### Chapter 6 PPT- Algebra

```6-1 Solving Systems by Graphing
Warm Up
Evaluate each expression for x = 1 and
y =–3.
1. x – 4y
2. –2x + y –5
13
Write each expression in slopeintercept form.
3. y – x = 1
y=x+1
4. 2x + 3y = 6 y =
x+2
5. 0 = 5y + 5x y = –x
Holt Algebra 1
6-1 Solving Systems by Graphing
Objectives
Identify solutions of linear equations in two
variables.
Solve systems of linear equations in two
variables by graphing.
Holt Algebra 1
6-1 Solving Systems by Graphing
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.
Holt Algebra 1
6-1 Solving Systems by Graphing
Example 1A: Identifying Systems of Solutions
Tell whether the ordered pair is a solution of the
given system.
(5, 2);
3x – y = 13
0
3x – y
13
3(5) – 2
13
Substitute 5 for x
and 2 for y in each
equation in the
system.
2–2 0
15 – 2 13
0 0
13 13 
The ordered pair (5, 2) makes both equations true.
(5, 2) is the solution of the system.
Holt Algebra 1
6-1 Solving Systems by Graphing
If an ordered pair does not satisfy the first
equation in the system, there is no reason to
check the other equations.
Holt Algebra 1
6-1 Solving Systems by Graphing
Example 1B: Identifying Systems of Solutions
Tell whether the ordered pair is a solution of
the given system.
x + 3y = 4
(–2, 2);
–x + y = 2
x + 3y = 4
–x + y = 2
–2 + 3(2) 4
–(–2) + 2
–2 + 6 4
4
4 4
2
2
Substitute –2 for x
and 2 for y in each
equation in the
system.
The ordered pair (–2, 2) makes one equation true but
not the other.
(–2, 2) is not a solution of the system.
Holt Algebra 1
6-1 Solving Systems by Graphing
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.
Holt Algebra 1
6-1 Solving Systems by Graphing
Sometimes it is difficult to tell exactly where the
lines cross when you solve by graphing. It is
into both equations.
Holt Algebra 1
6-1 Solving Systems by Graphing
Example 2A: Solving a System Equations by Graphing
y=x
Graph the system.
y = –2x – 3
The solution appears to
be at (–1, –1).
y=x
Check
Substitute (–1, –1) into
the system.
y = –2x – 3
y=x
•
(–1, –1)
y = –2x – 3
(–1)
–1
(–1)
–1

(–1) –2(–1) –3
–1
2–3
–1 – 1 
(–1, –1) is the solution of the system.
Holt Algebra 1
6-1 Solving Systems by Graphing
Example 2B: Solving a System Equations by Graphing
y=x–6
y+
Graph using a calculator and
then use the intercept
command.
x = –1
Rewrite the second equation in
slope-intercept form.
y+
−
x = –1
x
y=
Holt Algebra 1
−
x
y=x–6
6-1 Solving Systems by Graphing
Example 2B Continued
Check Substitute
into the system.
y=x–6
–6
y=x–6
+
–1
–1

–1
–1
Holt Algebra 1
– 1  The solution is
.
6-1 Solving Systems by Graphing
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?
Holt Algebra 1
6-1 Solving Systems by Graphing
Example 3 Continued
1
Understand the Problem
The answer will be the number of nights it
takes for the number of pages read to be the
same for both girls.
List the important information:
Wren on page 14
Jenni on page 6
Holt Algebra 1
6-1 Solving Systems by Graphing
Example 3 Continued
2
Make a Plan
Write a system of equations, one equation to
represent the number of pages read by each
girl. Let x be the number of nights and y be the
Total
pages
number
is
every
night plus
Wren
y
=
2
x
+
14
Jenni
y
=
3
x
+
6
Holt Algebra 1
6-1 Solving Systems by Graphing
Example 3 Continued
3
Solve
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.

(8, 30)
Nights
Holt Algebra 1
6-1 Solving Systems by Graphing
Example 3 Continued
4
Look Back
Check (8, 30) using both equations.
Number of days for Wren to read 30 pages.
2(8) + 14 = 16 + 14 = 30
Number of days for Jenni to read 30 pages.
3(8) + 6 = 24 + 6 = 30
Holt Algebra 1
6-2 Solving Systems by Substitution
Warm-Up
Tell whether the ordered pair is a solution of
the given system.
1. (–3, 1)
2. (2, –4)
3. Joy has 5 collectable stamps and will buy 2 more
each month. Ronald has 25 collectable stamps
and will sell 3 each month. After how many months
will they have the same number of stamps?
How many will that be?
Holt Algebra 1
6-2 Solving Systems by Substitution
Objective
Solve linear equations in two variables
by substitution.
Holt Algebra 1
6-2 Solving Systems by Substitution
Sometimes it is difficult to identify the
exact solution to a system by graphing. In
this case, you can use a method called
substitution.
The goal when using substitution is to
reduce the system to one equation that
has only one variable. Then you can
solve this equation by the methods
taught in Chapter 2.
Holt Algebra 1
6-2 Solving Systems by Substitution
Solving Systems of Equations by Substitution
Step 1
Solve for one variable in at least one
equation, if necessary.
Step 2
Substitute the resulting expression into the
other equation.
Step 3
Solve that equation to get the value of the
first variable.
Step 4
Substitute that value into one of the original
equations and solve.
Step 5
Write the values from steps 3 and 4 as an
ordered pair, (x, y), and check.
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 1A: Solving a System of Linear Equations by
Substitution
Solve the system by substitution.
y = 3x
y=x–2
Step 1 y = 3x
y=x–2
Both equations are solved for y.
Step 2
Substitute 3x for y in the second
equation.
Solve for x. Subtract x from both
sides and then divide by 2.
y= x–2
3x = x – 2
Step 3 –x
–x
2x =
–2
2x = –2
2
2
x = –1
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 1A Continued
Solve the system by substitution.
Step 4
Step 5
Write one of the original
equations.
Substitute –1 for x.
y = 3x
y = 3(–1)
y = –3
(–1, –3)
Write the solution as an
ordered pair.
Check Substitute (–1, –3) into both equations in the
system.
y = 3x
y=x–2
–3 3(–1)
–3 –1 – 2
–3
Holt Algebra 1
–3

–3
–3

6-2 Solving Systems by Substitution
Example 1B: Solving a System of Linear Equations by
Substitution
Solve the system by substitution.
y=x+1
4x + y = 6
Step 1 y = x + 1
Step 2 4x
4x
5x
Step 3
The first equation is solved for y.
+y=6
Substitute x + 1 for y in the
+ (x + 1) = 6
second equation.
Simplify. Solve for x.
+1=6
–1 –1
Subtract 1 from both sides.
5x
= 5
5x = 5
Divide both sides by 5.
5
5
x=1
Holt Algebra 1
6-2 Solving Systems by Substitution
Example1B Continued
Solve the system by substitution.
Step 4
Step 5
y=x+1
y=1+1
y=2
(1, 2)
Write one of the original
equations.
Substitute 1 for x.
Write the solution as an
ordered pair.
Check Substitute (1, 2) into both equations in the
system.
y=x+1
4x + y = 6
2 1+1
4(1) + 2 6
2 2 
6 6
Holt Algebra 1
6-2 Solving Systems by Substitution
Sometimes you substitute an
expression for a variable that has a
coefficient. When solving for the
second variable in this situation, you
can use the Distributive Property.
Holt Algebra 1
6-2 Solving Systems by Substitution
Caution
When you solve one equation for a variable, you
must substitute the value or expression into the
other original equation, not the one that had just
been solved.
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 2: Using the Distributive Property
Solve
y + 6x = 11
3x + 2y = –5
by substitution.
Step 1 y + 6x = 11
– 6x – 6x
y = –6x + 11
Solve the first equation for y
by subtracting 6x from each
side.
Step 2
3x + 2y = –5
3x + 2(–6x + 11) = –5
Substitute –6x + 11 for y in the
second equation.
3x + 2(–6x + 11) = –5
Distribute 2 to the expression
in parenthesis.
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 2 Continued
Solve
y + 6x = 11
3x + 2y = –5
by substitution.
Simplify. Solve for x.
Step 3 3x + 2(–6x) + 2(11) = –5
3x – 12x + 22 = –5
–9x + 22 = –5
– 22 –22 Subtract 22 from
–9x = –27
both sides.
–9x = –27 Divide both sides
by –9.
–9
–9
x=3
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 2 Continued
Solve
Step 4
y + 6x = 11
3x + 2y = –5
y + 6x = 11
y + 6(3) = 11
y + 18 = 11
–18 –18
by substitution.
Write one of the original
equations.
Substitute 3 for x.
Simplify.
Subtract 18 from each side.
y = –7
Step 5
Holt Algebra 1
(3, –7)
Write the solution as an
ordered pair.
6-2 Solving Systems by Substitution
Example 2: Consumer Economics 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.
Write an equation for each option. Let t represent
the total amount paid and m represent the number
of months.
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 2 Continued
Total
paid
is
signup fee
payment for each
plus amount month.
Option 1
t
=
\$50
+
\$20
m
Option 2
t
=
\$30
+
\$25
m
Step 1 t = 50 + 20m
t = 30 + 25m
Both equations are solved
for t.
Step 2 50 + 20m = 30 + 25m
Substitute 50 + 20m for t in
the second equation.
Holt Algebra 1
6-2 Solving Systems by Substitution
Example 2 Continued
Step 3 50 + 20m = 30 + 25m
–20m
– 20m
50
= 30 + 5m
–30
–30
20
=
5m
20 = 5m
5
5
m=4
Solve for m. Subtract 20m
from both sides.
Subtract 30 from both
sides.
Step 4 t = 30 + 25m
Write one of the original
equations.
Substitute 4 for m.
Simplify.
t = 30 + 25(4)
t = 30 + 100
t = 130
Holt Algebra 1
Divide both sides by 5.
6-2 Solving Systems by Substitution
Example 2 Continued
Step 5
(4, 130)
Write the solution as an
ordered pair.
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.
Holt Algebra 1
6-3 Solving Systems by Elimination
Warm-Up
Holt Algebra 1
6-3 Solving Systems by Elimination
Objectives
Solve systems of linear equations in
two variables by elimination.
Compare and choose an appropriate
method for solving systems of linear
equations.
Holt Algebra 1
6-3 Solving Systems by Elimination
Another method for solving systems of
equations is elimination. Like substitution, the
goal of elimination is to get one equation that
has only one variable. To do this by elimination,
you add the two equations in the system
together.
Holt Algebra 1
6-3 Solving Systems by Elimination
Solving Systems of Equations by
Elimination
Step 1
Write the system so that like
terms are aligned.
Step 2
Eliminate one of the variables and
solve for the other variable.
Step 3
Substitute the value of the variable
into one of the original equations
and solve for the other variable.
Step 4
Write the answers from Steps 2 and 3
as an ordered pair, (x, y), and check.
Holt Algebra 1
6-3 Solving Systems by Elimination
Solve
3x – 4y = 10
by elimination.
x + 4y = –2
Step 1
Step 2
3x – 4y = 10
x + 4y = –2
4x + 0 = 8
4x = 8
4x = 8
4
4
x=2
Holt Algebra 1
Write the system so that
like terms are aligned.
eliminate the y-terms.
Simplify and solve for x.
Divide both sides by 4.
6-3 Solving Systems by Elimination
Example 1 Continued
Step 3 x + 4y = –2
2 + 4y = –2
–2
–2
4y = –4
4y
–4
4
4
y = –1
Step 4 (2, –1)
Holt Algebra 1
Write one of the original
equations.
Substitute 2 for x.
Subtract 2 from both sides.
Divide both sides by 4.
Write the solution as an
ordered pair.
6-3 Solving Systems by Elimination
When two equations each contain
the same term, you can subtract
one equation from the other to
solve the system. To subtract an
equation add the opposite of each
term.
Holt Algebra 1
6-3 Solving Systems by Elimination
Example 2: Elimination Using Subtraction
Solve
2x + y = –5
by elimination.
2x – 5y = 13
Step 1
2x + y = –5
–(2x – 5y = 13)
Step 2
2x + y = –5
–2x + 5y = –13
0 + 6y = –18
6y = –18
y = –3
Holt Algebra 1
term in the second
equation.
Eliminate the x term.
Simplify and solve for y.
6-3 Solving Systems by Elimination
Example 2 Continued
Step 3 2x + y = –5
2x + (–3) = –5
2x – 3 = –5
+3
+3
Write one of the original
equations.
Substitute –3 for y.
2x
Simplify and solve for x.
= –2
x = –1
Step 4 (–1, –3)
Holt Algebra 1
Write the solution as an
ordered pair.
6-3 Solving Systems by Elimination
Remember!
into both original equations.
Holt Algebra 1
6-3 Solving Systems by Elimination
In some cases, you will first need to
multiply one or both of the equations by
a number so that one variable has
opposite coefficients. This will be the
new Step 1.
Holt Algebra 1
6-3 Solving Systems by Elimination
Example 3A: Elimination Using Multiplication First
Solve the system by elimination.
x + 2y = 11
–3x + y = –5
Step 1
Step 2
Holt Algebra 1
x + 2y = 11
–2(–3x + y = –5)
x + 2y = 11
+(6x –2y = +10)
7x + 0 = 21
7x = 21
x=3
Multiply each term in the
second equation by –2 to
get opposite y-coefficients.
the first equation.
Simplify and solve for x.
6-3 Solving Systems by Elimination
Example 3A Continued
Step 3 x + 2y = 11
3 + 2y = 11
–3
–3
2y = 8
y=4
Step 4
Holt Algebra 1
(3, 4)
Write one of the original
equations.
Substitute 3 for x.
Subtract 3 from each side.
Simplify and solve for y.
Write the solution as an
ordered pair.
6-3 Solving Systems by Elimination
Example 3B: Elimination Using Multiplication First
Solve the system by elimination.
–5x + 2y = 32
2x + 3y = 10
Step 1
2(–5x + 2y = 32)
5(2x + 3y = 10)
–10x + 4y = 64
+(10x + 15y = 50)
Step 2
Holt Algebra 1
19y = 114
y=6
Multiply the first equation
by 2 and the second
equation by 5 to get
opposite x-coefficients
Simplify and solve for y.
6-3 Solving Systems by Elimination
Example 3B Continued
Step 3
2x + 3y = 10
2x + 3(6) = 10
2x + 18 = 10
–18 –18
Step 4
Holt Algebra 1
2x = –8
x = –4
(–4, 6)
Write one of the original
equations.
Substitute 6 for y.
Subtract 18 from both sides.
Simplify and solve for x.
Write the solution as an
ordered pair.
6-3 Solving Systems by Elimination
Check It Out! Example 3b
Solve the system by elimination.
2x + 5y = 26
–3x – 4y = –25
Step 1
Step 2
Holt Algebra 1
3(2x + 5y = 26)
+(2)(–3x – 4y = –25)
Multiply the first equation
by 3 and the second
equation by 2 to get
opposite x-coefficients
6x + 15y = 78
+(–6x – 8y = –50) Add the new equations.
0 + 7y = 28
Simplify and solve for y.
y =4
6-3 Solving Systems by Elimination
Check It Out! Example 3b Continued
Step 3
2x + 5y = 26
2x + 5(4) = 26
2x + 20 = 26
–20 –20
2X
= 6
x=3
Step 4
(3, 4)
Holt Algebra 1
Write one of the original
equations.
Substitute 4 for y.
Subtract 20 from both
sides.
Simplify and solve for x.
Write the solution as an
ordered pair.
6-3 Solving Systems by Elimination
Example 4: Application
Paige has \$7.75 to buy 12 sheets of felt and
card stock for her scrapbook. The felt costs
\$0.50 per sheet, and the card stock costs
\$0.75 per sheet. How many sheets of each
Write a system. Use f for the number of felt
sheets and c for the number of card stock sheets.
0.50f + 0.75c = 7.75
f + c = 12
Holt Algebra 1
The cost of felt and card
stock totals \$7.75.
The total number of sheets
is 12.
6-3 Solving Systems by Elimination
Example 4 Continued
Step 1
0.50f + 0.75c = 7.75 Multiply the second
equation by –0.50 to get
+ (–0.50)(f + c) = 12
opposite f-coefficients.
0.50f + 0.75c = 7.75
+ (–0.50f – 0.50c = –6)
first equation to
0.25c = 1.75
Step 2
eliminate the f-term.
Simplify and solve for c.
c=7
Step 3
f + c = 12
f + 7 = 12
–7 –7
f
= 5
Holt Algebra 1
Write one of the original
equations.
Substitute 7 for c.
Subtract 7 from both sides.
6-3 Solving Systems by Elimination
Example 4 Continued
Step 4
(7, 5)
Write the solution as an
ordered pair.
Paige can buy 7 sheets of card stock and 5
sheets of felt.
Holt Algebra 1
6-3 Solving Systems by Elimination
Holt Algebra 1
6-4 Solving Special Systems
Warm-Up
Solve each system by elimination.
1.
2x + y = 25
3y = 2x – 13
2.
–3x + 4y = –18
x = –2y – 4
(2, –3)
3.
–2x + 3y = –15
3x + 2y = –23
(–3, –7)
(11, 3)
4. Harlan has \$44 to buy 7 pairs of socks. Athletic
socks cost \$5 per pair. Dress socks cost \$8 per
pair. How many pairs of each can Harlan buy?
4 pairs of athletic socks and 3 pairs of dress socks
Holt Algebra 1
6-4 Solving Special Systems
Objectives
Solve special systems of linear
equations in two variables.
Classify systems of linear equations and
determine the number of solutions.
Holt Algebra 1
6-4 Solving Special Systems
In Lesson 6-1, you saw that when two lines intersect
at a point, there is exactly one solution to the
system. Systems with at least one solution are called
consistent.
When the two lines in a system do not intersect
they are parallel lines. There are no ordered pairs
that satisfy both equations, so there is no solution.
A system that has no solution is an inconsistent
system.
Holt Algebra 1
6-4 Solving Special Systems
Example 1: Systems with No Solution
Solve
y=x–4
–x + y = 3
Method 1 Compare slopes and y-intercepts.
y=x–4
–x + y = 3
y = 1x – 4 Write both equations in slopeintercept form.
y = 1x + 3
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.
Holt Algebra 1
6-4 Solving Special Systems
If two linear equations in a system
have the same graph, the graphs are
coincident lines, or the same line.
There are infinitely many solutions of
the system because every point on the
line represents a solution of both
equations.
Holt Algebra 1
6-4 Solving Special Systems
Example 2A: Systems with Infinitely Many Solutions
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
y = 3x + 2
have the same slope and
the same y-intercept.
If this system were graphed, the graphs
would be the same line. There are infinitely
many solutions.
Holt Algebra 1
6-4 Solving Special Systems
Example 2A Continued
Solve
y = 3x + 2
3x – y + 2= 0
Method 2 Solve the system algebraically. Use
the elimination method.
y = 3x + 2
3x − y + 2= 0
y − 3x =
2
−y + 3x = −2
0 = 0
Write equations to line up
like terms.
True. The equation is an
identity.
There are infinitely many solutions.
Holt Algebra 1
6-4 Solving Special Systems
Caution!
0 = 0 is a true statement. It does not mean the
system has zero solutions or no solution.
Holt Algebra 1
6-4 Solving Special Systems
Holt Algebra 1
6-4 Solving Special Systems
Example 3A: Classifying Systems of Linear Equations
Classify the system. Give the number of solutions.
Solve
3y = x + 3
3y = x + 3
x+y=1
x+y=1
y=
x+1
Write both equations in
slope-intercept form.
y=
x+1
The lines have the same slope
and the same y-intercepts.
They are the same.
The system is consistent and dependent. It has
infinitely many solutions.
Holt Algebra 1
6-4 Solving Special Systems
Example 3B: Classifying Systems of Linear equations
Classify the system. Give the number of solutions.
Solve
x+y=5
4 + y = –x
x+y=5
y = –1x + 5
4 + y = –x
y = –1x – 4
Write both equations in
slope-intercept form.
The lines have the same
slope and different yintercepts. They are
parallel.
The system is inconsistent. It has no solutions.
Holt Algebra 1
6-5 Solving Linear Inequalities
Warm Up
Graph each inequality.
1. x > –5
3. Write –6x + 2y = –4
in slope-intercept form,
and graph.
y = 3x – 2
Holt Algebra 1
2. y ≤ 0
6-5 Solving Linear Inequalities
Objective
Graph and solve linear inequalities in
two variables.
Holt Algebra 1
6-5 Solving Linear Inequalities
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.
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 1A: Identifying Solutions of Inequalities
Tell whether the ordered pair is a solution of
the inequality.
(–2, 4); y < 2x + 1
y < 2x + 1
4 2(–2) + 1
4 –4 + 1
4 < –3 
(–2, 4) is not a solution.
Holt Algebra 1
Substitute (–2, 4) for (x, y).
6-5 Solving Linear Inequalities
Example 1B: Identifying Solutions of Inequalities
Tell whether the ordered pair is a solution of
the inequality.
(3, 1); y > x – 4
y>x−4
1
3–4
1> –1
Substitute (3, 1) for (x, y).

(3, 1) is a solution.
Holt Algebra 1
6-5 Solving Linear Inequalities
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.
Holt Algebra 1
6-5 Solving Linear Inequalities
Holt Algebra 1
6-5 Solving Linear Inequalities
Graphing Linear Inequalities
Step 1
Solve the inequality for y (slopeintercept form).
Step 2
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 >
Step 3 or ≥. Shade the half-plane below the line
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 2B: Graphing Linear Inequalities in Two
Variables
Graph the solutions of the linear inequality.
5x + 2y > –8
Step 1 Solve the inequality for y.
5x + 2y > –8
–5x
–5x
2y > –5x – 8
y>
x–4
Step 2 Graph the boundary line y =
dashed line for >.
Holt Algebra 1
x – 4. Use a
6-5 Solving Linear Inequalities
Example 2B Continued
Graph the solutions of the linear inequality.
5x + 2y > –8
Step 3 The inequality is >, so
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 2B Continued
Graph the solutions of the linear inequality.
5x + 2y > –8
Substitute ( 0, 0)
Check y >
x–4
for (x, y)
because it is
not on the
0
(0) – 4
boundary line.
0
–4
The point (0, 0)
0 > –4
satisfies the
inequality, so the
graph is correctly
Holt Algebra 1
6-5 Solving Linear Inequalities
The point (0, 0) is a good test point to use if it
does not lie on the boundary line.
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 2C: Graphing Linear Inequalities in two
Variables
Graph the solutions of the linear inequality.
4x – y + 2 ≤ 0
Step 1 Solve the inequality for y.
4x – y + 2 ≤ 0
–y
–1
≤ –4x – 2
–1
y ≥ 4x + 2
Step 2 Graph the boundary line y ≥= 4x + 2.
Use a solid line for ≥.
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 2C Continued
Graph the solutions of the linear inequality.
4x – y + 2 ≤ 0
Step 3 The inequality is ≥, so
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 2C Continued
Check
y ≥ 4x + 2
3
4(–3)+ 2
3
–12 + 2
3 ≥ –10 
Substitute ( –3, 3) for (x, y)
because it is not on the
boundary line.
The point (–3, 3) satisfies the
inequality, so the graph is
Holt Algebra 1
6-5 Solving Linear Inequalities
Check It Out! Example 2a
Graph the solutions of the linear inequality.
4x – 3y > 12
Step 1 Solve the inequality for y.
4x – 3y > 12
–4x
–4x
–3y > –4x + 12
y<
–4
Step 2 Graph the boundary line y =
Use a dashed line for <.
Holt Algebra 1
– 4.
6-5 Solving Linear Inequalities
Check It Out! Example 2a Continued
Graph the solutions of the linear inequality.
4x – 3y > 12
Step 3 The inequality is <, so
Holt Algebra 1
6-5 Solving Linear Inequalities
Check It Out! Example 2a Continued
Graph the solutions of the linear inequality.
4x – 3y > 12
Check
y<
–6
–6
–6 <
–4
(1) – 4
–4

Substitute ( 1, –6) for (x, y)
because it is not on the
boundary line.
Holt Algebra 1
The point (1, –6) satisfies the
inequality, so the graph is
6-5 Solving Linear Inequalities
Check It Out! Example 2c
Graph the solutions of the linear inequality.
Step 1 The inequality is
Step 2 Graph the boundary
line
=
. Use a solid line for
≥.
Step 3 The inequality is ≥,
Holt Algebra 1
6-5 Solving Linear Inequalities
Check It Out! Example 2c Continued
Graph the solutions of the linear inequality.
Substitute (0, 0) for (x, y) because it
is not on the boundary line.
Check
y≥
0
x+1
(0) + 1
0
0+1
0 ≥
1
A false statement means that the half-plane containing
(0, 0) should NOT be shaded. (0, 0) is not one of the
solutions, so the graph is shaded correctly.
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 3a: Application
necklace requires 40 beads, and a bracelet
Let x represent the number of necklaces and y the
number of bracelets.
Write an inequality. Use ≤ for “at most.”
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 3a Continued
Necklace
40x
plus
bracelet
is at
most
285
+
15y
≤
285
Solve the inequality for y.
40x + 15y ≤ 285
–40x
–40x
15y ≤ –40x + 285
Subtract 40x from
both sides.
Divide both sides by 15.
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 3b
b. Graph the solutions.
Step 1 Since Ada cannot make a
negative amount of jewelry, the
system is graphed only in
line
for ≤.
Holt Algebra 1
=
. Use a solid line
6-5 Solving Linear Inequalities
Example 3b Continued
b. Graph the solutions.
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
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 3c
c. Give two combinations of necklaces and
Two different combinations of
with 285 beads could be 2
necklaces and 8 bracelets or 5
necklaces and 3 bracelets.
(2, 8)


(5, 3)
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 4A: Writing an Inequality from a Graph
Write an inequality to represent the graph.
y-intercept: 1; slope:
Write an equation in slopeintercept form.
The graph is shaded above a
dashed boundary line.
Replace = with > to write the inequality
Holt Algebra 1
6-5 Solving Linear Inequalities
Example 4B: Writing an Inequality from a Graph
Write an inequality to represent the graph.
y-intercept: –5 slope:
Write an equation in slopeintercept form.
The graph is shaded below a
solid boundary line.
Replace = with ≤ to write the inequality
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Warm-Up
1. You can spend at most \$12.00 for drinks at a picnic. Iced
tea costs \$1.50 a gallon, and lemonade costs \$2.00 per
gallon. Write an inequality to describe the situation. Graph
the solutions, describe reasonable solutions, and then give
two possible combinations of drinks you could buy.
2. Write an inequality to represent
the graph.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Objective
Graph and solve systems of linear
inequalities in two variables.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
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.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Example 1A: 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.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Check It Out! Example 1b
Tell whether the ordered pair is a solution of
the given system.
y > –x + 1
(0, 0);
y>x–1
(0, 0)
y > –x + 1
0
–1(0) + 1
0
0+1
0 > 1
(0, 0)
y>x–1
0
0–1
0 ≥ –1
(0, 0) is not a solution to the system because it does
not satisfy both inequalities.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
System of Inequalities
Graph the
following
inequalities
on the same
graph:
y<2
x ≥ -1
y > x -2
Holt Algebra 1
The Graph
of the
system is the
overlap, or
intersection,
of the 3
half-planes
shown.
6-6 Solving Systems of Linear Inequalities
A Triangular Solution Region
A solution of
a system of
inequalities
is
an ordered
pair that is
a solution
of each
inequality
in the
Holt Algebra 1
(2,1) is a solution
because:
(2,1)
y<2
1<2
x ≥ -1
2 ≥ -1
y > x -2
1 > 2 -2
6-6 Solving Systems of Linear Inequalities
Solution Region Between Parallel
Lines
What is the
system of
inequalities
represente
d by this
graph?
Holt Algebra 1
The graph of one
inequality is the
half-plane
below
the line y = 3
The graph of the
other inequality
is the half-plane
above
the line y = 1
6-6 Solving Systems of Linear Inequalities
Solution Region Between Parallel
Lines
region of the
graph is the
horizontal
band that lies
between
the two
horizontal
lines, but not
on the lines.
Holt Algebra 1
The solution
of this system
of inequalities
can be
represented
by this
compound
inequality:
1<y<3
6-6 Solving Systems of Linear Inequalities
Example 2A: Solving a System of Linear Inequalities
by Graphing
Graph the system of linear inequalities. Give two
ordered pairs that are solutions and two that are
not solutions.
y≤3
y > –x + 5

(–1, 4)
Graph the system.
y≤3
y > –x + 5
(8, 1) and (6, 3) are solutions.
(–1, 4) and (2, 6) are not solutions.
Holt Algebra 1

(2, 6)

(6, 3)
(8, 1)

6-6 Solving Systems of Linear Inequalities
Example 2B: Solving a System of Linear Inequalities
by Graphing
Graph the system of linear inequalities. Give two
ordered pairs that are solutions and two that are
not solutions.
–3x + 2y ≥ 2
y < 4x + 3
–3x + 2y ≥ 2
2y ≥ 3x + 2
Holt Algebra 1
Write the first inequality in slopeintercept form.
6-6 Solving Systems of Linear Inequalities
Example 2B Continued
Graph the system.
y < 4x + 3
(2, 6) and (1, 3) are solutions.
(0, 0) and (–4, 5) are not solutions.
Holt Algebra 1
(–4, 5)


(2, 6)
 (1, 3)
(0, 0)

6-6 Solving Systems of Linear Inequalities
In Lesson 6-4, you saw that in systems of
linear equations, if the lines are parallel, there
are no solutions. With systems of linear
inequalities, that is not always true.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Example 3A: Graphing Systems with Parallel
Boundary Lines
Graph the system of linear inequalities.
y ≤ –2x – 4
y > –2x + 5
This system has
no solutions.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Example 3B: Graphing Systems with Parallel
Boundary Lines
Graph the system of linear inequalities.
y > 3x – 2
y < 3x + 6
The solutions are all points
between the parallel lines but
not on the dashed lines.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Check It Out! Example 3a
Graph the system of linear inequalities.
y>x+1
y≤x–3
This system has
no solutions.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
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 (\$)
Holt Algebra 1
Mowing
20
Raking
10
6-6 Solving Systems of Linear Inequalities
Example 4 Continued
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
y≤7
20x + 10y > 125
Holt Algebra 1
He can do at most 9
mowing jobs.
He can do at most 7
raking jobs.
He wants to earn more
than \$125.
6-6 Solving Systems of Linear Inequalities
Example 4 Continued
Step 2 Graph the system.
The graph should be in only the first quadrant
because the number of jobs cannot be negative.
Solutions
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Example 4 Continued
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
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
An ordered pair solution of the system need not
have whole numbers, but answers to many
application problems may be restricted to whole
numbers.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
1. Graph
Lesson Quiz: Part I
y<x+2
.
5x + 2y ≥ 10
Give two ordered pairs that are solutions and
two that are not solutions.
solutions: (4, 4), (8, 6);
not solutions: (0, 0), (–2, 3)
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Lesson Quiz: Part II
2. Dee has at most \$150 to spend on restocking
dolls and trains at her toy store. Dolls cost \$7.50
and trains cost \$5.00. Dee needs no more than
10 trains and she needs at least 8 dolls. Show
and describe all possible combinations of dolls
and trains that Dee can buy. List two possible
combinations.
Holt Algebra 1
6-6 Solving Systems of Linear Inequalities
Lesson Quiz: Part II Continued