Chapter 6: Systems of Equations
and Inequalities
Algebra I
Table of Contents
•
•
•
•
•
•
6.1 - Solving Systems by Graphing
6.2 - Solving Systems by Substitution
6.3 - Solving Systems by Elimination
6.4 - Solving Special Systems
6.5 - Solving Linear Inequalities
6.6 - Solving Systems of Linear Inequalities
6.1 - Solving Systems by Graphing
Algebra I
6.1
Algebra 1 (bell work)
Just Read
A system of linear equations is a set of two or more
linear equations containing two or more variables.
A solution of a system of linear equations with two
variables is an ordered pair that satisfies each equation
in the system.
-So, if an ordered pair is a solution, it will make both
equations true.
6.1
Example 1
Identifying Solutions of Systems
Tell whether the ordered pair is a solution of the given system.
(5, 2);
3x – y = 13
3x – y 13
0
2–2
0
0
0
3(5) – 2
13
15 – 2
13
13
13
The ordered pair (5, 2) makes both equations true.
(5, 2) is the solution of the system.

6.1
Tell whether the ordered pair is a solution of the given system.
2x + y = 5
–2x + y = 1
(1, 3);
2x + y = 5
–2x + y = 1
2(1) + 3
5
–2(1) + 3 1
2+3
5
–2 + 3
5
5

1
1
1

The ordered pair (1, 3) makes both equations true.
(1, 3) is the solution of the system.
6.1
Tell whether the ordered pair is a solution of the given system.
(2, –1);
x – 2y = 4
3x + y = 6
3(2) + (–1) 6
2 – 2(–1) 4
6–1
2+2 4
4
4
x – 2y = 4
3x + y = 6

5
6
6
The ordered pair (2, –1) makes one equation true, but not the other.
(2, –1) is not a solution of the system.
Optional
6.1
Math Joke
• Teacher: What is the name of the formula that
describes the phases of the moon?
• Student: The lunear (linear) equation
6.1
Just Read
All solutions of a linear equation are on its graph.
To find a solution of a system of linear equations, you need a point that each
line has in common.
In other words, you need their point of intersection.
y = 2x – 1
y = –x + 5
The point (2, 3) is where the two
lines intersect and is a solution of
both equations, so (2, 3) is the
solution of the systems.
6.1
Example 2
Solving a System of Linear Equations by Graphing
Solve the system by graphing. Check your answer.
y=x
y = –2x – 3
Check
Substitute (–1, –1) into
the system.
y=x
(–1) (–1)
–1 –1

y = –2x – 3
(–1) –2(–1) –3
–1 2 – 3
–1 – 1

6.1
Solve the system by graphing. Check your answer.
y=x–6
y+
x = –1
y + x = –1
− x
y=
−
x
6.1
Solve the system by graphing. Check your answer.
y = –2x – 1
y=x+5
Check Substitute (–2, 3)
into the system.
y = –2x – 1
3
–2(–2) – 1
3
4 –1
3
3

y=x+5
3
–2 + 5
3
3

Day 2
6.1
Solve the system by graphing. Check your answer.
2x + y = 4
2x + y = 4
–2x
– 2x
y = –2x + 4
6.1
Example 3
Problem Solving Application
Wren and Jenni are reading the same book.
Wren is on page 14 and reads 2 pages every night.
Jenni is on page 6 and reads 3 pages every night.
After how many nights will they have read the same
number of pages? How many pages will that be?
Total
pages
is
number
read
every
night
plus
already
read.
Wren
y
=
2
x
+
14
Jenni
y
=
3
x
+
6
6.1
Method 1 (Graph)

(8, 30)
Nights
Graph y = 2x + 14 and y = 3x + 6.
The lines appear to intersect at (8, 30).
So, the number of pages read will be the
same at 8 nights with a total of 30 pages.
Method 2 (Substitution)
6.1
Video club A charges $10 for membership and $3 per
movie rental.
Video club B charges $15 for membership and $2 per
movie rental.
For how many movie rentals will the cost be the same
at both video clubs? What is that cost?
Total
cost
is price
for each
rental
memberplus ship fee.
Club A
y
=
3
x
+
10
Club B
y
=
2
x
+
15
Optional
6.1
Method 1 (Graph)
Graph y = 3x + 10 and y = 2x +
15. The lines appear to
intersect at (5, 25). So, the
cost will be the same for 5
rentals and the total cost will
be $25.
Method 2 (Substitution)
HW pg. 400
• 6.1– Day 1: 1-7, 27, 33-42
– Day 2: 12-15, 8, 16, 17, 25, 26, 28
– For the story problems you can use method 1 or 2
– Ch: 18, 23, 24, 31, 32
– HW Guidelines or ½ off
– Start a new HW sheet for Ch 6
6.2 – Solving Systems by
Substitution
Algebra I
6.2
Algebra 1 (bell work)
Steps for solving by Substitution
1. Solve one equation for x= or y=
2. Substitute that equation into the other
equation
3. Use the new equation to solve for the value
of x or y
4. Plug the value back in to find other variable
(Back Substition)
6.2
Example 1
Solving a System of Linear Equations by Substitution
Solve the system by substitution.
y = 3x
y=x–2
Step 4
y = 3x
y = 3(–1)
y = –3
Step 1 y = 3x
Step 5
(–1, –3)
y=x–2
Step 2
y=
x–2
3x =
Step 3
–x
x–2
–x
2x =
–2
2x = –2
2
2
x=–
1
Solution is always an ordered point or pair
Solve the system by substitution.
6.2
y=x+1
4x + y = 6
Step 4
y=x+1
y=1+1
Step 1 y = x + 1
y=2
Step 5
Step 2 4x + y = 6
4x + (x + 1) = 6
5x + 1 = 6
Step 3
–1
5x
–1
=
5
5x = 5
5
5
x=1
(1, 2)
6.2
Math Joke
• Q: Why did the chef throw the unsolved
system of equations out of the restaurant?
• A: The menu said, “No substitution.”
Solve the system by substitution.
6.2
x + 2y = –1
x–y=5
Step 3
–3y – 1 = 5
+1 +1
–3y = 6
Step 1 x + 2y = –1
−2y
x
Step 2
–3y = 6
−2y
–3
= –2y – 1
y = –2
x–y=5
(–2y – 1) – y = 5
–3y – 1 = 5
Step 4
–3
x–y=5
x – (–2) = 5
x+2=5
–2
–2
x =3
Step 5 (3, –2)
Optional
6.2
Example 2
Using the Distributive Property
Solve the system by substitution.
y + 6x = 11
3x + 2y = –5
Step 3
3x – 12x + 22 = –5
–9x + 22 = –5
Step 1 y + 6x = 11
– 6x
3x + 2(–6x) + 2(11) = –5
– 22
– 6x
–22
–9x = –27
y = –6x + 11
–9x = –27
Step 2
3x + 2y = –5
–9
–9
x=3
3x + 2(–6x + 11) = –5
Step 4
y + 6x = 11
y + 6(3) = 11
3x + 2(–6x + 11) = –5
y + 18 = 11
–18 –18
y = –7
Step 5
(3, –7)
Solve the system by substitution.
6.2
–2x + y = 8
3x + 2y = 9
Step 3
7x + 16 = 9
–16
+2x
–16
7x = –7
y = 2x + 8
Step 2
3x + 2(2x) + 2(8) = 9
3x + 4x + 16 = 9
Step 1 –2x + y = 8
+ 2x
Day 2
7x = –7
3x + 2y = 9
7
3x + 2(2x + 8) = 9
7
x = –1
Step 4
3x + 2(2x + 8) = 9
–2x + y = 8
–2(–1) + y = 8
y+2=8
–2 –2
Step 5
y
=6
(–1, 6)
6.2
Example 3
Application
Jenna is deciding between two cell-phone plans.
The first plan has a $50 sign-up fee and costs $20 per month.
The second plan has a $30 sign-up fee and costs $25 per month.
After how many months will the total costs be the same? What will the
costs be? If Jenna has to sign a one-year contract, which plan will be
cheaper? Explain.
Total
paid
Option 1
Option 2
is
sign-up
fee
plus
payment
amount
for each
month.
t
=
$50
+
$20
m
t
=
$30
+
$25
m
Step 1 t = 50 + 20m
t = 30 + 25m
Step 2
50 + 20m = 30 + 25m
6.2
Total
paid
is
sign-up
fee
plus
payment
amount
for each
month.
Option 1
t
=
$50
+
$20
m
Option 2
t
=
$30
+
$25
m
Step 3
50 + 20m =
–20m
30 + 25m
– 20m
50 = 30 + 5m
–30
20
–30
=
5m
20 = 5m
5
5
m=4
Step 4
t = 30 + 25m
t = 30 + 25(4)
t = 30 + 100
t = 130
6.2
After how many months will the total costs be the same? What
will the costs be?
If Jenna has to sign a one-year contract, which plan will be
cheaper? Explain.
Step 5
(4, 130)
In 4 months, the total cost for each option would be the same $130.
If Jenna has to sign a one-year contract, which plan will be
cheaper? Explain.
Option 1: t = 50 + 20(12) = 290
Option 2: t = 30 + 25(12) = 330
Jenna should choose the first plan because it costs $290 for the year and the
second plan costs $330.
6.2
One cable television provider has a $60 setup fee and
$80 per month, and the second has a $160 equipment
fee and $70 per month.
Optional
a. In how many months will the cost be the
same? What will that cost be.
Total
paid
is
fee
payment for each
plus amount month.
Option 1
t
=
$60
+
$80
m
Option 2
t
=
$160
+
$70
m
Step 1 t = 60 + 80m
t = 160 + 70m
Step 2 60 + 80m = 160 + 70m
6.2
Step 3 60 + 80m = 160 + 70m
–70m
–70m
60 + 10m = 160
–60
–60
10m = 100
10
10
m = 10
Step 4 t = 160 + 70m
t = 160 + 70(10)
t = 160 + 700
t = 860
6.2
Step 5 (10, 860)
In 10 months, the total cost for each option
would be the same, $860.
b. If you plan to move in 6 months, which is
the cheaper option? Explain.
Option 1: t = 60 + 80(6) = 540
Option 2: t = 160 + 270(6) = 580
The first option is cheaper for the first six months.
HW pg. 408
• 6.2– Day 1: 1-5 (Odd), 9-15 (Odd), 45-47
– Day 2: 7, 17, 19, 26, 32, 33
– Ch: 24, 25, 27
– HW Guidelines or ½ off
6.3 - Solving Systems by
Elimination
Algebra I
6.3
Algebra 1 (bell work)
Steps for solving by Elimination
1. Line x and y up vertically
2. Eliminate one variable and solve for the other
3. Take solved x or y and use back substitution
4. Check your answer
6.3
Example 1
Solve
Step 1
Elimination Using Addition
3x – 4y = 10
by elimination.
x + 4y = –2
3x – 4y = 10
Step 3
x + 4y = –2
x + 4y = –2
Step 2
4x + 0 =
4x = 8
8
2 + 4y = –2
–2
–2
4x = 8
4y = –4
4
4y
–4
4
4
4
x=2
y = –1
Step 4 (2, –1)
6.3
Solve
Step 1
y + 3x = –2
2y – 3x = 14
y + 3x = –2
2y – 3x = 14
Step 2
3y + 0
= 12
by elimination.
Step 3 y + 3x = –2
4 + 3x = –2
–4
–4
3x = –6
3y = 12
3x = –6
3
3
x = –2
y=4
Step 4
(–2, 4)
6.3
Example 2
Solve
Step 1
Elimination Using Subtraction
2x + y = –5 by elimination.
2x – 5y = 13
2x + y = –5
Step 3
–(2x – 5y = 13)
2x + (–3) = –5
2x + y = –5
2x – 3 = –5
–2x + 5y = –13
Step 2
2x + y = –5
+3
0 + 6y = –18
2x
6y = –18
y = –3
+3
= –2
x = –1
Step 4
(–1, –3)
6.3
Math Joke
• Teacher: What’s your solution of the system of
equations?
• Student: Nothing – I used elimination and got
rid of the whole thing!
Solve the system by elimination.
6.3
x + 2y = 11
–3x + y = –5
Step 3
Step 1
x + 2y = 11
3 + 2y = 11
–2(–3x + y = –5)
x + 2y =
–3
11
Step 2
–3
2y = 8
+(6x –2y = +10)
7x +
x + 2y = 11
y=4
0 = 21
7x = 21
x=3
Step 4
(3, 4)
6.3
Example 3
Elimination Using Multiplication First
Day 2
Solve the system by elimination.
–5x + 2y = 32
2x + 3y = 10
Step 3
Step 1
2(–5x + 2y = 32)
5(2x + 3y
= 10)
–10x + 4y = 64
2x + 3y = 10
2x + 3(6) = 10
2x + 18 = 10
–18
+(10x + 15y = 50)
Step 2
2x = –8
19y = 114
y=6
–18
x = –4
Step 4
(–4, 6)
Solve the system by elimination.
6.3
2x + 5y = 26
–3x – 4y = –25
Step 1
3(2x + 5y = 26)
Step 3
+(2)(–3x – 4y = –25)
2x + 5(4) = 26
6x + 15y = 78
2x + 20 = 26
+(–6x – 8y = –50)
Step 2
0
2x + 5y = 26
–20
+ 7y = 28
2X
y =4
–20
= 6
x=3
Step 4
(3, 4)
6.3
Example 4
Application
Sally spent $14.85 to buy 13 flowers.
She bought lilies, which cost $1.25 each, and tulips, which
cost $0.90 each.
How many of each flower did Sally buy?
Write a system. Use l for the number of lilies and t for the number of tulips.
1.25l + 0.90t = 14.85
l +
t = 13
6.3
Step 1
1.25l + .90t = 14.85
+ (–.90)(l + t) = 13
1.25l + 0.90t = 14.85
+ (–0.90l – 0.90t = –11.70)
0.35l = 3.15
Step 2
l=9
Step 3
l + t = 13
Step 4
9 + t = 13
–9
–9
t =
4
(9, 4)
Sally bought 9 lilies and 4 tulips.
HW pg. 415
• 6.3– Day 1: 1-6, 15, 25, 27, 41-47 (Odd)
– Day 2: 7-10, 17, 19, 20
– Ch: 21, 30, 31
– HW Guidelines or ½ off
6.4 – Solving Special Systems
Algebra I
6.4
Algebra 1 (bell work)
Write down the following definitions
1. Systems with at least one solution are called
consistent.
2. A system that has no solution is an inconsistent
system. (Parallel Lines)
3. An independent system has exactly one solution.
4. A dependent system has infinitely many solutions.
(Same Line, which means same slope and yintercept)
6.4
1. One solution, consistent, independent
2. All real, consistent, dependent.
3. No solution, inconsistent
Pg. 421
6.4
Solve
y=x–4
–x + y = 3
Method 1 Compare slopes and y-intercepts.
y=x–4
y = 1x – 4
–x + y = 3
y = 1x + 3
Write both equations in slopeintercept form.
The lines are parallel because they
have the same slope and
different y-intercepts.
This system has no solution so it is an inconsistent
system.
6.4
Solve
y=x–4
–x + y = 3
Method 2 Solve the system algebraically.
–x + (x – 4) = 3
–4 = 3 
This system has no solution so it is an inconsistent
system.
6.4
Solve
y = –2x + 5
2x + y = 1
Method 1 Compare slopes and y-intercepts.
y = –2x + 5
2x + y = 1
y = –2x + 5
y = –2x + 1
Write both equations in slope-intercept
form.
The lines are parallel because they have
the same slope and different yintercepts.
This system has no solution so it is an inconsistent
system.
6.4
Solve
y = –2x + 5
2x + y = 1
Method 2 Solve the system algebraically.
2x + (–2x + 5) = 1
5 = 1
This system has no solution so it is an inconsistent
system.
6.4
Solve
y = 3x + 2
3x – y + 2= 0
Method 1 Compare slopes and y-intercepts.
y = 3x + 2
3x – y + 2= 0
y = 3x + 2
Write both equations in slopeintercept form. The lines have the
same slope and the same yintercept.
y = 3x + 2
If this system were graphed, the graphs would be
the same line. There are infinitely many solutions,
consistent, dependent.
6.4
Solve
y = 3x + 2
3x – y + 2= 0
Method 2 Solve the system algebraically.
y = 3x + 2
3x − y + 2= 0
y − 3x = 2
−y + 3x = −2
0 = 0
There are infinitely many solutions, consistent,
dependent.
Classify the system. Give the number of solutions.
6.4
Solve
3y = x + 3
x+y=1
The system is consistent and dependent. It has
infinitely many solutions.
Day 2
Classify the system. Give the number of solutions.
6.4
Solve
x+y=5
4 + y = –x
The system is inconsistent. It has no solutions.
Classify the system. Give the number of solutions.
6.4
Solve
y = 4(x + 1)
y–3=x
The system is consistent and independent. It has
one solution.
6.4
Example 4
Application
Jared and David both started a savings account in January.
If the pattern of savings in the table continues, when will the
amount in Jared’s account equal the amount in David’s
account?
6.4
Total
saved
is
Jared
y
=
$25
+
$5
x
David
y
=
$40
+
$5
x
start
amount plus
amount
saved
for each
month.
y = 5x + 25
y = 5x + 40
y = 5x + 25
y = 5x + 40
The graphs of the two equations are parallel lines, so there is no
solution.
If the patterns continue, the amount in Jared’s account will never
be equal to the amount in David’s account.
6.4
Optional
Matt has $100 in a checking account and deposits $20 per month.
Ben has $80 in a checking account and deposits $30 per month.
Will the accounts ever have the same balance? Explain.
y = 20x + 100
y = 30x + 80
y = 20x + 100
y = 30x + 80
The accounts will have the same balance.
The graphs of the two equations have different slopes so they
intersect.
HW pg. 423
• 6.4– Day 1: 2-7, 15, 19, 39-43 (Odd)
– Day 2: 8-11, 21, 23, 24, 29, 31
– Ch: 28, 30
– HW Guidelines or ½ off
6.5 - Solving Linear Inequalities
Algebra I
6.5
Algebra 1 (bell work)
Copy Definitions
A linear inequality is similar to a linear equation, but
the equal sign is replaced with an inequality symbol.
A solution of a linear inequality is any ordered pair
that makes the inequality true.
6.5
Identifying Solutions of Inequalities
Tell whether the ordered pair is a solution of the inequality.
(–2, 4); y < 2x + 1
(3, 1); y > x – 4
y < 2x + 1
y>x−4
4 < 2(–2) + 1
4 < –4 + 1
4 < –3 
1 >
3–4
1 >
–1
(–2, 4) is not a solution.

(3, 1) is a solution.
6.5
Don’t Copy Down
A linear inequality describes a region of a coordinate plane called
a half-plane.
All points in the region are solutions of the linear inequality.
The boundary line of the region is the graph of the related
equation.
6.5
Don’t Copy Down
6.5
Steps to Graphing Linear Inequalities
Graphing Linear Inequalities
Step 1
Step 2
Step 3
Solve the inequality for y (slope-intercept form).
Graph the boundary line. Use a solid line for ≤ or ≥. Use
a dashed line for < or >.
Shade the half-plane above the line for y > or ≥. Shade the
half-plane below the line for y < or y ≤. Check your
answer.
6.5
Example 2
Graph the
solutions of the
linear inequality.
y  2x – 3
Check ( 0, 0 )
Graphing Linear Inequalities in Two Variables
6.5
Math Joke
• Q: Why did the math student name her
boundary line Hope?
• A: Because it was dashed.
6.5
Graph the
solutions of the
linear inequality.
5x + 2y > –8
Check ( 0, 0 )
6.5
Graph the
solutions of the
linear inequality.
4x – 3y > 12
Check ( 0, 0 )
6.5
Optional
Graph the
solutions of the
linear inequality.
2x – y – 4 > 0
Check ( 0, 0 )
6.5
Day 2
Graph the
solutions of the
linear inequality.
4x – y + 2 ≤ 0
Check ( 0, 0 )
6.5
Example 4
Application
Ada has at most 285 beads to make jewelry. A necklace requires 40
beads, and a bracelet requires 15 beads.
Let x represent the number of necklaces and y the number of
bracelets.
Write an inequality. Use ≤ for “at most.”
Necklace
beads
40x
plus
bracelet
beads
is at
most
285
beads.
+
15y
≤
285
Solve the inequality for y.
40x + 15y ≤ 285
–40x
–40x
15y ≤ –40x + 285
6.5
b. Graph the solutions.
Step 1 Since Ada cannot make a negative
amount of jewelry, the system is graphed
only in Quadrant I.
Graph the boundary line
Use a solid line for ≤.
6.5
b. Graph the solutions.
Step 2 Shade below the line. Ada can only
make whole numbers of jewelry. All points
on or below the line with whole number
coordinates are the different
combinations of bracelets and necklaces
that Ada can make.
6.5
c. Give two combinations of necklaces and bracelets that Ada
could make.
Two different combinations of
jewelry that Ada could make with 285
beads could be 2 necklaces and 8
bracelets or 5 necklaces and 3
bracelets.
(2, 8)


(5, 3)
6.5
Example 4
Application
Dirk is going to bring two types of olives to the Honor Society
induction and can spend no more than $6.
Green olives cost $2 per pound and black olives cost $2.50
per pound.
a. Write a linear inequality to describe the situation.
b. Graph the solutions.
c. Give two combinations of olives that Dirk could buy.
6.5
Write an inequality. Use ≤ for “no more than.”
Green
olives
2x
plus
black
olives
+
2.50y
Solve the inequality for y.
2x + 2.50y ≤ 6
–2x
–2x
2.50y ≤ –2x + 6
2.50y ≤ –2x + 6
2.50
2.50
is no
more
than
≤
total
cost.
6
6.5
y ≤ –0.80x + 2.4
Step 1 Since Dirk cannot buy
negative amounts of olive, the
system is graphed only in
Quadrant I. Graph the boundary
line for y = –0.80x + 2.4. Use a
solid line for≤.
Black Olives
b. Graph the solutions.
Green Olives
6.5
Two different combinations of
olives that Dirk could purchase
with $6 could be 1 pound of green
olives and 1 pound of black olives
or 0.5 pound of green olives and 2
pounds of black olives.
Black Olives
c. Give two combinations of olives that
Dirk could buy.

(0.5, 2)

(1, 1)
Green Olives
6.5
Example 4
Writing an Inequality from a Graph
Write an inequality to represent the graph.
y-intercept: 0 slope: –1
Write an equation in slope-intercept
form.
y = mx + b y= – 1x
The graph is shaded below a dashed
boundary line.
Replace = with < to write the inequality y < –x.
6.5
Write an inequality to represent the graph.
y-intercept: 1; slope:
Write an equation in slope-intercept
form.
The graph is shaded above a dashed
boundary line.
Replace = with > to write the inequality
HW pg. 432
• 6.5– Day 1: 2-8, 15-18, 51-65 (Odd)
– Day 2: 9-11, 19-21, 37 (Equation Only), 38
– Ch: 22, 27, 28, 40-42
6.6 - Solving Systems of Linear
Inequalities
Algebra I
6.6
Algebra 1 (bell work)
A system of linear inequalities is a set of two or
more linear inequalities containing two or more
variables.
The solutions of a system of linear inequalities
consists of all the ordered pairs that satisfy all the
linear inequalities in the system.
6.6
Example 1
Identifying Solutions of Systems of Linear Inequalities
Tell whether the ordered pair is a solution of the given system.
(–1, –3);
y ≤ –3x + 1
y < 2x + 2
(–1, –3)
y ≤ –3x + 1
–3
–3(–1) + 1
–3
3+1
–3 ≤ 4 
(–1, –3)
y < 2x + 2
–3
2(–1) + 2
–3
–2 + 2
–3 < 0 
(–1, –3) is a solution to the system because it
satisfies both inequalities.
6.6
Optional
Tell whether the ordered pair is a solution of the given system.
(–1, 5);
y < –2x – 1
y≥x+3
(–1, 5)
y < –2x – 1
5
–2(–1) – 1
5
2–1
5 < 1 
(–1, 5)
y≥x+3
5
–1 + 3
5 ≥ 2
(–1, 5) is not a solution to the system because it does
not satisfy both inequalities.
6.5
Example 2
y≤3
y > –x + 5
Solving a System of Linear Inequalities by Graphing
Math Joke
• Q: Why did the math student wear two pairs
of sunglasses?
• A: He wanted to have overlapping shades
6.5
y>x–7
3x + 6y ≤ 12
6.5
–3x + 2y ≥ 2
y < 4x + 3
6.5
Example 3
Graph and Describe the
Solutions
y ≤ –2x – 4
y > –2x + 5
Graphing Systems with Parallel Boundary Lines
Day 2
6.5
Graph and Describe the
Solutions
y > 3x – 2
y < 3x + 6
6.6
Example 4
Application
In one week, Ed can mow at most 9 times and rake at most 7
times. He charges $20 for mowing and $10 for raking. He needs to
make more than $125 in one week.
Show and describe all the possible combinations of mowing and
raking that Ed can do to meet his goal. List two possible
combinations.
Earnings per Job ($)
Mowing
20
Raking
10
Step 1 Write a system of inequalities.
Let x represent the number of mowing jobs
and y represent the number of raking jobs.
6.6
In one week, Ed can mow at most 9 times and rake at most 7
times. He charges $20 for mowing and $10 for raking. He needs
to make more than $125 in one week. Show and describe all the
possible combinations of mowing and raking that Ed can do to
meet his goal. List two possible combinations.
Step 1 Write a system of inequalities.
Let x represent the number of mowing jobs and y represent the number of
raking jobs.
x≤9
He can do at most 9 mowing jobs.
y≤7
He can do at most 7 raking jobs.
20x + 10y > 125
He wants to earn more than $125.
6.6
Step 2 Graph the system.
Solutions
The graph should be in only the first quadrant
because the number of jobs cannot be negative.
6.6
Step 3 Describe all possible combinations.
All possible combinations represented by ordered pairs of
whole numbers in the solution region will meet Ed’s
requirement of mowing, raking, and earning more than $125
in one week. Answers must be whole numbers because he
cannot work a portion of a job.
Step 4 List the two possible combinations.
Two possible combinations are:
7 mowing and 4 raking jobs
8 mowing and 1 raking jobs
Solutions
6.6
At her party, Alice is serving pepper jack cheese and cheddar cheese. She
wants to have at least 2 pounds of each. Alice wants to spend at most $20
on cheese. Show and describe all possible combinations of the two cheeses
Alice could buy. List two possible combinations.
Price per Pound ($)
Pepper Jack
4
Cheddar
2
Step 1 Write a system of inequalities.
Let x represent the pounds of pepper jack
and y represent the pounds of cheddar.
x≥2
y≥2
4x + 2y ≤ 20
She wants at least 2 pounds of pepper jack.
She wants at least 2 pounds of cheddar.
She wants to spend no more than $20.
6.6
Step 2 Graph the system.
Solutions
The graph should be in only the first quadrant
because the amount of cheese cannot be negative.
6.6
Step 3 Describe all possible combinations.
All possible combinations within the gray region will
meet Alice’s requirement of at most $20 for cheese
and no less than 2 pounds of either type of cheese.
Answers need not be whole numbers as she can buy
fractions of a pound of cheese.
Step 4 Two possible
combinations are (3, 2)
and (2.5, 4).
3 pepper jack, 2 cheddar or
2.5 pepper jack, 4
cheddar.
HW pg. 438
• 6.6– Day 1: 2-8, 19, 21, 55, 57
– Day 2: 9-15, 25, 29-30 (Equations Only)
– Ch: 38-42