Linear
Inequalities
in
2-5
2-5 Linear Inequalities in Two Variables
Two Variables
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
2-5
Linear Inequalities in Two Variables
Warm Up
Find the intercepts of each line.
1. 3x + 2y = 18 (0, 9), (6, 0)
2. 4x – y = 8
(0, –8), (2, 0)
3. 5x + 10 = 2y (0, 5), (–2, 0)
Write the function in slope-intercept form.
Then graph.
4. 2x + 3y = –3
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Objectives
Graph linear inequalities on the
coordinate plane.
Solve problems using linear inequalities.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Vocabulary
linear inequality
boundary line
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Linear functions form the basis of linear
inequalities. A linear inequality in two variables
relates two variables using an inequality symbol,
such as y > 2x – 4. Its graph is a region of the
coordinate plane bounded by a line. The line is a
boundary line, which divides the coordinate
plane into two regions.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
For example, the line
y = 2x – 4, shown at
right, divides the
coordinate plane into
two parts: one where
y > 2x – 4 and one
where y < 2x – 4. In
the coordinate plane
higher points have
larger y values, so the
region where
y > 2x – 4 is above
the boundary line
where y = 2x – 4.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
To graph y ≥ 2x – 4,
make the boundary
line solid, and shade
the region above the
line. To graph
y > 2x – 4, make the
boundary line dashed
because y-values
equal to 2x – 4 are
not included.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Helpful Hint
Think of the underlines in the symbols ≤ and ≥
as representing solid lines on the graph.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Example 1A: Graphing Linear Inequalities
Graph the inequality
The boundary line is
y-intercept of 2 and a slope of
Draw the boundary line
dashed because it is not
part of the solution.
Then shade the region above
the boundary line to show
.
Holt Algebra 2
.
which has a
.
2-5
Linear Inequalities in Two Variables
Example 1A Continued
Check Choose a point in the solution region, such
as (3, 2) and test it in the inequality.
?
?
2>1
The test point satisfies the
inequality, so the solution
region appears to be correct.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Example 1B: Graphing Linear Inequalities
Graph the inequality y ≤ –1.
Recall that y= –1 is a
horizontal line.
Step 1 Draw a solid line for
y=–1 because the boundary
line is part of the graph.
Step 2 Shade the region
below the boundary line to
show where y < –1.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Example 1B Continued
Check The point (0, –2) is a
solution because –2 ≤ –1.
Note that any point on or
below y = –1 is a solution,
regardless of the value of x.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Check It Out! Example 1a
Graph the inequality y ≥ 3x –2.
The boundary line is y = 3x – 2 which has a
y–intercept of –2 and a slope of 3.
Draw a solid line because it
is part of the solution.
Then shade the region
above the boundary line to
show y > 3x – 2.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Check It Out! Example 1a Continued
Check Choose a point in the solution region, such as
(–3, 2) and test it in the inequality.
y ≥ 3x –2
?
2 ≥ 3(–3) –2
?
2 ≥ (–9) –2
?
2 > –11 
The test point satisfies the
inequality, so the solution
region appears to be correct.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Check It Out! Example 1b
Graph the inequality y < –3.
Recall that y = –3 is a
horizontal line.
Step 1 Draw the boundary
line dashed because it is not
part of the solution.
Step 2 Shade the region
below the boundary line to
show where y < –3.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Check It Out! Example 1b Continued
Check The point (0, –4) is a
solution because –4 < –3.
Note that any point below
y < –4 is a solution,
regardless of the value of x.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
If the equation of the boundary line is not in
slope-intercept form, you can choose a test point
that is not on the line to determine which region
to shade. If the point satisfies the inequality,
then shade the region containing that point.
Otherwise, shade the other region.
Helpful Hint
The point (0, 0) is the easiest point to test if it is
not on the boundary line.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Example 2: Graphing Linear Inequalities Using
Intercepts
Graph 3x + 4y ≤ 12 using intercepts.
Step 1 Find the intercepts.
Substitute x = 0 and y = 0 into 3x + 4y = 12 to
find the intercepts of the boundary line.
y-intercept
x-intercept
3x + 4y = 12
3(0) + 4y = 12
3x + 4y = 12
3x + 4(0) = 12
4y = 12
y=3
3x = 12
x=4
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Example 2 Continued
Step 2 Draw the boundary line.
The line goes through (0, 3) and
(4, 0). Draw a solid line for the
boundary line because it is part of
the graph.
Step 3 Find the correct region
to shade.
Substitute (0, 0) into the
inequality. Because 0 + 0 ≤ 12
is true, shade the region that
contains (0, 0).
Holt Algebra 2
(0, 3)
(4, 0)
2-5
Linear Inequalities in Two Variables
Check It Out! Example 2
Graph 3x – 4y > 12 using intercepts.
Step 1 Find the intercepts.
Substitute x = 0 and y = 0 into 3x – 4y = 12 to
find the intercepts of the boundary line.
y-intercept
3x – 4y = 12
3(0) – 4y = 12
x-intercept
3x – 4y = 12
3x – 4(0) = 12
– 4y = 12
y=–3
3x = 12
x=4
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Check It Out! Example 2
Step 2 Draw the boundary line.
The line goes through (0, –3) and
(4, 0). Draw the boundary line
dashed because it is not part of
the solution.
Step 3 Find the correct region
to shade.
Substitute (0, 0) into the
inequality. Because 0 + 0 >12
is false, shade the region that
does not contain (0, 0).
Holt Algebra 2
(4, 0)
(0, –3)
2-5
Linear Inequalities in Two Variables
Many applications of inequalities in two variables
use only nonnegative values for the variables.
Graph only the part of the plane that includes
realistic solutions.
Caution
Don’t forget which variable represents which
quantity.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Example 3: Problem-Solving Application
A school carnival charges $4.50 for
adults and $3.00 for children. The
school needs to make at least $135 to
cover expenses.
A. Using x as the adult ticket price and y as
the child ticket price, write and graph an
inequality for the amount the school
makes on ticket sales.
B. If 25 child tickets are sold, how many adult
tickets must be sold to cover expenses?
Holt Algebra 2
2-5
1
Linear Inequalities in Two Variables
Understand the Problem
The answer will be in two parts: (1) an
inequality graph showing the number of each
type of ticket that must be sold to cover
expenses (2) the number of adult tickets that
must be sold to make at least $135 if 25 child
tickets are sold.
List the important information:
• The school sells tickets at $4.50 for adults
and $3.00 for children.
• The school needs to make at least $135.
Holt Algebra 2
2-5
2
Linear Inequalities in Two Variables
Make a Plan
Let x represent the number of adult tickets
and y represent the number of child tickets
that must be sold. Write an inequality to
represent the situation.
Adult
price
4.50
times
number of
adult
tickets
•
x
plus
child
price
+
3.00
times
number
of child
tickets
is at
least
total.
•
y

135
An inequality that models the problem is 4.5x + 3y ≥ 135.
Holt Algebra 2
2-5
3
Linear Inequalities in Two Variables
Solve
Find the intercepts of the boundary line.
4.5(0) + 3y = 135
4.5x + 3(0) = 135
y = 45
Graph the boundary line
through (0, 45) and (30, 0)
as a solid line.
Shade the region above the
line that is in the first
quadrant, as ticket sales
cannot be negative.
Holt Algebra 2
x = 30
2-5
Linear Inequalities in Two Variables
If 25 child tickets are sold,
4.5x + 3(25) ≥ 135
4.5x + 75 ≥ 135
_
4.5x ≥ 60, so x ≥ 13.3
Substitute 25 for y in 4.5x + 3y ≥ 135.
Multiply 3 by 25.
A whole number of tickets must be sold.
At least 14 adult tickets must be sold.
4
Look Back
14($4.50) + 25($3.00) = $138.00, so the answer is
reasonable.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Check It Out! Example 3
A café gives away prizes. A large prize
costs the café $125, and the small
prize costs $40. The café will not spend
more than $1500. How many of each
prize can be awarded? How many small
prizes can be awarded if 4 large prizes
are given away?
Holt Algebra 2
2-5
1
Linear Inequalities in Two Variables
Understand the Problem
The answer will be in two parts: (1) an
inequality graph showing the number of each
type of prize awarded not too exceed a
certain amount (2) the number of small
prizes awarded if 4 large prizes are awarded.
List the important information:
• The café awarded large prizes valued at
$125 and $40 for small prizes.
• The café will not spend over $1500.
Holt Algebra 2
2-5
2
Linear Inequalities in Two Variables
Make a Plan
Let x represent the number of small prizes and y
represent the number of large prizes, the total not
too exceed $1500. Write an inequality to represent
the situation.
Small
prize
times
40
number
awarded
plus
large
prize
x
+
125
times
number
awarded
is less
than
total.
y
≤
1500
An inequality that models the problem is 40x + 125y ≤ 135.
Holt Algebra 2
2-5
3
Linear Inequalities in Two Variables
Solve
Find the intercepts of the boundary line.
40(0) + 125y = 1500
y = 12
40x + 125(0) = 1500
x = 37.5
Graph the boundary line
through (0, 12) and (37.5, 0)
as a solid line.
Shade the region below the
line that is in the first
quadrant, as prizes awarded
cannot be negative.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
If 4 large prizes are awarded,
40x + 125(4) ≤ 1500
40x + 500 ≤ 1500
40x ≥ 1000, so x ≤ 25
Substitute 4 for y in 40x + 125y ≤ 135.
Multiply 125 by 4.
A whole number of small prizes must
be awarded.
No more than 25 small prizes can be awarded.
4
Look Back
$40(25) + $125(4) = $1500, so the answer is
reasonable.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
You can graph a linear
inequality that is solved
for y with a graphing
calculator. Press
and use the left arrow
key to move to the left
side.
Each time you press
you will see one of the
graph styles shown here.
You are already familiar
with the line style.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Example 4: Solving and Graphing Linear Inequalities
Solve
for y. Graph the solution.
Multiply both sides by
8x – 2y > 8
–2y > –8x + 8 Subtract 8x from both sides.
y < 4x – 4
Holt Algebra 2
Divide by –2, and reverse the
inequality symbol.
2-5
Linear Inequalities in Two Variables
Example 4 Continued
Use the calculator option to shade below the line
y < 4x – 4.
Note that the graph is shown in
the STANDARD SQUARE
window.
(
by
6:ZStandard followed
5:ZSquare).
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Check It Out! Example 4
Solve 2(3x – 4y) > 24 for y. Graph the solution.
3x – 4y > 12
–4y > –3x + 12
Divide both sides by 2.
Subtract 3x from both sides.
Divide by –4, and reverse the
inequality symbol.
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Check It Out! Example 4 Continued
Use the calculator option to shade below the line
.
Note that the graph is shown in
the STANDARD SQUARE
window.
(
by
6:ZStandard followed
5:ZSquare).
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Lesson Quiz: Part I
1. Graph 2x –5y  10
using intercepts.
2. Solve –6y < 18x – 12 for y.
Graph the solution.
y > –3x + 2
Holt Algebra 2
2-5
Linear Inequalities in Two Variables
Lesson Quiz: Part II
3. Potatoes cost a chef $18 a box, and carrots cost
$12 a box. The chef wants to spend no more
than $144. Use x as the number of boxes of
potatoes and y as the number of boxes of
carrots.
a. Write an inequality for the number of boxes
the chef can buy.
18x + 12y ≤ 144
b. How many boxes of potatoes can the chef
order if she orders 4 boxes of carrot?
no more than 5
Holt Algebra 2