Transcript 09-Systems of Linear Equations and Inequalities

CHAPTER
9
9.1
9.2
9.3
9.4
9.5
9.6
9.7
Systems of Linear Equations and
Inequalities
Solving Systems of Linear Equations Graphically
Solving Systems of Linear Equations by Substitution;
Applications
Solving Systems of Linear Equations by Elimination;
Applications
Solving Systems of Linear Equations in Three Variables;
Applications
Solving Systems of Linear Equations Using Matrices
Solving Systems of Linear Equations Using Cramer’s
Rule
Solving Systems of Linear Inequalities
Copyright © 2011 Pearson Education, Inc.
9.1
Solving Systems of Linear Equations
Graphically
1. Determine whether 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.
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System of equations: A group of two or more
equations.
 x y 5

3x  4 y  8
(Equation 1)
(Equation 2)
Solution for a system of equations: An ordered set of
numbers that makes all equations in the system.
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Slide 9- 3
Checking a Solution to a System of Equations
To 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.
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Slide 9- 4
Example 1
Determine whether each ordered pair is a solution to
the system of equations.
x  y  7

 y  3x  2
a. (3, 2)
Solution
a. (3, 2)
(Equation 1)
(Equation 2)
b) (3, 7)
x+y=7
y = 3x – 2
3 + 2 = 7
2 = 3(3) – 2
1 = 7
2 = 11
False
False
Because (3, 2) does not satisfy both equations, it is not a
solution for the system.
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Slide 9- 5
continued
b. (3, 4)
x+y=7
3+4=7
7=7
True
y = 3x – 2
4 = 3(3) – 2
4=7
False
Because (3, 4) does not satisfy both equations, it is not
a solution to the system of equations.
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Slide 9- 6
A system of two linear equations in two variables can
have one solution, no solution, or an infinite number of
solutions.
x  y  5

 y  2x  4
The graphs intersect
at a single point.
There is one
solution.
 y  3x  1

 y  3x  2
The equations have
the same slope, the
graphs are parallel.
There is no solution.
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x  2 y  4

2 x  4 y  8
The graphs are
identical. There are
an infinite number of
solutions.
Slide 9- 7
Solving Systems of Equations Graphically
To 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, there is no solution.
c. If the lines are identical, there are an infinite
number of solutions, which are the coordinates of
all the points on that line.
2. Check your solution.
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Slide 9- 8
Example 2a
Solve the system of equations graphically.
y  2  x

2 x  4 y  12
(Equation 1)
(Equation 2)
Solution
Graph each equation.
The lines intersect at a single
point, (2, 4).
We can check the point in each
equation to verify and we will
leave that to you.
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(2, 4)
2x + 4y = 12
y=2–x
Slide 9- 9
Example 2b
Solve the system of equations graphically.
3

y  x 3
4


3x  4 y  4
(Equation 1)
(Equation 2)
Solution
Graph each equation.
The lines appear to be parallel,
which we can verify by
comparing the slopes. 3x  4 y  4
The slopes are the
4 y  3x  4
same, so the lines are
3
parallel. The system
y  x 1
has no solution
4
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Slide 9- 10
Example 2c
Solve the system of equations graphically.
4 x  2 y  2

 y  2x 1
(Equation 1)
(Equation 2)
Solution
Graph each equation.
The lines appear to be identical.
4x  2 y  2
2 y  2  4 x
y  1  2 x
y  2x 1
The equations are identical. All
ordered pairs along the line are
solutions.
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Slide 9- 11
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.
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Slide 9- 12
Classifying Systems of Equations
To 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 with
independent equations:
The system has a single
solution at the point of
intersection.
The graphs are different.
They have different slopes.
Consistent system with
dependent equations: The
system has an infinite
number of solutions.
The graphs are identical.
They have the same slope
and same y-intercept.
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Inconsistent system: The
system has no solution.
The graphs are parallel
lines. They have the same
slope, but different yintercepts.
Slide 9- 13
Example 3
For each of the systems of equations in Example 2,
determine whether the system is consistent with
independent equations, consistent with dependent
equations, or inconsistent. How many solutions does
the system have?
(Equation 1)
2a.  y  2  x

2 x  4 y  12
(Equation 2)
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).
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Slide 9- 14
Example 3
For each of the systems of equations in Example 2,
determine whether the system is consistent with
independent equations, consistent with dependent
equations, or inconsistent. How many solutions does
the system have?
2b.  y  3 x  3
(Equation 1)
4


3x  4 y  4
(Equation 2)
The graphs were parallel lines. The system is
inconsistent and has no solutions.
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Slide 9- 15
Example 3
For each of the systems of equations in Example 2,
determine whether the system is consistent with
independent equations, consistent with dependent
equations, or inconsistent. How many solutions does
the system have?
2c. 4 x  2 y  2
(Equation 1)

 y  2x 1
(Equation 2)
The graphs coincide. The system is consistent (it
has a solution) with dependent equations (same
graph) and has an infinite number of solutions.
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Slide 9- 16
Which set of points is a solution to the
system?
2 x  4 y  2

3x  2 y  1
a) (–1, 1)
b) (–1, –1)
c) (0, 2)
d) (–3, 7)
9.1
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Slide 5- 17
Which set of points is a solution to the
system?
2 x  4 y  2

3x  2 y  1
a) (–1, 1)
b) (–1, –1)
c) (0, 2)
d) (–3, 7)
9.1
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Slide 5- 18
9.2
Solving Systems of Linear Equations by
Substitution; Applications
1. Solve systems of linear equations using substitution.
2. Solve applications involving two unknowns using a
system of equations.
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Solving Systems of Two Equations Using Substitution
To 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.
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Slide 9- 20
Example 1
Solve the system of equations using substitution.
4 x  5 y  8

x  1 y
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.
4x  5 y  8
4(1  y )  5 y  8
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Slide 9- 21
4 x  5 y  8

x  1 y
continued
Step 3: Solve for y.
4(1  y )  5 y  8
4  4y  5y  8
4 y 8
y4
Step 4: Solve for x by substituting 4 for y in one of
the original equations.
x  1 y
x  1 4
x  3
The solution is (3, 4).
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Slide 9- 22
Example 2
Solve the system of equations using substitution.
2 x  y  7

 x  y  1
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.
x  y  1
x  (7  2 x)  1
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Slide 9- 23
2 x  y  7

 x  y  1
continued
Step 3: Solve for x.
x  (7  2 x)  1
x  7  2x  1
3x  7  1
3x  6
x2
Step 4: Solve for y by substituting 2 for x in one of
the original equations.
2x  y  7
2(2)  y  7
4 y  7
The solution is (2, 3).
y3
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Slide 9- 24
Inconsistent Systems of Equations
The 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 Equations
The 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.
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Slide 9- 25
Example 4
Solve the system of equations using substitution.
3x  y  4

 y  3x  5
Solution
Substitute y = 3x – 5 into the first equation.
3 x  y  4
3 x  (3 x  5)  4
3x  3x  5  4
5  4
False statement. The
system is inconsistent
and has no solution.
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Slide 9- 26
Example 5
Solve the system of equations using substitution.
 y  4  3x

6 x  2 y  8
Solution
Substitute y = 4 – 3x into the second equation.
6 x  2 y  8
6 x  2(4  3x)  8
6x  8  6x  8
6x  6x  8  8
8  8
True statement. The
number of solutions is
infinite.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 27
Solving Applications Using a System of Equations
To 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.
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Slide 9- 28
Example 6
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.
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Slide 9- 29
continued
Execute Let l represent the length and w the width.
Relationship 1: The length is twice the width.
Translation: l = 2w
Relationship 2: The perimeter is 216 inches.
Translation: 2l + 2w = 216
System:
l  2w

2l  2w  216
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Slide 9- 30
continued
Solve.
l  2w

2l  2w  216
2(2w)  2w  216
4w  2w  216
6w  216
w  36
Substitute 2w for l.
Combine like terms.
Divide both sides by 6 to isolate w.
Now find the value of l. l = 2w
= 2(36)
= 72
Answer The length should be 72 inches and the width
36 inches.
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Slide 9- 31
Example 8
Anita and Ernesto are traveling north in separate cars
on the same highway. Anita is traveling at 55 miles per
hour and Ernesto is traveling at 70 miles per hour.
Anita passes Exit 54 at 1:30 p.m. Ernesto passes the
same exit at 1:45 p.m. At what time will Ernesto catch
up with Anita?
Understand To determine what time Ernesto will
catch up with Anita, we need to calculate the amount
of time it will take him to catch up to her. We can then
add the amount to 1:45.
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Slide 9- 32
continued
Plan and Execute
Let x = Anita’s travel time after passing the exit.
Let y = Ernesto’s travel time after passing the exit.
Category
Anita
Rate
55
Time
x
Distance
55x
Ernesto
70
y
70y
Relationship 1: Ernesto passes the exit 15 minutes after Anita;
Anita will have traveled 15 minutes longer.
Translation: x = y + ¼ (1/4 of an hour)
Relationship 2: When Ernesto catches up, they will have
traveled the same distance.
Translation: 55x = 70y
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Slide 9- 33
continued
1

x  y 
Our system: 
4
55 x  70 y
Use substitution:
1
x y
4
55 x  70 y
1

55  y    70 y
4

55
55 y 
 70 y
4
13.75  15 y
1
Substitute to find x. x  y 
4
x  0.92  0.25
0.92  y
x  1.17
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Slide 9- 34
continued
Answer
Ernesto will catch up to Anita in a little over 1 hour
(1.17, which is 1 hour 10 minutes). The time will be
1:45 p.m. + 1 hr 10 minutes = 2:55 p.m.
Check
Verify both given relationships.
1
x y
4
1.17  0.92  0.25
1.17  1.17
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55 x  70 y
55(1.17)  70(0.92)
64.35  64.4
Slide 9- 35
Solve the system.
x  y  4

2 x  y  6
a) (10, 3)
b) (1, 4)
c) (10, 14)
d) (3, 6)
9.2
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Slide 5- 36
Solve the system.
x  y  4

2 x  y  6
a) (10, 3)
b) (1, 4)
c) (10, 14)
d) (3, 6)
9.2
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Slide 5- 37
Solve the system.
 x  3 y  16

7 x  4 y  10
a) (2, 3)
b) (2, 6)
c) (2, 6)
d) (3, 2)
9.2
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Slide 5- 38
Solve the system.
 x  3 y  16

7 x  4 y  10
a) (2, 3)
b) (2, 6)
c) (2, 6)
d) (3, 2)
9.2
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Slide 5- 39
9.3
Solving Systems of Linear Equations by
Elimination; Applications
1. Solve systems of linear equations using elimination.
2. Solve applications using elimination.
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x  y  9

3x  y  7
Example 1
Solve the system of equations using elimination.
Solution We can add the equations.
x y 9
3x  y  7
4x  0  16
4x  16
x4
Notice that y is eliminated,
so we can easily solve for
the value of x.
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=9
The solution is (4, 5).
y=5
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Slide 9- 41
Example 2
x  y  8
Solve the system of equations. 3x  4 y  11

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.
x y 8
Multiply the first equation by 4.
4 x  4 y  32
3 x  4 y  11
7 x  0  21
7 x  21
x3
4  x  4  y  4 8
4 x  4 y  32
Solve for y.
x+y=8
3+y=8
y=5
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The solution
is (3, 5).
Slide 9- 42
Example 3
4 x  5 y  19

3 x  2 y  9
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 x  5 y  19
3x  2 y  9
Multiply by 2.
Multiply by 5.
8 x  10 y  38
15 x  10 y  45
Add the equations to
eliminate y.
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7 x  0  7
7 x  7
x 1
Slide 9- 43
4 x  5 y  19

3 x  2 y  9
continued
Substitute x = 1 into either of the original equations.
4x – 5y = 19
4(1) – 5y = 19
4 – 5y = 19
–5y = 15
y = 3
The solution is (1, –3).
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Slide 9- 44
Example 4
Solve the system of equations.
0.03 x  0.02 y  0.03

2
2
4
 5 x  5 y  5
Solution
To clear the decimals in Equation 1, multiply by 100.
To clear the fractions in Equation 2, multiply by 5.
0.03x  0.02 y  0.03
4
2
2
x y 
5
5
5
Multiply by 100.
Multiply by 5.
Multiply equation 2 by 1
then combine the
equations.
3x  2 y  3
4x  2 y  2
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3x  2 y  3
4x  2 y  2
3x  2 y  3
4 x  2 y  2
Slide 9- 45
continued
3x  2 y  3
4 x  2 y  2
x  0  1
x  1
x  1
Substitute to find y.
3x  2 y  3
3(1)  2 y  3
3  2 y  3
2 y  6
y  3
The solution is (1, 3).
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Slide 9- 46
Solving Systems of Two Linear Equations Using Elimination
To 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
(optional).
3. If necessary, 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.
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Slide 9- 47
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. There are an infinite number of
solutions.
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Slide 9- 48
Example 5a
Solve the system of equations.
 3 x  y  4

3x  y  5
Solution
Notice that the left side of the equations are additive
inverses. Adding the equations will eliminate both
variables.
3 x  y  4
3 x  y   5
0  9
False statement. The
system is inconsistent
and has no solution.
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Slide 9- 49
Example 5b
Solve the system of equations.
 3x  y  4

6 x  2 y  8
Solution
To eliminate y multiply the first equation by 2.
3x  y  4
6 x  2 y  8
Multiply by 2.
6x  2 y  8
6 x  2 y  8
00
True statement. The equations are dependent.
There are an infinite number of solutions.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 9- 50
Example 7
Mia sells concessions at a movie theatre. In one hour,
she sells 78 popcorns. The popcorn sizes are small,
which sell for $4 each, and large, which sell for $6
each. If her sales totaled $420, then how many of each
size popcorn did she sell?
Understand The unknowns are the number of each
size popcorn.
One relationship involves the number of popcorns (78
total), and the other relationship involves the total sales
in dollars ($420).
Copyright © 2011 Pearson Education, Inc.
Slide 9- 51
continued
Plan and Execute
Let x represent the number of small popcorns.
Let y represent the number of large popcorns.
Category
Small
Large
Price
$4
$6
Number
x
y
Revenue
4x
6y
Relationship 1: The total number sold is 78.
Translation: x + y = 78
Relationship 2: The total sales revenue is $420.
Translation: 4x + 6y = 420
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Slide 9- 52
continued
 x  y  78
Our system: 
4 x  6 y  420
Use elimination; choose x.
x  y  78 Multiply by 4.
4 x  6 y  420

4 x  4 y  312
4 x  6 y  420
2 y  108
y  54
Substitute to find x.
x  y  78
x  54  78
x  24
Copyright © 2011 Pearson Education, Inc.
Slide 9- 53
continued
Answer
Mia sold 54 large popcorns and 24 small popcorns.
Check
Verify both given relationships.
x  y  78
24  54  78
78  78
4 x  6 y  420
4(24)  6(54)  420
96  324  420
420  420
Copyright © 2011 Pearson Education, Inc.
Slide 9- 54
Example 8
How many milliliters of a 20% HCl solution and 50%
HCl solution must be mixed together to make 500
milliliters of 35% HCl solution?
Understand The unknowns are the volumes of 20%
and 50% solution that are mixed.
One relationship involves the concentrations of each
solution in the mixture and the other relationship
involves the total volume of the final mixture (500 ml).
Copyright © 2011 Pearson Education, Inc.
Slide 9- 55
continued
Plan and Execute
Let x and y represent the two amounts to be mixed.
Solution
Concentration
Volume
Amount of HCl
20%
0.20
x
0.20x
50%
0.50
y
0.50y
35%
0.35
x+y
0.35(500)
Relationship 1: The total volume is 500 ml.
Translation: x + y = 500
Relationship 2: The combined volumes of HCl in the
two mixtures is to be 35% of the total mixture.
Translation: 0.20x + 0.50y = 0.35(500)
Copyright © 2011 Pearson Education, Inc.
Slide 9- 56
continued
 x  y  500
Our system: 
0.20 x  0.50 y  175
Use elimination; choose x.
x  y  500 Multiply by 0.20.
0.20 x  0.50 y  175

0.20 x  0.20 y  100
0.20 x  0.50 y  175
0.30 y  75
y  250
Substitute to find x.
x  y  500
x  250  500
x  250
Copyright © 2011 Pearson Education, Inc.
Slide 9- 57
continued
Answer
Mixing 250 ml of 20% solution with 250 ml of 50%
solution gives 500 ml of 35% solution.
Check
Verify both given relationships.
x  y  500
250  250  500
500  500
0.20 x  0.50 y  175
0.20(250)  0.50(250)  175
50  125  175
175  175
Copyright © 2011 Pearson Education, Inc.
Slide 9- 58
Solve the system.
x  3y  7

 x  5 y  13
a) (2, 3)
b) (7, 0)
c) (–2, 3)
d) (5, 5)
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 59
Solve the system.
x  3y  7

 x  5 y  13
a) (2, 3)
b) (7, 0)
c) (–2, 3)
d) (5, 5)
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 60
Solve the
5 x  5 y  50
system. 
 x  y  2.5
a) (0, 10)
b) (1, 5)
c) (–2, 3)
d) no solution
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 61
Solve the
5 x  5 y  50
system. 
 x  y  2.5
a) (0, 10)
b) (1, 5)
c) (–2, 3)
d) no solution
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 62
Solve the
5 x  6 y  11
system. 
2 x  4 y  2
a) (1, 1)
b) (1, 5)
c) (–1, 1)
d) no solution
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 63
Solve the
5 x  6 y  11
system. 
2 x  4 y  2
a) (1, 1)
b) (1, 5)
c) (–1, 1)
d) no solution
9.3
Copyright © 2011 Pearson Education, Inc.
Slide 5- 64
9.4
Solving Systems of Linear Equations in
Three Variables; Applications
1. Determine whether 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.
4. Solve application problems that translate to a system of
three linear equations.
Copyright © 2011 Pearson Education, Inc.
Example 1
Determine whether (2, –1, 3) is a solution of the
x  y  z  4,
system
2 x  2 y  z  3,
4 x  y  2 z  3.
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
2x – 2y – z = 3
– 4x + y + 2z = –3
2(2) – 2(–1) – 3 = 3 – 4(2) + (–1) + 2(3) = –3
3=3
–3 = –3
Because (2, 1, 3) satisfies all three equations in
TRUE
TRUE
TRUE
the system, it is a solution for the system.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 66
Types of Solution Sets
A Single Solution:
If the planes intersect at a
single point, that ordered
triple is the solution to the
system.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 67
Types of Solution Sets
Infinite Number of Solutions:
If the three planes intersect
along a line, the system has an
infinite number of solutions,
which are the coordinates of
any point along that line.
Infinite Number of Solutions:
If all three graphs are the same
plane, the system has an
infinite number of solutions,
which are the coordinates of
any point in the plane.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 68
Types of Solution Sets
No Solution:
If all of the planes are
parallel, the system has
no solution.
No Solution:
Pairs of planes also can
intersect, as shown.
However, because all three
planes do not have a
common intersection, the
system has no solution.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 69
Example 2a
Solve the system using elimination.
x  y  z  6,
x  2 y  z  2,
x  y  3z  8.
(1)
(2)
(3)
Solution
We select any two of the three equations and
work to get one equation in two variables. Let’s
add equations (1) and (2):
x yz6
(1)
(2)
x  2y  z  2
Adding to
(4)
2x + 3y
=8
eliminate z
Copyright © 2011 Pearson Education, Inc.
Slide 9- 70
Next, we select a different pair of equations and
eliminate the same variable. Let’s use (2) and (3) to
again eliminate z.
x  2y  z  2
x  y  3z  8
Multiplying
equation (2)
by 3
3x + 6y – 3z = 6
x – y + 3z = 8
4x + 5y
= 14.
(5)
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,
(4)
2x + 3y = 8
(5)
4x + 5y = 14
Copyright © 2011 Pearson Education, Inc.
Slide 9- 71
We multiply both sides of equation (4) by –2 and
then add to equation (5):
–4x – 6y = –16,
4x + 5y = 14
–y = –2
y=2
Substituting into either equation (4) or (5) we find
that x = 1.
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.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 72
Let’s use equation (1) and substitute our two
numbers in it:
x+y+z=6
1+2+z=6
z = 3.
We have obtained the ordered triple (1, 2, 3). It
should check in all three equations.
The solution is (1, 2, 3).
Copyright © 2011 Pearson Education, Inc.
Slide 9- 73
Example 2b
Solve the system using elimination.
3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
x  y  z  2 (3)
Solution
The equations are in standard form.
Eliminate z from equations (2) and (3).
2 x  y  z  2 (2)
x  y  z  2 (3)
(4)
3x + 2y = 4
Adding
Copyright © 2011 Pearson Education, Inc.
Slide 9- 74
3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
Eliminate z from equations (1) and (2). x  y  z  2 (3)
continued
3x  9 y  6 z  3
2x  y  z  2
Multiplying
equation (2) by 6
3x  9 y  6 z  3
12 x  6 y  6 z  12
15x + 15y = 15
Adding
Eliminate x from equations (4) and (5).
3x + 2y = 4
15x + 15y = 15
Multiplying top
by 5
15x – 10y = 20
15x + 15y = 15
5y = 5
y = 1
Adding
Using y = 1, find x from equation 4 by substituting.
3x + 2y = 4
3x + 2(1) = 4
x=2
Copyright © 2011 Pearson Education, Inc.
Slide 9- 75
3x  9 y  6 z  3 (1)
2 x  y  z  2 (2)
x  y  z  2 (3)
continued
Substitute x = 2 and y = 1 to find z.
x+y+z=2
2–1+z=2
1+z=2
z=1
The solution is the ordered triple (2, 1, 1).
Copyright © 2011 Pearson Education, Inc.
Slide 9- 76
Solving Systems of Three Linear Equations
Using Elimination
To solve a system of three linear equations with
three unknowns 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.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 77
continued
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 original
equations.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 78
Example 3
Solve the system using elimination.
x  3y  z  1
(1)
(2)
2x  y  2z  2
x  2 y  3 z  1 (3)
Solution
The equations are in standard form.
Eliminate x from equations (1) and (2).
2 x  6 y  2 z  2 (1)
2 x  y  2 z  2 (2)
(4)
5y  4z = 0
Adding
Copyright © 2011 Pearson Education, Inc.
Slide 9- 79
x  3 y  z  1 (1)
2 x  y  2 z  2 (2)
Eliminate x from equations (1) and (3). x  2 y  3z  1(3)
continued
x  3y  z  1
 x  2 y  3z  1
5y + 4z = 2
(5)
Eliminate y from equations (4) and (5).
5y  4z = 0
5y + 4z = 2
0=2
All variables are eliminated and the resulting equation
is false, which means that this system has no solution;
it is inconsistent.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 80
Example 4
At 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.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 81
continued
Execute: p = popcorn, d = drink, and c = candy
Relationship 1: one popcorn, two drinks and two candy
bars, cost $12
Translation: p + 2d + 2c = 12
Relationship 2: two popcorns, three drinks, and one
candy bar cost $17
Translation: 2p + 3d + c = 17
Relationship 3: one popcorn, one drink and three
candy bars cost $11
Translation: p + d + 3c = 11
Copyright © 2011 Pearson Education, Inc.
Slide 9- 82
continued
p  2d  2c  12
Our system: 2 p  3d  c  17
p  d  3c  11
(Equation 1)
(Equation 2)
(Equation 3)
Choose to eliminate p: Start with equations 1 and 3.
p  2d  2c  12
p  d  3c  11 Multiply by
1
p  2d  2c  12
 p  d  3c  11
d c 1
(Equation 4)
Choose to eliminate p: Start with equations 1 and 2.
p  2d  2c  12 Multiply by
2 p  3d  c  17
2
2 p  4d  4d  24
2 p  3d  c  17
d  3c  7
(Equation 5)
Copyright © 2011 Pearson Education, Inc.
Slide 9- 83
continued
Use equations 4 and 5 to eliminate d.
d c 1
(Equation 4)
d  3c  7
(Equation 5)
4c  6
c  1.5
Substitute for c in d – c = 1
d c 1
d  1.5  1
d  2.5
Substitute
The for
costc of
and
one
d in
candy
p + dbar
+ is
3c$1.50.
= 11
The cost of one
$2.50.= 11
p +drink
2.5 +is3(1.5)
The cost of one popcorn is $4.00.
p + 7 = 11
p =4
Copyright © 2011 Pearson Education, Inc.
Slide 9- 84
Determine if (2, –5, 3) is a solution to
the given system.
x  5 y  2z  7

 3 y  5 z  16

z  5

a) Yes
b) No
9.4
Copyright © 2011 Pearson Education, Inc.
Slide 4- 85
Determine if (2, –5, 3) is a solution to
the given system.
x  5 y  2z  7

 3 y  5 z  16

z  5

a) Yes
b) No
9.4
Copyright © 2011 Pearson Education, Inc.
Slide 4- 86
Solve the system.
2 x  4 y  z  1

5 x  2 y  z  9
3x  y  2 z  14

a) (–2, 2, –5)
b) (–5, 2, –2)
c) (–2, 5, 2)
d) no solution
9.4
Copyright © 2011 Pearson Education, Inc.
Slide 4- 87
Solve the system.
2 x  4 y  z  1

5 x  2 y  z  9
3x  y  2 z  14

a) (–2, 2, –5)
b) (–5, 2, –2)
c) (–2, 5, 2)
d) no solution
9.4
Copyright © 2011 Pearson Education, Inc.
Slide 4- 88
9.5
Solving Systems of Linear Equations
Using Matrices
1. Write a system of equations as an augmented matrix.
2. Solve a system of linear equations by transforming its
augmented matrix into row echelon form.
Copyright © 2011 Pearson Education, Inc.
Matrix: A rectangular array of numbers.
The following are examples of matrices:
1  1 6 9 1/ 2 0 
1 2    

7
6
7 5  ,  4 ,  4 3 1


   
 0   6 5 9 1 2 
The individual numbers are called elements.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 90
The rows of a matrix are horizontal, and the
columns are vertical.
1
4

 6
9
1
9
0

6

2 
column 1
column 2
column 3
Copyright © 2011 Pearson Education, Inc.
row 1
row 2
row 3
Slide 9- 91
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.
Let’s write this system as an augmented matrix:
3x  y  7

 x  3 y  1
3 1 7 
 1 3 1 


Copyright © 2011 Pearson Education, Inc.
Slide 9- 92
Example 1
2 x  y  z  8

Write as an augmented matrix.  x  y  z  1
 x  2 y  z  2

Solution
Write the augmented matrix.
1 1 8 
2
1

1 1 1


 1 2 1 2 
Copyright © 2011 Pearson Education, Inc.
Slide 9- 93
Row Operations
The 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.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 94
Row echelon form: An augmented matrix whose
coefficient portion has 1’s on the diagonal from
upper left to lower right and 0s below the 1’s.
1 1 1 1 
0 1
1 2 


0 0 1
1
Copyright © 2011 Pearson Education, Inc.
Slide 9- 95
Example 2
Solve the following linear system by transforming
its augmented matrix into row echelon form.
3x  y  7

Solution
 x  3 y  1
We write the augmented matrix. 3 1 7 
 1 3 1 


We perform row operations to transform the
matrix into echelon form.
3R2 + R1
3 1 7 
0 10 10 


Copyright © 2011 Pearson Education, Inc.
Slide 9- 96
3 1 7 
0 10 10 


continued
3 1 7 
0 1 1 


3x  y  7

The resulting matrix represents the system: 
R2  10

y 1
Since y = 1, we can solve for x using substitution.
3x + y = 7
3x + 1 = 7
3x = 6
x=2
The solution is (2, 1).
Copyright © 2011 Pearson Education, Inc.
Slide 9- 97
Example 3
Use the row echelon method to solve
2 x  y  z  8

.
x  y  z  1
Solution
 x  2 y  z  2

Write the augmented matrix:
1 1 8 
2
1
1 1 1  .


 1 2 1 2 
Copyright © 2011 Pearson Education, Inc.
Slide 9- 98
continued
Our goal is to transform
1 8 
2 1
1
1 1 1  into the form


 1 2 1
2 
a b c
0 e f

0 0 h
d
g .

i 
1 1 1 
1
Interchange Row 1 and Row 2
2 1

1 8


 1 2 1
2 
1 1 1 
1
0 3 3 6 


 1 2 1
2 
–2R1 + R2
Copyright © 2011 Pearson Education, Inc.
Slide 9- 99
continued
1 1 1 1 
0 1
1 2 


0 1
2 1
(1/3)R2
R1 + R3
1 1 1 1 
0 1
1 2 


0 0 3 3
–R2 + R3
1 1 1 1 
0 1
1 2 


0 0 1
1
(1/3)R3
Copyright © 2011 Pearson Education, Inc.
Slide 9- 100
continued
z = –1.
Substitute z into y – z = 2
y –(1) = 2
y+1=2
y=1
Substitute y and z into x – y + z = 1
x–1–1=1
x–2=1
x=3
The solution is (3, 1, –1).
Copyright © 2011 Pearson Education, Inc.
Slide 9- 101
Replace R2 in
9.5
 8 6 4 
 2 4 0 


a)
 10 10 4 
 10 10 4 


c)
 8 14 2 
 2 6 0 


b)
d)
Copyright © 2011 Pearson Education, Inc.
with R1 + R2.
 8 6 4 
 10 10 4 


 10 10 4 
 2 4 0 


Slide 9- 102
Replace R2 in
9.5
 8 6 4 
 2 4 0 


a)
 10 10 4 
 10 10 4 


c)
 8 14 2 
 2 6 0 


b)
d)
Copyright © 2011 Pearson Education, Inc.
with R1 + R2.
 8 6 4 
 10 10 4 


 10 10 4 
 2 4 0 


Slide 9- 103
Solve by transforming the
augmented matrix into row
echelon form.
9.5
x  y  z  0

 x  y  4 z  11
4 x  y  z  6

a) (2, 1, 3)
b) (3, 1, 2)
c) (3, 2, 1)
d) No solution
Copyright © 2011 Pearson Education, Inc.
Slide 4- 104
Solve by transforming the
augmented matrix into row
echelon form.
9.5
x  y  z  0

 x  y  4 z  11
4 x  y  z  6

a) (2, 1, 3)
b) (3, 1, 2)
c) (3, 2, 1)
d) No solution
Copyright © 2011 Pearson Education, Inc.
Slide 4- 105
9.6
Solving Systems of Linear Equations
Using Cramer’s Rule
1. Evaluate determinants of 2  2 matrices.
2. Evaluate determinants of 3  3 matrices.
3. Solve systems of equations using Cramer’s Rule.
Copyright © 2011 Pearson Education, Inc.
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
 a1
If A  
 a2
a1
b1
a2
b2
a1
b1 
, then det(A) =

a2
b2 
b1
b2
 a1b2  a2b1.
 a1b2  a2b1
Copyright © 2011 Pearson Education, Inc.
Slide 9- 107
Example 1a
Find the determinant of the following matrix.
 3 2
A

7
1


Solution
3
7
2
 (3)(1) (7)( 2)
1
 3  14
 17
Copyright © 2011 Pearson Education, Inc.
Slide 9- 108
Example 1b
Find the determinant of the following matrix.
 4 3 
B

5

3


Solution
4
5
3
 (4)(3) (5)(3)
3
 12 15
 3
Copyright © 2011 Pearson Education, Inc.
Slide 9- 109
Example 1c
Find the determinant of the following matrix.
1 2 
K 

4
8


Solution
1
4
2
 (1)(8)  (2)(4)
8
 88
0
Copyright © 2011 Pearson Education, Inc.
Slide 9- 110
Minor of an element of a matrix: The determinant of
the remaining matrix when the row and column in
which the element is located are ignored.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 111
Example 2
 3 4 5
Find the minor of 4 in  2 6 1 .


 4 3 1 
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 
 (2)(1)  (4)(1)
 2  4
2
Copyright © 2011 Pearson Education, Inc.
Slide 9- 112
Evaluating the Determinant of a 3  3 Matrix
a1
A  a2
a3
b1
b2
b3
c1
 minor 
 minor 
 minor 
c2  a1 

  a2 
  a3 
of
a
of a1 
of a2 
3 



c3
b2
 a1
b3
c2
b1
 a2
c3
b3
Copyright © 2011 Pearson Education, Inc.
c1
b1
 a3
c3
b2
c1
c2
Slide 9- 113
Example 3
 3 4 5
 2 6 1 .
Find the determinant of 

 4 3 1 
Solution Use the rule for expanding by minors along
the first column, we have the following:
3 4 5
 minor 
 minor 
 minor 
2 6 1  3  of 3   (4)  of  4   (5)  of  5 






4 3 1
6 1
2 1
2 6
3
3
1
 (4)
4
1
 (5)
4
3
 3(6  3)  4(2  4)  5(6  24)
 9  8  90  107
Copyright © 2011 Pearson Education, Inc.
Slide 9- 114
Cramer’s Rule
The solution to the system of linear equations
a1 x  b1 y  c1
is

a2 x  b2 y  c2
c1 b1
c2 b2 Dx
x

and
a1 b1
D
a2 b2
a1
a2
y
a1
a2
c1
c2 Dy

b1
D
b2
Copyright © 2011 Pearson Education, Inc.
Slide 9- 115
The solution to the system of linear equations
a1 x  b1 y  c1 z  d1

a2 x  b2 y  c2 z  d 2 is
a x  b y  c z  d
3
3
3
 3
d1 b1
d 2 b2
d3 b3
x
a1 b1
a2 b2
a3 b3
c1
c2
c3 Dx

,
c1
D
c2
a1
c3
a2
a3
z
a1
a2
a3
b1
b2
b3
b1
b2
b3
a1 d1 c1
a2 d 2 c2
a3 d3 c3 Dy
y

and
a1 b1 c1
D
a2 b2 c2
d1
a3 b3 c3
d2
d3 Dz

c1
D
c2
c3
Copyright © 2011 Pearson Education, Inc.
Slide 9- 116
Example 4
6 x  y  2
.
Use Cramer’s Rule to solve 
2 x  3 y  2
Solution First, we find D, Dx, and Dy.
6 1
2 1
6 2
D
Dx 
Dy 
2 2
2 3
2 3
 6(2)  (2)(2)
 6(3)  (2)(1)
 2(3)  (2)(1)
 12  4
 18  2
 6  2
 20
 16
 4
Dx 4 1
x


D 20 5
Dy 16 4
y


D 20 5
The solution is (1/5, 4/5).
Copyright © 2011 Pearson Education, Inc.
Slide 9- 117
Example 5
2 x  y  z  8

.
Use Cramer’s Rule to solve  x  y  z  1
 x  2 y  z  2

Solution
We need to find D, Dx, Dy, and Dz.
2 1 1
1 1
1 1
1 1
D  1 1 1  (2)
 (1)
 (1)
2 1
2 1
1 1
1 2 1
 (2)(1  2)  (1)(1  2)  (1)(1  1)
 6  3  0
 9
Copyright © 2011 Pearson Education, Inc.
Slide 9- 118
2 x  y  z  8

x  y  z  1
 x  2 y  z  2

continued
8 1 1
1 1
1 1
1 1
 (1)
 (2)
Dx  1 1 1  (8)
2 1
2 1
1 1
2 2 1
 (8)(1  2)  (1)(1  2)  (2)(1  1)
 24  3  0  27
2 8 1
1 1
8 1
8 1
Dy  1 1 1  (2) 2 1  (1) 2 1  (1) 1 1
1 2 1
 (2)(1  2)  (1)(8  2)  (1)(8  1)
 6  6  9  9
Copyright © 2011 Pearson Education, Inc.
Slide 9- 119
2 x  y  z  8

x  y  z  1
 x  2 y  z  2

continued
2 1 8
1 1
1 8
1 8
 (2)
 (1)
 (1)
2 2
2 2
1 1
Dz  1 1 1
1 2 2  (2)(2  2)  (1)(2  16)  (1)(1  8)
 0  18  9  9
Dy 9
Dx 27
Dz
9
x

 3, y 

 1, z 

 1.
D
9
D 9
D 9
The solution is (3, 1, –1). The check is left to the
viewer.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 120
9 3 

Find the determinant.
 0 8


a) 75
b) 72
c) 27
d) 72
9.6
Copyright © 2011 Pearson Education, Inc.
Slide 4- 121
9 3 

Find the determinant.
 0 8


a) 75
b) 72
c) 27
d) 72
9.6
Copyright © 2011 Pearson Education, Inc.
Slide 4- 122
Solve using Cramer’s Rule.
 x  5 y  19

2 x  6 y  22
a) (1, 4)
b) (1, 3)
c) (2, 3)
d) No solution
9.6
Copyright © 2011 Pearson Education, Inc.
Slide 4- 123
Solve using Cramer’s Rule.
 x  5 y  19

2 x  6 y  22
a) (1, 4)
b) (1, 3)
c) (2, 3)
d) No solution
9.6
Copyright © 2011 Pearson Education, Inc.
Slide 4- 124
9.7
Solving Systems of Linear Inequalities
1. Graph the solution set of a system of linear inequalities.
2. Solve applications involving a system of linear
inequalities.
Copyright © 2011 Pearson Education, Inc.
Solving a System of Linear Inequalities in Two Variables
To solve a system of linear inequalities in two
variables, 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.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 126
Example 1a
Graph the solution set for the system of inequalities.
x  2 y  5

 y  3x  2
Solution
Solution
Graph the inequalities on the
same grid.
Both lines will be dashed.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 127
Example 1b
Graph the solution set for the system of inequalities.
y  3

3x  y  6
Solution
Graph the inequalities on the
same grid.
Solution
Copyright © 2011 Pearson Education, Inc.
Slide 9- 128
Example 2—Inconsistent System
Graph the solution set for the system of inequalities.
2

y  x 3
3

 2 x  3 y  3
Solution
Graph the inequalities on the
same grid.
The slopes are equal, so the lines
are parallel. Since the shaded
regions do not overlap there is no
solution for the system.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 129
Example 3
Mr. Reynolds is landscaping his yard with 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.
Copyright © 2011 Pearson Education, Inc.
Slide 9- 130
continued
Plan and Execute
Let 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
Copyright © 2011 Pearson Education, Inc.
Slide 9- 131
continued
Answer
Since Mr. Reynolds cannot purchase negative plants,
the solution set is confined to Quadrant 1.
x  y  3

40 x  20 y  300
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)
Copyright © 2011 Pearson Education, Inc.
Slide 9- 132
Graph.
9.7
x  y  2

x  y  4
a)
b)
c)
d)
Copyright © 2011 Pearson Education, Inc.
Slide 5- 133
Graph.
9.7
x  y  2

x  y  4
a)
b)
c)
d)
Copyright © 2011 Pearson Education, Inc.
Slide 5- 134