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Solving Linear Equations and
2-1
2-1 Inequalities
Solving Linear Equations and Inequalities
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
2-1
Solving Linear Equations and Inequalities
Warm Up
Simplify each expression.
2. –(w – 2) –w + 2
1. 2x + 5 – 3x –x + 5
3. 6(2 – 3g) 12 – 18g
Graph on a number line.
4. t > –2
–4 –3 –2 –1
0
1
2
3
4
5
5. Is 2 a solution of the inequality –2x < –6? Explain.
No; when 2 is substituted for x, the inequality is false:
–4 < –6
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Objectives
Solve linear equations using a variety
of methods.
Solve linear inequalities.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Vocabulary
equation
solution set of an equation
linear equation in one variable
identify
contradiction
inequality
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
An equation is a mathematical statement that two
expressions are equivalent. The solution set of an
equation is the value or values of the variable that
make the equation true. A linear equation in one
variable can be written in the form ax = b, where a
and b are constants and a ≠ 0.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Linear Equations in
One variable
4x = 8
3x –
= –9
2x – 5 = 0.1x +2
Nonlinear
Equations
+ 1 = 32
+ 1 = 41
3 – 2x = –5
Notice that the variable in a linear equation is not
under a radical sign and is not raised to a power other
than 1. The variable is also not an exponent and is not
in a denominator.
Solving a linear equation requires isolating the
variable on one side of the equation by using the
properties of equality.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
To isolate the variable, perform the inverse or
opposite of every operation in the equation on
both sides of the equation. Do inverse
operations in the reverse order of operations.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 1: Consumer Application
The local phone company charges
$12.95 a month for the first 200 of air
time, plus $0.07 for each additional
minute. If Nina’s bill for the month was
$14.56, how many additional minutes
did she use?
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 1 Continued
Let m represent the number of additional minutes that
Nina used.
Model
monthly
plus
charge
12.95
Holt Algebra 2
+
additional
minute
charge
times
0.07
*
number of
additional = total
charge
minutes
m
=
14.56
2-1
Solving Linear Equations and Inequalities
Example 1 Continued
Solve.
12.95 + 0.07m = 14.56
–12.95
–12.95
0.07m =
0.07
1.61
0.07
Subtract 12.95 from both
sides.
Divide both sides by 0.07.
m = 23
Nina used 23 additional minutes.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 1
Stacked cups are to be placed in a
pantry. One cup is 3.25 in. high and
each additional cup raises the stack
0.25 in. How many cups fit between
two shelves 14 in. apart?
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 1 Continued
Let c represent the number of additional cups needed.
Model
additional
cup
one cup plus
height
3.25
Holt Algebra 2
+
0.25
times
*
number of
total
additional = height
cups
c
=
14.00
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 1 Continued
Solve.
3.25 + 0.25c = 14.00
–3.25
–3.25
0.25c = 10.75
0.25
0.25
Subtract 3.25 from both
sides.
Divide both sides by 0.25.
c = 43
44 cups fit between the 14 in. shelves.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 2: Solving Equations with the Distributive
Property
Solve 4(m + 12) = –36
Method 1
The quantity (m + 12) is multiplied by 4, so divide
by 4 first.
4(m + 12) = –36
4
4
m + 12 = –9
–12 –12
m = –21
Holt Algebra 2
Divide both sides by 4.
Subtract 12 from both sides.
2-1
Solving Linear Equations and Inequalities
Example 2 Continued
Check
4(m + 12) = –36
4(–21 + 12)
4(–9)
–36
Holt Algebra 2
–36
–36
–36 
2-1
Solving Linear Equations and Inequalities
Example 2 Continued
Solve 4(m + 12) = –36
Method 2
Distribute before solving.
4m + 48 = –36
–48 –48
Distribute 4.
Subtract 48 from both sides.
4m = –84
4m –84
=
4
4
m = –21
Holt Algebra 2
Divide both sides by 4.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2a
Solve 3(2 –3p) = 42.
Method 1
The quantity (2 – 3p) is multiplied by 3, so divide
by 3 first.
3(2 – 3p) = 42
3
3
2 – 3p = 14
–2
–2
–3p = 12
–3
–3
p = –4
Holt Algebra 2
Divide both sides by 3.
Subtract 2 from both sides.
Divide both sides by –3.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2a Continued
Check
Holt Algebra 2
3(2 – 3p) =
3(2 + 12)
6 + 36
42
42
42
42
42 
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2a Continued
Solve 3(2 – 3p) = 42 .
Method 2
Distribute before solving.
6 – 9p = 42
–6
–6
–9p = 36
–9p 36
=
–9 –9
p = –4
Holt Algebra 2
Distribute 3.
Subtract 6 from both sides.
Divide both sides by –9.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2b
Solve –3(5 – 4r) = –9.
Method 1
The quantity (5 – 4r) is multiplied by –3, so divide by
–3 first.
–3(5 – 4r) = –9
–3
–3
5 – 4r = 3
–5
–5
–4r = –2
Holt Algebra 2
Divide both sides by –3.
Subtract 5 from both sides.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2b Continued
Solve –3(5 – 4r) = –9.
Method 1
–4r –2
–4
–4
r=
Divide both sides by –4.
=
Check
–3(5 –4r) = –9
–3(5 – 4• ) –9
–3(5 – 2)
–3(3)
–9
Holt Algebra 2
–9
–9
–9 
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2b Continued
Solve –3(5 – 4r) = –9.
Method 2
Distribute before solving.
–15 + 12r = –9
+15
+15
12r = 6
12r
6
=
12
12
r=
Holt Algebra 2
Distribute 3.
Add 15 to both sides.
Divide both sides by 12.
2-1
Solving Linear Equations and Inequalities
If there are variables on both sides of the
equation, (1) simplify each side. (2) collect all
variable terms on one side and all constants
terms on the other side. (3) isolate the
variables as you did in the previous problems.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 3: Solving Equations with
Variables on Both Sides
Solve 3k– 14k + 25 = 2 – 6k – 12.
Simplify each side by combining
–11k + 25 = –6k – 10
like terms.
+11k
+11k
Collect variables on the right side.
25 = 5k – 10
+10
+ 10
35 = 5k
5
5
7=k
Holt Algebra 2
Add.
Collect constants on the left side.
Isolate the variable.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 3
Solve 3(w + 7) – 5w = w + 12.
–2w + 21 =
+2w
w + 12
+2w
21 =
–12
9 =
3
3=
Holt Algebra 2
3w + 12
–12
3w
3
w
Simplify each side by combining
like terms.
Collect variables on the right side.
Add.
Collect constants on the left side.
Isolate the variable.
2-1
Solving Linear Equations and Inequalities
You have solved equations that have a
single solution. Equations may also have
infinitely many solutions or no solution.
An equation that is true for all values of the
variable, such as x = x, is an identity. An
equation that has no solutions, such as
3 = 5, is a contradiction because there
are no values that make it true.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 4A: Identifying Identities and Contractions
Solve 3v – 9 – 4v = –(5 + v).
3v – 9 – 4v = –(5 + v)
–9 – v = –5 – v
+v
+v
–9 ≠ –5
x
Simplify.
Contradiction
The equation has no solution. The solution set
is the empty set, which is represented by the
symbol .
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 4B: Identifying Identities and Contractions
Solve 2(x – 6) = –5x – 12 + 7x.
2(x – 6) = –5x – 12 + 7x
Simplify.
2x – 12 = 2x – 12
–2x
–2x
–12 = –12 
Identity
The solutions set is all real number, or .
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 4a
Solve 5(x – 6) = 3x – 18 + 2x.
5(x – 6) = 3x – 18 + 2x
5x – 30 = 5x – 18
–5x
–5x
–30 ≠ –18 x
Simplify.
Contradiction
The equation has no solution. The solution set
is the empty set, which is represented by the
symbol .
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 4b
Solve 3(2 –3x) = –7x – 2(x –3).
3(2 –3x) = –7x – 2(x –3)
6 – 9x = –9x + 6
Simplify.
+ 9x
+9x
6=6 
Identity
The solutions set is all real numbers, or .
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
An inequality is a statement that compares two
expressions by using the symbols <, >, ≤, ≥, or ≠.
The graph of an inequality is the solution set, the set
of all points on the number line that satisfy the
inequality.
The properties of equality are true for inequalities,
with one important difference. If you multiply or
divide both sides by a negative number, you must
reverse the inequality symbol.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
These properties also apply to inequalities
expressed with >, ≥, and ≤.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Helpful
Hint
To check an inequality, test
• the value being compared with x
• a value less than that, and
• a value greater than that.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 5: Solving Inequalities
Solve and graph 8a –2 ≥ 13a + 8.
8a – 2 ≥ 13a + 8
–13a
–13a
–5a – 2 ≥ 8
+2
+2
–5a ≥ 10
–5a ≤ 10
–5
–5
a ≤ –2
Holt Algebra 2
Subtract 13a from both sides.
Add 2 to both sides.
Divide both sides by –5 and
reverse the inequality.
2-1
Solving Linear Equations and Inequalities
Example 5 Continued
Solve and graph 8a – 2 ≥ 13a + 8.
Check Test values in
the original inequality.
Test x = –4
•
–10 –9
Test x = –2
–8 –7 –6 –5 –4
–3 –2 –1
Test x = –1
8(–4) – 2 ≥ 13(–4) + 8 8(–2) – 2 ≥ 13(–2) + 8 8(–1) – 2 ≥ 13(–1) + 8
–34 ≥ –44 
So –4 is a
solution.
Holt Algebra 2
–18 ≥ –18 
So –2 is a
solution.
–10 ≥ –5 x
So –1 is not
a solution.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 5
Solve and graph x + 8 ≥ 4x + 17.
x + 8 ≥ 4x + 17
–x
–x
8 ≥ 3x +17
–17
–17
–9 ≥ 3x
–9 ≥ 3x
3
3
–3 ≥ x or x ≤ –3
Holt Algebra 2
Subtract x from both sides.
Subtract 17 from both sides.
Divide both sides by 3.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 5 Continued
Solve and graph x + 8 ≥ 4x + 17.
Check Test values in the
original inequality.
Test x = –6
•
–6 –5
Test x = –3
–6 + 8 ≥ 4(–6) + 17 –3 +8 ≥ 4(–3) + 17
2 ≥ –7 
So –6 is a
solution.
Holt Algebra 2
5≥5

So –3 is a
solution.
–4 –3 –2 –1
0
1
2
3
Test x = 0
0 +8 ≥ 4(0) + 17
8 ≥ 17 x
So 0 is not
a solution.
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part I
1. Alex pays $19.99 for cable service each month.
He also pays $2.50 for each movie he orders
through the cable company’s pay-per-view
service. If his bill last month was $32.49, how
many movies did Alex order?
5 movies
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part II
Solve.
x=6
2. 2(3x – 1) = 34
3. 4y – 9 – 6y = 2(y + 5) – 3 y = –4
4. r + 8 – 5r = 2(4 – 2r)
all real numbers, or 
5. –4(2m + 7) =
no solution, or
Holt Algebra 2
(6 – 16m)
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part III
5. Solve and graph.
12 + 3q > 9q – 18
q<5
°
–2 –1
Holt Algebra 2
0
1
2
3
4
5
6
7
Linear Equations
and Inequalities
2-1
2-2 Solving
Proportional
Reasoning
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
2-1
Solving Linear Equations and Inequalities
Warm Up
Write as a decimal and a percent.
1.
0.4; 40%
2.
1.875; 187.5%
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Warm Up Continued
Graph on a coordinate plane.
3. A(–1, 2)
4. B(0, –3)
A(–1, 2)
B(0, –3)
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Warm Up Continued
5. The distance from Max’s house to the park is
3.5 mi. What is the distance in feet?
(1 mi = 5280 ft)
18,480 ft
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Objective
Apply proportional relationships to
rates, similarity, and scale.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Vocabulary
ratio
proportion
rate
similar
indirect measurement
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Recall that a ratio is a comparison of two
numbers by division and a proportion is an
equation stating that two ratios are equal.
In a proportion, the cross products are
equal.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
If a proportion contains a variable, you can
cross multiply to solve for the variable.
When you set the cross products equal, you
create a linear equation that you can solve
by using the skills that you learned in
Lesson 2-1.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Reading Math
In a ÷ b = c ÷ d, b and c are the means, and a
and d are the extremes. In a proportion, the
product of the means is equal to the product of
the extremes.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 1: Solving Proportions
Solve each proportion.
A. 16 = 24
p
12.9
B.
16
24
=
p
12.9
206.4 = 24p
206.4 24p
=
24
24
8.6 = p
Holt Algebra 2
14
c
=
88
132
14
c
=
88
132
Set cross products
equal.
Divide both sides.
88c = 1848
88c
1848
=
88
88
c = 21
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 1
Solve each proportion.
A.
y
77
=
12
84
B.
y
77
=
12
84
924 = 84y
924
84y
=
84
84
11 = y
Holt Algebra 2
15
2.5
=
x
7
15
2.5
=
x
7
Set cross
products equal.
Divide both sides.
2.5x =105
2.5x
=
2.5
105
2.5
x = 42
2-1
Solving Linear Equations and Inequalities
Because percents can be expressed as
ratios, you can use the proportion
to solve percent problems.
Remember!
Percent is a ratio that means per hundred.
For example:
30% = 0.30 = 30
100
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 2: Solving Percent Problems
A poll taken one day before an election
showed that 22.5% of voters planned to vote
for a certain candidate. If 1800 voters
participated in the poll, how many indicated
that they planned to vote for that candidate?
You know the percent and the total number of
voters, so you are trying to find the part of the
whole (the number of voters who are planning
to vote for that candidate).
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 2 Continued
Method 1 Use a
proportion.
Method 2 Use a percent
equation.
22.5%  0.225
Divide the percent
by 100.
Percent (as decimal)  whole = part
0.225  1800 = x
22.5(1800) = 100x Cross
multiply.
405 = x
Solve for x.
x = 405
So 405 voters are planning to vote for that candidate.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2
At Clay High School, 434 students, or 35% of
the students, play a sport. How many students
does Clay High School have?
You know the percent and the total number of
students, so you are trying to find the part of
the whole (the number of students that Clay
High School has).
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 2 Continued
Method 1 Use a
proportion.
Method 2 Use a percent
equation.
35% = 0.35
Divide the percent
by 100.
Percent (as decimal)  whole = part
100(434) = 35x Cross
multiply.
0.35x = 434
x = 1240
Solve for x.
x = 1240
Clay High School has 1240 students.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
A rate is a ratio that involves two different
units. You are familiar with many rates,
such as miles per hour (mi/h), words per
minute (wpm), or dollars per gallon of
gasoline. Rates can be helpful in solving
many problems.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 3: Fitness Application
Ryan ran 600 meters and counted 482
strides. How long is Ryan’s stride in inches?
(Hint: 1 m ≈ 39.37 in.)
Use a proportion to find the length of his stride in
meters.
600 m
xm
=
482 strides
1 stride
600 = 482x
x ≈ 1.24 m
Holt Algebra 2
meters
Write both ratios in the form strides .
Find the cross products.
2-1
Solving Linear Equations and Inequalities
Example 3: Fitness Application continued
Convert the stride length to inches.
39.37 in.
is the conversion factor.
1m
1.24 m

1 stride length
39.37 in.
49 in.
≈
1m
1 stride length
Ryan’s stride length is approximately 49 inches.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 3
Luis ran 400 meters in 297 strides. Find his
stride length in inches.
Use a proportion to find the length of his stride in
meters.
400 m
xm
=
297 strides
1 stride
400 = 297x
x ≈ 1.35 m
Holt Algebra 2
meters
Write both ratios in the form strides .
Find the cross products.
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 3 Continued
Convert the stride length to inches.
39.37 in.
is the conversion factor.
1m
1.35 m
39.37 in.
53 in.

≈
1 stride length
1m
1 stride length
Luis’s stride length is approximately 53 inches.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Similar figures have the same shape but not
necessarily the same size. Two figures are similar
if their corresponding angles are congruent and
corresponding sides are proportional.
Reading Math
The ratio of the corresponding side lengths of
similar figures is often called the scale factor.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 4: Scaling Geometric Figures in the
Coordinate Plane
∆XYZ has vertices X(0, 0), Y(–6, 9) and Z(0, 9).
∆XAB is similar to ∆XYZ with a vertex at B(0, 3).
Graph ∆XYZ and ∆XAB on the same grid.
Step 1 Graph ∆XYZ. Then draw XB.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 4 Continued
Step 2 To find the width of ∆XAB, use a proportion.
height of ∆XAB
width of ∆XAB
=
height of ∆XYZ
width of ∆XYZ
3
9
=
x
6
9x = 18, so x = 2
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example 4 Continued
Step 3
To graph ∆XAB, first
find the coordinate
of A.
The width is 2 units,
and the height is 3
units, so the
coordinates of A are
(–2, 3).
Holt Algebra 2
Z
Y
A
B
X
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 4
∆DEF has vertices D(0, 0), E(–6, 0) and F(0, –4).
∆DGH is similar to ∆DEF with a vertex at G(–3, 0).
Graph ∆DEF and ∆DGH on the same grid.
Step 1 Graph ∆DEF. Then draw DG.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 4 Continued
Step 2 To find the height of ∆DGH, use a proportion.
width of ∆DGH
=
width of ∆DEF
height of ∆DGH
height of ∆DEF
3
x
=
6
4
6x = 12, so x = 2
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 4 Continued
Step 3
To graph ∆DGH,
first find the
coordinate of H.
The width is 3 units,
and the height is 2
units, so the
coordinates of H are
(0, –2).
Holt Algebra 2
E(–6, 0) G(–3, 0)
●
●
● D(0, 0)
●H(0, –2)
●
F(0,–4)
2-1
Solving Linear Equations and Inequalities
Example 5: Nature Application
The tree in front of Luka’s house casts a 6-foot
shadow at the same time as the house casts a
22-fot shadow. If the tree is 9 feet tall, how tall
is the house?
Sketch the situation. The triangles formed by
9 ft
using the shadows are similar, so Luka can use
a proportion to find h the height of the house.
6 ft
6
22
=
9
h
Shadow of tree
Shadow of house
=
Height of tree
Height of house
h ft
6h = 198
h = 33
The house is 33 feet high.
Holt Algebra 2
22 ft
2-1
Solving Linear Equations and Inequalities
Check It Out! Example 5
A 6-foot-tall climber casts a 20-foot long
shadow at the same time that a tree casts a
90-foot long shadow. How tall is the tree?
Sketch the situation. The triangles formed by
using the shadows are similar, so the climber can
use a proportion to find h the height of the tree.
20
90
=
6
h
6 ft
20 ft
Shadow of climber Shadow of tree
=
Height of climber
Height of tree
20h = 540
h ft
h = 27
The tree is 27 feet high.
Holt Algebra 2
90 ft
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part I
Solve each proportion.
1.
k=8
2.
g = 42
3. The results of a recent survey showed that
61.5% of those surveyed had a pet. If 738
people had pets, how many were surveyed?
1200
4. Gina earned $68.75 for 5 hours of tutoring.
Approximately how much did she earn per
minute? $0.23
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part II
5. ∆XYZ has vertices, X(0, 0), Y(3, –6), and Z(0, –6).
∆XAB is similar to ∆XYZ, with a vertex at B(0, –4).
Graph ∆XYZ and ∆XAB on the same grid.
X
B
Z
Holt Algebra 2
A
Y
2-1
Solving Linear Equations and Inequalities
Lesson Quiz: Part III
6. A 12-foot flagpole casts a 10 foot-shadow. At the
same time, a nearby building casts a 48-foot
shadow. How tall is the building?
57.6 ft
Holt Algebra 2