallel, then there is no solution.c. If the lines are identical, then there are an infinite

number of solutions, which are the coordinates of

all the points on that line.2. Check your solution.


ExampleSolve the system of equations graphically.

SolutionGraph each equation.The lines intersect at a singlepoint, (2, 4).We can check the point in eachequation to verify and will leave that to you.

2 (Equation 1)

2 4 12 (Equation 2)

y x

x y

y = 2 – x

2x + 4y = 12(2, 4)


ExampleSolve the system of equations graphically.

SolutionGraph each equation.The lines appear to be parallel, which we can verify by comparing the slopes.

(Equation 1)

(Equation 2)3 4

33

44x

y x

y

3 4 4

4 3 4

31

4

x y

y x

y x

The slopes are the same, so the lines are parallel.


Objective 3

Classify systems of linear equations in two unknowns.


Consistent system of equations: A system of equations that has at least one solution.Inconsistent system of equations: A system of equations that has no solution.

Dependent linear equations in two unknowns: Equations with identical graphs.Independent linear equations in two unknowns: Equations that have different graphs.


Classifying Systems of EquationsTo classify a system of two linear equations in two unknowns, write the equations in slope-intercept form and compare the slopes and y-intercepts. Consistent system: The system has a single solution at the point of intersection.

Independent equations: The graphs are different and intersect at one point. They have different slopes.

Consistent system: The system has an infinite number of solutions.

Dependent equations: The graphs are identical. They have the same slope and same y-intercept.

Inconsistent system: The system has no solution.

Independent equations: The graphs are different and are parallel. Though they have the same slopes, they have different y-intercepts.


ExampleReturning to a previous example (1) is the system consistent or inconsistent, are the equations dependent or independent, and how many solutions does the system have?2 4 12 (Equation 1)

2 (Equation 2)

x y

y x

The graphs intersected at a single point.

The system is consistent.

The equations are independent (different graphs), and the system has one solution: (2, 4).


Which set of points is a solution to the system?

a) (–1, 1)

b) (–1, –1)

c) (0, 2)

d) (–3, 7)

2 4 2

3 2 1

x y

x y


Solving Systems of Linear Equations by Substitution9.29.2

1. Solve systems of linear equations using substitution.2. Solve applications involving two unknowns using a

system of equations.


Objective 1

Solve systems of linear equations using substitution.


Solving Systems of Two Equations Using SubstitutionTo find the solution of a system of two linear equations using the substitution method:1. Isolate one of the variables in one of the equations.2. In the other equation, substitute the expression you found in step 1 for that variable.3. Solve this new equation. (It will now have only one variable.)4. Using one of the equations containing both variables, substitute the value you found in step 3 for that variable and solve for the value of the other variable.5. Check the solution in the original equations.


ExampleSolve the system of equations using substitution.

Solution Step 1: Isolate a variable in one equation. The second equation is solved for x.

Step 2: Substitute x = 1 – y for x in the first equation.

4 5 8

1

x y

x y

(1

4 5 8

4 5 8)

x y

yy


continued

Step 3: Solve for y. (1 )4 5 8yy

4 4 5 8y y

4 8y

4y

Step 4: Solve for x by substituting 4 for y in one of the original equations.

4 5 8

1

x y

x y

1x y 1 4x

3x The solution is (3, 4).



Solution Step 1: Isolate a variable in one equation. Use either equation. 2x + y = 7

y = 7 – 2x

Step 2: Substitute y = 7 – 2x for y in the second equation.

2 7

1

x y

x y

1

(7 2 ) 1x x

x y


continuedStep 3: Solve for x.

Step 4: Solve for y by substituting 2 for x in one of the original equations.

2 7x y 2(2) 7y

4 7y The solution is (2, 3).

7 2( ) 1xx

2 7

1

x y

x y

7 2 1x x 3 7 1x

3 6x 2x

3y


Inconsistent Systems of EquationsThe system has no solution because the graphs of the equations are parallel lines.You will get a false statement such as 3 = 4.

Consistent Systems with Dependent EquationsThe system has an infinite number of solutions because the graphs of the equations are the same line.You will get a true statement such as 8 = 8.



SolutionSubstitute y = 4 – 3x into the second equation.

4 3

6 2 8

y x

x y

6 2 8

6 2( )4 3 8

x

x

y

x

6 8 6 8x x 6 6 8 8x x

8 8 True statement. The number of solutions is infinite.


Objective 2

Solve applications involving two unknowns using a system of equations.


Solving Applications Using a System of EquationsTo solve a problem with two unknowns using a system of equations:1. Select a variable for each of the unknowns.2. Translate each relationship to an equation. 3. Solve the system.


Example

Terry is designing a table so that the length is twice the width. The perimeter is to be 216 inches. Find the length and width of the table.

Understand We are given two relationships and are to find the length and width.

Plan Select a variable for the length and another variable for the width, translate the relationships to a system of equations, and then solve the system.


continued

Execute Let l represent the length and w the width.Relationship 1: The length is twice the width.

Translation: l = 2wRelationship 2: The perimeter is 216 inches.

Translation: 2l + 2w = 216

System: 2

2 2 216

l w

l w


continuedSolve.

Now find the value of l. l = 2w = 2(36) = 72

Answer The length should be 72 inches and the width 36 inches.

2

2 2 216

l w

wl

2( ) 2 162 2w w 4 2 216w w

6 216w 36w

Substitute 2w for l.

Combine like terms.

Divide both sides by 6 to isolate w.


Solve the system.

a) (10, 3)

b) (1, 4)

c) (10, 14)

d) (3, 6)

4

2 6

x y

x y


Solve the system.

a) (2, 3)

b) (2, 6)

c) (2, 6)

d) (3, 2)

3 16

7 4 10

x y

x y


Solving Systems of Linear Equations by Elimination9.39.3

1. Solve systems of linear equations using elimination.2. Solve applications using elimination.


Objective 1

Solve systems of linear equations using elimination.


ExampleSolve the system of equations.Solution We can add the equations.

9

3 7

x y

x y

9

3 7

x y

x y

4 0 16x

Notice that y is eliminated, so we can easily solve for the value of x.

4x 4 16x

Divide both sides by 4 to isolate x.

Now that we have the value of x, we can find y by substituting 4 for x in one of the original equations. x + y = 9

4 + y = 9y = 5

The solution is (4, 5).


Example—Multiplying Solve the system of equations.Solution Because no variables are eliminated if we

add, we rewrite one of the equations so that it has a term that is the additive inverse of one of the terms in the other equation.

Multiply the first equation by 4.

8

3 4 11

x y

x y

8x y 4 4 4 8x y

4 4 32x y 4 4 32

3 4 11

x y

x y

7 21x 3x

7 0 21x Solve for y.x + y = 83 + y = 8 y = 5



Example—Multiplying Solve the system of equations.Solution Choose a variable to eliminate, y, then

multiply both equations by numbers that make the y terms additive inverses.

Multiply the first equation by 2.Multiply the second equation by 5.

4 5 19

3 2 9

x y

x y

4 5 19

3 2 9

x y

x y

8 10 38

15 10 45

x y

x y

Multiply by 2.

Multiply by 5.

Add the equations to eliminate y.

7 0 7x 7 7x

1x


continued

Substitute x = 1 into either of the original equations. 4x – 5y = 194(1) – 5y = 19 4 – 5y = 19 –5y = 15

y = 3

The solution is (1, –3).

4 5 19

3 2 9

x y

x y


Example—Fractions or DecimalsSolve the system of equations.

Solution To clear the decimals in Equation 1, multiply by 100.To clear the fractions in Equation 2, multiply by 5.

0.03 0.02 0.03

4 2 2

5 5 5

x y

x y

0.03 0.02 0.03

4 2 2

5 5 5

x y

x y

Multiply by 100.

Multiply by 5.

3 2 3

4 2 2

x y

x y

Multiply equation 2 by 1 then combine the equations.

3 2 3

4 2 2

x y

x y

3 2 3

4 2 2

x y

x y


continued

3 2 3

4 2 2

x y

x y

1x 1x

0 1x

Substitute to find y. 3 2 3x y ( 2 313 ) y

3 2 3y 2 6y

3y The solution is (1, 3).


Solving Systems of Equations Using EliminationTo solve a system of two linear equations using the elimination method:1. Write the equations in standard form (Ax + By = C).2. Use the multiplication principle to clear fractions or decimals.3. Multiply one or both equations by a number (or numbers) so that they have a pair of terms that are additive inverses.4. Add the equations. The result should be an equation in terms of one variable.5. Solve the equation from step 4 for the value of that variable.6. Using an equation containing both variables, substitute the value you found in step 5 for the corresponding variable and solve for the value of the other variable.7. Check your solution in the original equations.


Inconsistent Systems and Dependent Equations

When both variables have been eliminated and the resulting equation is false, such as 0 = 5, there is no solution. The system is inconsistent.

When both variables have been eliminated and the resulting equation is true, such as 0 = 0, the equations are dependent, so there are an infinite number of solutions.


Objective 2

Solve applications using elimination.


ExampleThe length of Richard’s garden is 4 meters greater than 3 times the width. The perimeter of his garden is 72 meters. What are the dimensions of the garden?Understand We are to find the dimensions of the garden. The formula for the perimeter is P = 2l + 2w.Plan Translate the relationships to a system of equations, and then solve the system.Execute Let l = the length

Let w = the width


continued

Relationship 1 The length is 4 m more than 3 times the widthTranslation l = 3w + 4.

Relationship 2 The perimeter of the garden is 72 mTranslation 2l + 2w = 72.

Solve the system: 3 4

2 2 72

l w

l w

3 4

2 2 72

l w

l w

rewrite

Multiply by 2. 2 6 8

2 2 72

l w

l w

3 4

2 2 72

l w

l w


continued

2 6 8

2 2 72

l w

l w

8w 8 64w

Now we can find l using l = 3w + 4 l = 3(8) + 4 l = 28

Answer The garden has a length of 28 meters and a width of 8 meters.

Check 2l + 2w = 72 2(28) + 2(8) = 72 56 + 16 = 72 True


Solve the system.

a) (2, 3)

b) (7, 0)

c) (–2, 3)

d) (5, 5)

3 7

5 13

x y

x y


Solve the system.

a) (0, 10)

b) (1, 5)

c) (–2, 3)

d) no solution

5 5 50

2.5

x y

x y


Solve the system.

a) (1, 1)

b) (1, 5)

c) (–1, 1)

d) no solution

5 6 11

2 4 2

x y

x y


Solving Systems of Linear Equations in Three Variables9.49.4

1. Determine if an ordered triple is a solution for a system of equations.

2. Understand the types of solution sets for systems of three equations.

3. Solve a system of three linear equations using the elimination method.


ExampleDetermine whether (2, –1, 3) is a solution of the system 4,

2 2 3,

4 2 3.

x y z

x y z

x y z

Solution In all three equations, replace x with 2, y with –1, and z with 3.

x + y + z = 4

2 + (–1) + 3 = 4

4 = 4

TRUE

3 = 3

TRUE

2x – 2y – z = 3

2(2) – 2(–1) – 3 = 3

– 4x + y + 2z = –3

– 4(2) + (–1) + 2(3) = –3

–3 = –3

TRUEBecause (2, 1, 3) satisfies all three equations in the system, it is a solution for the system.


Example

Solution

Solve the system using elimination.

6,

2 2,

3 8.

x y z

x y z

x y z

We select any two of the three equations and work to get one equation in two variables. Let’s add equations (1) and (2):

(1)

(2)

(3)

6

2 2

x y z

x y z

(1)

(2)

(4)2x + 3y = 8Adding to eliminate z


Next, we select a different pair of equations and eliminate the same variable. Let’s use (2) and (3) to again eliminate z.

2 2

3 8

x y z

x y z

(5) x – y + 3z = 8

Multiplying equation (2) by 3

4x + 5y = 14.

3x + 6y – 3z = 6

Now we solve the resulting system of equations (4) and (5). That will give us two of the numbers in the solution of the original system,

4x + 5y = 14

2x + 3y = 8(5)

(4)


We multiply both sides of equation (4) by –2 and then add to equation (5):

Substituting into either equation (4) or (5) we find that x = 1.

4x + 5y = 14–4x – 6y = –16,

–y = –2 y = 2

Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.


Let’s use equation (1) and substitute our two numbers in it:

We have obtained the ordered triple (1, 2, 3). It should check in all three equations.

x + y + z = 6

1 + 2 + z = 6z = 3.

The solution is (1, 2, 3).


Solving Systems of Three Equations Using Elimination

1. Write each equation in the form Ax + By+ Cz = D.

2. Eliminate one variable from one pair of equations using the elimination method.

3. If necessary, eliminate the same variable from another pair of equations.

4. Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method.

5. To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the third variable.

6. Check the ordered triple in all three of the original equations.


Example

Solution

Solve the system using elimination.3 9 6 3

2 2

2

x y z

x y z

x y z

The equations are in standard form.

(1)

(2)

(3)

2 2

2

x y z

x y z

(2)

(3)

(4)3x + 2y = 4 Adding

Eliminate z from equations (2) and (3).


Eliminate z from equations (1) and (2).Multiplying equation (2) by 6

3 9 6 3

2 2

2

x y z

x y z

x y z

(1)

(2)

(3)

3 9 6 3

2 2

x y z

x y z

3 9 6 3

12 6 6 12

x y z

x y z

Adding15x + 15y = 15

Eliminate x from equations (4) and (5).

3x + 2y = 415x + 15y = 15

Multiplying top by 5

15x – 10y = 2015x + 15y = 15

Adding5y = 5y = 1

Using y = 1, find x from equation 4 by substituting.

3x + 2y = 43x + 2(1) = 4

x = 2

continued


continued

Substitute x = 2 and y = 1 to find z.x + y + z = 22 – 1 + z = 2

1 + z = 2 z = 1

The solution is the ordered triple (2, 1, 1).

3 9 6 3

2 2

2

x y z

x y z

x y z

(1)

(2)

(3)


ExampleAt a movie theatre, Kara buys one popcorn, two drinks and 2 candy bars, all for $12. Rebecca buys two popcorns, three drinks, and one candy bar for $17. Leah buys one popcorn, one drink and three candy bars for $11. Find the individual cost of one popcorn, one drink and one candy bar.Understand We have three unknowns and three relationships, and we are to find the cost of each.Plan Select a variable for each unknown, translate the relationship to a system of three equations, and then solve the system.


continuedExecute: p = popcorn, d = drink, and c = candyRelationship 1: one popcorn, two drinks and two candy

bars, cost $12Translation: p + 2d + 2c = 12

Relationship 2: two popcorns, three drinks, and one candy bar cost $17

Translation: 2p + 3d + c = 17Relationship 3: one popcorn, one drink and three

candy bars cost $11Translation: p + d + 3c = 11


continuedOur system:

Choose to eliminate p: Start with equations 1 and 3.

Choose to eliminate p: Start with equations 1 and 2.

2 2 12 (Equation 1)

2 3 17 (Equation 2)

3 11 (Equation 3)

p d c

p d c

p d c

Multiply by 1

2 2 12 2 2 12

3 11 3 11

p d c p d c

p d c p d c

��

1 (Equation 4)d c

Multiply by 22 2 12 2 4 4 24

2 3 17 2 3 17

p d c p d d

p d c p d c

��

3 7 (Equation 5)d c


continuedUse equations 4 and 5 to eliminate d.

1 (Equation 4)d c 3 7 (Equation 5)d c 4 6

1.5

c

c

Substitute for c in d – c = 11.5

.

1

2 5

1

cd

d

d

Substitute for c and d in p + d + 3c = 11

p + 2.5 + 3(1.5) = 11

p + 7 = 11

p = 4

The cost of one candy bar is $1.50.The cost of one drink is $2.50.The cost of one popcorn is $4.00.


Determine if (2, –5, 3) is a solution to the given system.

a) Yes

b) No

5 2 7

3 5 16

5

x y z

y z

z


Solve the system.

a) (–2, 2, –5)

b) (–5, 2, –2)

c) (–2, 5, 2)

d) no solution

2 4 1

5 2 9

3 2 14

x y z

x y z

x y z


A company makes 3 types of cable. Cable A requires 3 black, 3 white, and 2 red wires. B requires 1 black, 2 white, and 1 red. C requires 2 black, 1 white, and 2 red. They used 100 black, 110 white and 90 red wires. How many of each cable were made?

a) 10 cable A, 30 cable B, 20 cable C

b) 20 cable A, 30 cable B, 10 cable C

c) 10 cable A, 103 cable B, 20 cable C

d) 10 cable A, 30 cable B, 93 cable C


Solving Systems of Linear Equations Using Matrices9.59.5

1. Write a system of equations as an augmented matrix.2. Solve a system of linear equations by transforming its

augmented matrix to echelon form.


Matrix: A rectangular arrow of numbers.

The following are examples of matrices:

1 1 6 9 1/ 2 01 2

, 4 , 4 3 1 7 67 5

0 6 5 9 1 2

The individual numbers are called elements.


The rows of a matrix are horizontal, and the columns are vertical.

1 9 0

4 1 6

6 9 2

column 1 column 2 column 3

row 1

row 2

row 3


Augmented Matrix: A matrix made up of the coefficients and the constant terms of a system. The constant terms are separated from the coefficients by a dashed vertical line.

3 7

3 1

x y

x y

3 1 7

1 3 1

Let’s write this system as an augmented matrix:


Row OperationsThe solution of a system is not affected by the following row operations in its augmented matrix: 1. Any two rows may be interchanged. 2. The elements of any row may be multiplied (or divided) by any nonzero real number.3. Any row may be replaced by a row resulting from

adding the elements of that row (or multiples of that row) to a multiple of the elements of any other row.


Echelon form: An augmented matrix whose coefficient portion has 1s on the diagonal from upper left to lower right and 0s below the 1s.

1 1 1 1

0 1 1 2

0 0 1 1


Solution

Example

3 7

3 1

x y

x y

Solve the following linear system by transforming its augmented matrix into echelon form.

We write the augmented matrix. 3 1 7

1 3 1

We perform row operations to transform the matrix into echelon form.

3R2 + R13 1 7

0 10 10


The resulting matrix represents the system:

R2 10 3 1 7

0 1 1

continued3 1 7

0 10 10

3 7

1

x y

y

Since y = 1, we can solve for x using substitution.


3x + 1 = 7

3x + y = 7

3x = 6

x = 2


Example 2 8

1

2 2

x y z

x y z

x y z

SolutionWrite the augmented matrix.

2 1 1 8

1 1 1 1

1 2 1 2

Use the echelon method to solve


Our goal is to transform

into the form

2 1 1 8

1 1 1 1

1 2 1 2

0 .

0 0

a b c d

e f g

h i

Interchange Row 1 and Row 21 1 1 1

2 1 1 8

1 2 1 2

1 1 1 1

0 3 3 6

1 2 1 2

–2R1 + R2

continued


1 1 1 1

0 1 1 2

0 1 2 1

1 1 1 1

0 1 1 2

0 0 3 3

–R2 + R3

(1/3)R2

R1 + R3

1 1 1 1

0 1 1 2

0 0 1 1

(1/3)R3

continued


z = –1.

Substitute z into y – z = 2 y –(1) = 2

y + 1 = 2 y = 1

continued

Substitute y and z into x – y + z = 1 x – 1 – 1 = 1

x – 2 = 1 x = 3

The solution is (3, 1, –1).


Replace R2 in with R1 + R2.

a) b)

c) d)

8 6 4

2 4 0

10 10 4

10 10 4

8 6 4

10 10 4

8 14 2

2 6 0

10 10 4

2 4 0


Solve by transforming the augmented matrix into echelon form.

a) (2, 1, 3) b) (3, 1, 2)

c) (3, 2, 1) d) No solution

0

4 11

4 6

x y z

x y z

x y z


Solving Systems of Linear Equations Using Cramer’s Rule9.69.6

1. Evaluate determinants of 2 2 matrices.2. Evaluate determinants of 3 3 matrices.3. Solve systems of equations using Cramer’s Rule.


Square matrix: A matrix that has the same number of rows and columns.

Every square matrix has a determinant.

Determinant of a 2 2 Matrix

If then det(A) = 1 1

2 2

,a b

Aa b

1 1

1 2 2 12 2

.a b

a b a ba b

12 1

11

22

2

aa bb

b

ba

a


ExampleFind the determinant of the following matrix.

3 2

7 1A

Solution

3

1

2

7

3 14 17

(3)(1) (7)( 2)


Minor: The determinant of the remaining matrix when the row and column in which the element is located are ignored.


ExampleFind the minor of 4 in

Solution To find the minor of 4, we ignore its row and column (shown in blue) and evaluate the determinant of the remaining matrix (shown in red).

3 4 5

2 6 1 .

4 3 1

3 4 5

2 6 1

4 3 1

( 2)(1) (4)( 1)

2 4

2


Determinant of a 3 3 Matrix

If1 1 1

2 2 2 1 2 331 2

3 3 3

minorminor minor

of of of

a b c

A a b c a a aaa a

a b c

2 2 1 1 1 11 2 3

3 3 3 3 2 2

b c b c b ca a a

b c b c b c


Cramer’s Rule

The solution to the system of linear equations

1 1 1

2 2 2

isa x b y c

a x b y c

1 1

2 2

1 1

2 2

andx

c b

c b Dx

a b D

a b

1 1

2 2

1 1

2 2

y

a c

Da cy

a b D

a b


The solution to the system of linear equations

1 1 1 1

2 2 2 2

3 3 3 3

is

a x b y c z d

a x b y c z d

a x b y c z d

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

,x

d b c

d b c

d b c Dx

a b c D

a b c

a b c

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

andy

a d c

a d c

Da d cy

a b c D

a b c

a b c

1 1 1

2 2 2

3 3 3

1 1 1

2 2 2

3 3 3

z

a b d

a b d

a b d Dz

a b c D

a b c

a b c


Solution

Example 6 2,

2 3 2.

x y

x y

Use Cramer’s Rule to solve

First, we find D, Dx, and Dy.6 1

2 3D

6( 3) (2)(1)

18 2

20

2 1

2 3xD

2( 3) ( 2)(1)

6 2

4

6 2

2 2yD

6( 2) (2)(2)

12 4

16

4 1

20 5xD

xD

16 4

20 5yD

yD

The solution is (1/5, 4/5).


Solution

Example 2 8

1

2 2

x y z

x y z

x y z

Use Cramer’s Rule to solve

2 1 1

1 1 1

1 2 1

D

We need to find D, Dx, Dy, and Dz.

1 1 1 1 1 1(2) (1) ( 1)

2 1 2 1 1 1

(2)( 1 2) (1)(1 2) ( 1)(1 1)

6 3 0

9


continued

2 8 1

1 1 1

1 2 1yD

8 1 1

1 1 1

2 2 1xD

2 8

1

2 2

x y z

x y z

x y z

1 1 1 1 1 1(8) (1) ( 2)

2 1 2 1 1 1

(8)( 1 2) (1)(1 2) ( 2)(1 1)

24 3 0 27

1 1 8 1 8 1(2) (1) ( 1)

2 1 2 1 1 1

(2)(1 2) (1)(8 2) ( 1)(8 1)

6 6 9 9


continued

2 8

1

2 2

x y z

x y z

x y z

2 1 8

1 1 1

1 2 2zD

27 9 93, 1, 1.

9 9 9yx z

DD Dx y z

D D D

The solution is (3, 1, –1). The check is left to the student.

1 1 1 8 1 8(2) (1) ( 1)

2 2 2 2 1 1

(2)(2 2) (1)( 2 16) ( 1)(1 8)

0 18 9 9


Find the determinant.

a) 75

b) 72

c) 27

d) 72

9 3

0 8


Solve using Cramer’s Rule.

a) (1, 4)

b) (1, 3)

c) (2, 3)

d) No solution

5 19

2 6 22

x y

x y


Solving Systems of Linear Inequalities9.79.7

1. Graph the solution set of a system of linear inequalities.2. Solve applications involving a system of linear

inequalities.


Objective 1

Graph the solution set of a system of linear inequalities.


Solving a System of Linear Inequalities

To solve a system of linear inequalities, graph all of the inequalities on the same grid. The solution set for the system contains all ordered pairs in the region where the inequalities’ solution sets overlap along with ordered pairs on the portion of any solid line that touches the region of overlap.


ExampleGraph the solution set for the system of inequalities.

SolutionGraph the inequalities on thesame grid.Both lines will be dashed.

3 2

2 5y

y x

x

Solution


ExampleGraph the solution set for the system of inequalities.

SolutionGraph the inequalities on thesame grid.

3 6

3

x y

y

Solution


Example—Inconsistent SystemGraph the solution set for the system of inequalities.

SolutionGraph the inequalities on thesame grid.

2 3

23

33x y

y x

The slopes are equal, so the lines are parallel. Since the shaded regions do not overlap there is no solution for the system.


Objective 2

Solve applications involving a system of linear inequalities.


ExampleMr. Reynolds is landscaping his yard. He is looking at some trees and bushes. He would like to purchase at least 3 plants. The trees cost $40 and the bushes cost $20. He cannot spend more than $300 for the plants. Write a system of inequalities that describes what Mr. Reynolds could purchase, then solve the system by graphing.

Understand We must translate to a system of inequalities, and then solve the system.


continuedPlan and ExecuteLet x represent the trees and y represent the bushes.

Relationship 1: Mr. Reynolds would like to purchase at least 3 plants.

x + y 3

Relationship 2: Mr. Reynolds cannot spend more than $300.

40x + 20y 300


continuedAnswerSince Mr. Reynolds cannot purchase negative plants, the solution set is confined to Quadrant 1.

40 20

3

300x y

x y

Any ordered pair in the overlapping region is a solution. Assuming that only whole trees and bushes can be purchased, only whole numbers would be in the solution set. For example: (4, 2); (3, 4)


Graph.

a) b)

c) d)

2

4

x y

x y

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