Holt Algebra 2 2-1 Solving Linear Equations and Inequalities 2-1 Solving Linear Equations and...

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities 2-1Solving Linear Equations and Inequalities

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Warm Up

1. 2x + 5 – 3x –x + 5 2. –(w – 2)

3. 6(2 – 3g)

–w + 2

12 – 18g

Graph on a number line.

4. t > –2

5. Is 2 a solution of the inequality –2x < –6? Explain. –4 –3 –2 –1 0 1 2 3 4 5

Simplify each expression.

No; when 2 is substituted for x, the inequality is false: –4 < –6

Holt Algebra 2


Solve linear equations using a variety of methods.Solve linear inequalities.

Objectives

Holt Algebra 2


equationsolution set of an equationlinear equation in one variableidentifycontradictioninequality

Vocabulary

Holt Algebra 2


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


Linear Equations in One variable

Nonlinear Equations

4x = 8

3x – = –9

2x – 5 = 0.1x +2

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.

+ 1 = 32

+ 1 = 41

3 – 2x = –5

Holt Algebra 2


Holt Algebra 2


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


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?

Example 1: Consumer Application

Holt Algebra 2


Example 1 Continued

monthly charge plus

additional minute charge

times

12.95 0.07

number of additional minutes

total charge

+

=

Let m represent the number of additional minutes that Nina used.

m 14.56* =

Model

Holt Algebra 2


Solve.

12.95 + 0.07m = 14.56

0.07m = 1.610.07 0.07

m = 23

Subtract 12.95 from both sides.

Divide both sides by 0.07.

Nina used 23 additional minutes.

Example 1 Continued

–12.95 –12.95

Holt Algebra 2


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


Check It Out! Example 1 Continued

Let c represent the number of additional cups needed.

one cup plusadditional

cup height

times

3.25 0.25

number of additional

cups

total height

+

=

c 14.00* =

Model

Holt Algebra 2



3.25 + 0.25c = 14.00

0.25c = 10.75

0.25 0.25

c = 43

44 cups fit between the 14 in. shelves.

Solve.

Subtract 3.25 from both sides.

Divide both sides by 0.25.

–3.25 –3.25

Holt Algebra 2


Example 2: Solving Equations with the Distributive Property

Solve 4(m + 12) = –36

Divide both sides by 4.

Method 1

The quantity (m + 12) is multiplied by 4, so divide by 4 first.

4(m + 12) = –364 4

m + 12 = –9

m = –21

–12 –12 Subtract 12 from both sides.

Holt Algebra 2


Check 4(m + 12) = –36

4(–21 + 12) –36

4(–9) –36–36 –36

Example 2 Continued

Holt Algebra 2


Example 2 Continued

Distribute 4.

Distribute before solving.

4m + 48 = –36

4m = –84


Divide both sides by 4.=4m –84 4 4

m = –21

Solve 4(m + 12) = –36

Method 2

Holt Algebra 2


Divide both sides by 3.

Method 1

The quantity (2 – 3p) is multiplied by 3, so divide by 3 first.

3(2 – 3p) = 42

3 3

Check It Out! Example 2a

Solve 3(2 –3p) = 42.

Subtract 2 from both sides.

–3p = 12

2 – 3p = 14 –2 –2

–3 –3 Divide both sides by –3.

p = –4

Holt Algebra 2


Check 3(2 – 3p) = 42

3(2 + 12) 426 + 36 42

42 42

Check It Out! Example 2a Continued

Holt Algebra 2


Distribute 3.

Method 2


6 – 9p = 42

–9p = 36


Divide both sides by –9.=–9p 36–9 –9

p = –4

Check It Out! Example 2a Continued

Solve 3(2 – 3p) = 42 .

Holt Algebra 2


Divide both sides by –3.

Method 1

The quantity (5 – 4r) is multiplied by –3, so divide by –3 first.

–3(5 – 4r) –9–3 –3

=

Check It Out! Example 2b

Solve –3(5 – 4r) = –9.

Subtract 5 from both sides.

–4r = –2

5 – 4r = 3 –5 –5

Holt Algebra 2


Check It Out! Example 2b Continued

Divide both sides by –4.=–4 –4–4r –2

r =

–9 –9

Check –3(5 –4r) = –9

–3(5 – 4• ) –9

–3(5 – 2) –9

–3(3) –9

Solve –3(5 – 4r) = –9.

Method 1

Holt Algebra 2


Distribute 3.


–15 + 12r = –9

12r = 6

+15 +15Add 15 to both sides.

Divide both sides by 12.=12r 612 12

Check It Out! Example 2b Continued

r =

Solve –3(5 – 4r) = –9.

Method 2

Holt Algebra 2


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


Example 3: Solving Equations with Variables on Both Sides

Simplify each side by combining like terms.

–11k + 25 = –6k – 10

Collect variables on the right side.

Add.

Collect constants on the left side.

Isolate the variable.

+11k +11k

25 = 5k – 10

35 = 5k

5 5

7 = k

+10 + 10

Solve 3k– 14k + 25 = 2 – 6k – 12.

Holt Algebra 2



Solve 3(w + 7) – 5w = w + 12.

Simplify each side by combining like terms.

–2w + 21 = w + 12

Collect variables on the right side.

Add.

Collect constants on the left side.

Isolate the variable.

+2w +2w

21 = 3w + 12

9 = 3w3 33 = w

–12 –12

Holt Algebra 2


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


Solve 3v – 9 – 4v = –(5 + v).

Example 4A: Identifying Identities and Contractions

3v – 9 – 4v = –(5 + v)

Simplify.–9 – v = –5 – v + v + v

–9 ≠ –5 x Contradiction

The equation has no solution. The solution set is the empty set, which is represented by the symbol .

Holt Algebra 2


Solve 2(x – 6) = –5x – 12 + 7x.

Example 4B: Identifying Identities and Contractions

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


Solve 5(x – 6) = 3x – 18 + 2x.

The equation has no solution. The solution set is the empty set, which is represented by the symbol .

Check It Out! Example 4a

5(x – 6) = 3x – 18 + 2x

Simplify.5x – 30 = 5x – 18

–5x –5x

–30 ≠ –18 x Contradiction

Holt Algebra 2


Solve 3(2 –3x) = –7x – 2(x –3).

3(2 –3x) = –7x – 2(x –3)

Simplify.6 – 9x = –9x + 6

+ 9x +9x

6 = 6 Identity

The solutions set is all real numbers, or .

Check It Out! Example 4b

Holt Algebra 2


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


These properties also apply to inequalities expressed with >, ≥, and ≤.

Holt Algebra 2


To check an inequality, test• the value being compared with x • a value less than that, and• a value greater than that.

Helpful Hint

Holt Algebra 2


Solve and graph 8a –2 ≥ 13a + 8.

Example 5: Solving Inequalities

Subtract 13a from both sides.8a – 2 ≥ 13a + 8

–13a –13a

–5a – 2 ≥ 8Add 2 to both sides. +2 +2

–5a ≥ 10Divide both sides by –5 and reverse the inequality.

–5 –5

–5a ≤ 10

a ≤ –2

Holt Algebra 2


Example 5 Continued

Check Test values in the original inequality. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

•

Test x = –4 Test x = –2 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.

So –1 is not a solution.


–18 ≥ –18 –10 ≥ –5 x

Solve and graph 8a – 2 ≥ 13a + 8.

Holt Algebra 2


Solve and graph x + 8 ≥ 4x + 17.

Subtract x from both sides.x + 8 ≥ 4x + 17

–x –x

8 ≥ 3x +17Subtract 17 from both sides.–17 –17

–9 ≥ 3xDivide both sides by 3.

3 3

–9 ≥ 3x

–3 ≥ x or x ≤ –3


Holt Algebra 2


Check Test values in the original inequality.

Test x = –6 Test x = –3 Test x = 0

–6 + 8 ≥ 4(–6) + 17 –3 +8 ≥ 4(–3) + 17 0 +8 ≥ 4(0) + 17

2 ≥ –7


So 0 is not a solution.


5 ≥ 5 8 ≥ 17 x


Solve and graph x + 8 ≥ 4x + 17.

–6 –5 –4 –3 –2 –1 0 1 2 3

•

Holt Algebra 2


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


Lesson Quiz: Part II

y = –4

x = 6

all real numbers, or

Solve.

2. 2(3x – 1) = 34

3. 4y – 9 – 6y = 2(y + 5) – 3

4. r + 8 – 5r = 2(4 – 2r)

5. –4(2m + 7) = (6 – 16m)

no solution, or

Holt Algebra 2


Lesson Quiz: Part III

5. Solve and graph.

12 + 3q > 9q – 18 q < 5

–2 –1 0 1 2 3 4 5 6 7

°

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities 2-2 Proportional Reasoning

Holt Algebra 2

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 2


Warm UpWrite as a decimal and a percent.

1.

2.

0.4; 40%

1.875; 187.5%

Holt Algebra 2


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


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


Apply proportional relationships to rates, similarity, and scale.

Objective

Holt Algebra 2


ratioproportionratesimilarindirect measurement

Vocabulary

Holt Algebra 2


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


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


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.

Reading Math

Holt Algebra 2


Solve each proportion.

Example 1: Solving Proportions

A.

206.4 = 24p Set cross products equal.

=

=16 24 p 12.9

16 24 p 12.9

206.4 24p 24 24 Divide both sides.

8.6 = p

14 c 88 132

=

=

=

B.

14 c 88 132

88c = 1848

=88c 184888 88

c = 21

Holt Algebra 2


Solve each proportion.

A.

924 = 84y Set cross products equal.

=

=

y 77 12 84

y 77 12 84

Divide both sides.

11 = y

15 2.5 x 7

=

924 84y 84 84

=

=

B.

15 2.5 x 7

2.5x =105

=2.5x 1052.5 2.5

x = 42


Holt Algebra 2


Percent is a ratio that means per hundred.

For example:

30% = 0.30 =

Remember!

30100

Because percents can be expressed as ratios, you can use the proportion

to solve percent problems.

Holt Algebra 2


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?

Example 2: Solving Percent Problems

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


Example 2 Continued

Method 1 Use a proportion.

Cross multiply.

Solve for x.

So 405 voters are planning to vote for that candidate.

Method 2 Use a percent equation.22.5% 0.225 Divide the percent

by 100.

Percent (as decimal) whole = part

0.225 1800 = x

405 = x

x = 405

22.5(1800) = 100x

Holt Algebra 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



Method 1 Use a proportion.

Cross multiply.

Solve for x.

Clay High School has 1240 students.

Method 2 Use a percent equation.

Divide the percent by 100.

0.35x = 434

35% = 0.35

x = 1240

Percent (as decimal) whole = part

x = 1240

100(434) = 35x

Holt Algebra 2


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


Ryan ran 600 meters and counted 482 strides. How long is Ryan’s stride in inches? (Hint: 1 m ≈ 39.37 in.)

Example 3: Fitness Application

Use a proportion to find the length of his stride in meters.

Find the cross products.

600 m 482 strides

x m 1 stride=

600 = 482x

x ≈ 1.24 m

Write both ratios in the form .

metersstrides

Holt Algebra 2


Example 3: Fitness Application continued

Convert the stride length to inches.

Ryan’s stride length is approximately 49 inches.

is the conversion factor. 39.37 in.1 m

≈ 1.24 m1 stride length

39.37 in.1 m

49 in.1 stride length

Holt Algebra 2


Use a proportion to find the length of his stride in meters.


Luis ran 400 meters in 297 strides. Find his stride length in inches.

x ≈ 1.35 m

400 = 297x Find the cross products.

400 m 297 strides

x m 1 stride=

Write both ratios in the form .

metersstrides

Holt Algebra 2


Convert the stride length to inches.

Luis’s stride length is approximately 53 inches.


is the conversion factor. 39.37 in.1 m

≈ 1.35 m1 stride length

39.37 in.1 m

53 in.1 stride length

Holt Algebra 2


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.

The ratio of the corresponding side lengths of similar figures is often called the scale factor.

Reading Math

Holt Algebra 2


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


Example 4 Continued

= height of ∆XAB width of ∆XAB

height of ∆XYZ width of ∆XYZ

=3 x

9 6

9x = 18, so x = 2

Step 2 To find the width of ∆XAB, use a proportion.

Holt Algebra 2


Example 4 Continued

The width is 2 units, and the height is 3 units, so the coordinates of A are (–2, 3).

BA

X

YZ

Step 3 To graph ∆XAB, first find the coordinate of A.

Holt Algebra 2


∆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


= width of ∆DGH height of ∆DGH

width of ∆DEF height of ∆DEF


Step 2 To find the height of ∆DGH, use a proportion.

6x = 12, so x = 2

=36 4

x

Holt Algebra 2


The width is 3 units, and the height is 2 units, so the coordinates of H are (0, –2).


G(–3, 0)D(0, 0)

H(0, –2)

● ●

●

●●

●

E(–6, 0)

F(0,–4)

Step 3

To graph ∆DGH, first find the coordinate of H.

Holt Algebra 2


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 using the shadows are similar, so Luka can use a proportion to find h the height of the house.

=6

9 h22

=Shadow of tree Height of tree

Shadow of house Height of house

6h = 198

h = 33

The house is 33 feet high.

9 ft

6 ft

h ft

22 ft

Holt Algebra 2


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 6 h

90=

Shadow of climber Height of climber

Shadow of treeHeight of tree

20h = 540

h = 27

The tree is 27 feet high.

6 ft

20 ft

h ft

90 ft


Holt Algebra 2


Lesson Quiz: Part ISolve each proportion.

1. 2.

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?

4. Gina earned $68.75 for 5 hours of tutoring. Approximately how much did she earn per minute?

k = 8 g = 42

$0.23

1200

Holt Algebra 2


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.

YZ

AB

X

Lesson Quiz: Part II

Holt Algebra 2


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

Lesson Quiz: Part III

Date post:	17-Dec-2015
Category:	Documents
Upload:	amice-gibbs
View:	226 times
Download:	3 times

Holt Algebra 2 2-1 Solving Linear Equations and Inequalities 2-1 Solving Linear Equations and...

Documents