section 2.1

Section 2.1

The Addition & Subtraction Properties of Equality

OBJECTIVESDetermine whether a number satisfies an equation.

A

OBJECTIVES

Use the addition and subtraction properties of equality to solve equations.

B

OBJECTIVESUse both properties together to solve an equation.

C

SOLUTIONSThe replacements of the variable that make the equation a true statement.

“Solved the equation.”

SOLUTIONSA number that is a solution of an equation satisfies the equation.

SOLUTIONS

3 satisfies the equation.

For example

2 3 = 6

3 is a solution of 2x = 6 because:

PROCEDURE:

1. Simplify both sides if necessary.

Solving Equations by Adding or Subtracting

PROCEDURE:Solving Equations by Adding or Subtracting

2. Add or subtract the same numbers on both sides so that one side contains only variables.

PROCEDURE:

3. Add or subtract the same expressions on both sides so that the other side contains only numbers.

Solving Equations by Adding or Subtracting

Section 2.1Exercise #1

Chapter 2Equations, Problem Solving, and Inequalities

Does the number 3 satisfy the equation6 = 9 – x ?

6 = 9 – x

Does the number 3 satisfy the equation6 = 9 – x ?

6 = 9 – 3

Yes.

Solve x – 2

7 = 3

7.

Add 2

7:

x – 2

7 + 2

7 = 3

7 + 2

7

Simplify: x = 5

7

Solve – 2x + 7

8 + 3x – 5

8 = 5

8.

Simplify: x + 2

8 = 5

8

Subtract 2

8:

x + 2

8 – 2

8 = 5

8 – 2

8

x = 3

8

2 = 3x – 3 + 5 – 2x

Solve 2 = 3 x – 1 + 5 – 2x.

Use Distributive Property:

Simplify: 2 = x + 2

2 – 2 = x + 2 – 2 Subtract 2:

0 = x

Solve 2 + 5 x + 1 = 8 + 5x.


Simplify: 7 + 5x = 8 + 5x

7 + 5x – 5x = 8 + 5x – 5x

2 + 5x + 5 = 8 + 5x

Subtract 5x:

7 = 8Impossible. No solution.

Solve – 3 – 2 x – 1 = – 1 – 2x.


Simplify: – 1 – 2x = – 1 – 2x

– 3 – 2x + 2 = – 1 – 2x

Identity: All real numbers.

Section 2.2

The Multiplication and Division Properties of Equality

OBJECTIVESUse the multiplication and division properties of equality to solve equations.

A

OBJECTIVES

Multiply by reciprocals to solve equations.

B

OBJECTIVESMultiply by LCMs to solve equations.

C

OBJECTIVESSolve applications involving percents.

D

PROCEDURE:

Multiply both sides of the equation by the LCM of the denominators.

Clearing Fractions

Or, multiply each term by the LCM.

Solve 2

3x = – 4

3 • 2

3x = – 4 • 3

x = – 6

2x = – 12 2 2

Solve – 2

3x = – 6

– 32

• – 23

x = – 6 • – 32

x = 9

x = – 6 • – 3

2 3

1

Solve x

4 + 2x

3 = 11

LCD = 12

3x + 8x = 132

12 x

4

+ 12 2x3

= 11 • 12

11x = 132

x = 12

3 4

1 1

Solve x

3 – x

5 = 2

LCD = 15

5x – 3x = 30

•15 – 15 = 15 23 5x x

2x = 30

x = 15

5 3

1 1

Solve. x – 2

5 – x + 1

8 = 0

8 – 2 – 5 + 1 = 0x x

• – 2 + 140 – 40 = 40 05 8

x x

8x – 16 – 5x – 5 = 0

3x = 21

8 5

1 1

3x – 21 = 0

x = 7

LCD = 40

What percent of 55 is 11?

55x = 11

x • 55 = 11

x = 0.20

x = 20%

Nine is 36% percent of what number?

9 = 0.36x

9 = 0.36 • x

25 = x

The number is 25.

Section 2.3

Linear Equations

OBJECTIVESSolve linear equations in one variable.

A

OBJECTIVES

Solve a literal equation for one of the unknowns.

B

PROCEDURE:

1. Clear fractions (Multiply by the LCM of the denominator).

Solving Linear Equations

1 7 – = – 24 6 12x x

PROCEDURE:Solving Linear Equations

12 12 11 7 – = – 24 6 122

x x

1 7 – = – 24 6 12x x

PROCEDURE:Solving Linear Equations2. Remove parentheses and

simplify.

3x – 2 = 7(x – 2)

3x – 2 = 7x – 14

PROCEDURE:Solving Linear Equations3. Isolate the variable.

3x – 2 + 2 = 7x – 14 + 2 3x – 2 = 7x – 14

3x = 7x – 12


3x = 7x – 12

3x – 7x = 7x – 7x – 12

– 4x = – 12

PROCEDURE:Solving Linear Equations4. If coefficient is NOT 1, divide

both sides by coefficient – 4x = – 12

– 4x – 4

= – 12 – 4


x = 3

– 4x – 4

= – 12 – 4


3 1 7 – = 3 – 24 6 12

5. Check your answer in the original equation.


9

12 – 212 = 7

12

3 1 7 – = 3 – 24 6 12

7

12 = 712

Literal EquationAn equation that contains several variables.

D = RT and I = Prt are examples.


Section 2.3

231 – = 5 3

Solve. 15

+ 3x x LCD = 15

– 28x = 112

3 – 5 = 23 + 5x x

3 – 5x = 23x + 115

– 5x = 23x + 112

x = – 4

Solve for h in S = 1

3r 2h

h = 3S

r 2

3S = 1r 2h

3S1r 2

= h

Section 2.4

Problem Solving: Integer, General, and Geometry Problems

OBJECTIVES

Integer problemsAUse RSTUV method to solve:

General word problemsB

Geometry word problemsC

PROCEDURE:

1. Read the problem and decide what is asked for (the unknown).

RSTUV Method for Solving Word Problems

PROCEDURE:RSTUV Method for Solving Word Problems2. Select a variable to

represent this unknown.

PROCEDURE:RSTUV Method for Solving Word Problems

3. Think of a plan to help you write an equation.

PROCEDURE:

4. Use algebra to solve the resulting equation.


PROCEDURE:RSTUV Method for Solving Word Problems5. Verify the answer.

The sum of two numbers is 75. If one of the numbers is 15 more than the other, what are the numbers?

Let x = smaller number

x + 15 = larger number

+ + 15 = 75x x

x + x + 15 = 75

2x + 15 = 75

2x = 60

The sum of two numbers is 75. If one of the numbers is 15 more than the other, what are the numbers?

Let x = smaller number

x + 15 = larger number

2x = 60

x = 30

x + 15 = 45

The numbers are 30 and 45.

A man invested a certain amount of money in stocks and bonds. His annual return from these investments is $840. If the stocks produce $230 more inreturns than the bonds, how much money does he receive annually fromeach investment?

Let x = return from stocks

840 – x = return from bonds

= 840 – + 230x x

x = 840 – x + 230

x = 1070 – x

2x = 1070

x = 535

840 – x = 305

$535 from stocks and $305 from bonds.

= 840 – + 230x x

Find the measure of an angle whose supplement is 50° less than 3 times its complement.

Let x = measure of the angle90 – x = complement of the angle

180 – x = supplement of the angle

180 – = 3 90 – – 50x x

Find the measure of an angle whose supplement is 50° less than 3 times its complement.

180 – = 3 90 – – 50x x

180 – x = 270 – 3x – 50

180 – x = 220 – 3x

180 + 2x = 220

2x = 40

x = 20

The angle measures 20°.

Section 2.5

Problem Solving: Motion, Mixture, and Investment Problems

OBJECTIVES

Motion problemsAUse the RSTUV method to solve:

Mixture problemsB

Investment problemsC

PROCEDURE:

1. Read the problem and decide what is asked for (the unknown).


PROCEDURE:RSTUV Method for Solving Word Problems

2. Select a variable to represent this unknown.

PROCEDURE:RSTUV Method for Solving Word Problems3. Think of a plan to help you

write an equation.

PROCEDURE:RSTUV Method for Solving Word Problems4. Use algebra to solve the

resulting equation.

PROCEDURE:RSTUV Method for Solving Word Problems5. Verify the answer.

A freight train leaves a station traveling at 30 milesper hour. Two hours later, a passenger train leaves the same station traveling in the same direction at 42 miles per hour. How long does it take for the passenger train to catch the freight train?

Let x = time of the passenger trainx + 2 = time of the freight train

D = 42x

= 30 + 2D x

r = 42 r = 30

D = r t

30 + 2 = 42x x

30x + 60 = 42x

30x = 42x – 60

– 12x = – 60

x = 5

The passenger train catchesthe freight train in 5 hours.

D = 42x = 30 + 2D x

An investor bought some municipal bonds yielding 5%annually and some certificates of deposit yielding 7%annually. If her total investment amounts to$20,000 and her annual return is $1160, howmuch money is invested in bonds and howmuch in certificates of deposit?

Total interest = 1160

Let x = amount invested in bonds at 5% 20,000 – x = amount invested in CD's at 7%

I = RT interest = 0.05x

interest = 0.07 20,000 – x

Total interest = 1160

0.05 + 0.07 20,000 – = 1160x x

5 + 7 20,000 – = 116,000x x

5x + 140,000 – 7x = 116,000

140,000 – 2x = 116,000


x = 12,000

20,000 – x = 8,000

$12,000 is invested in bonds at 5%,$8000 is invested in CD's at 7%

140,000 – 2x = 116,000

– 2x = – 24,000


I = RT

Section 2.6

Formulas and Geometry Applications

OBJECTIVES

Solve a formula for one variable and use the result to solve a problem.

A

OBJECTIVES

Solve problems involving geometric formulas.

B

OBJECTIVES

Solve geometric problems involving angle measurement.

C

OBJECTIVES

Solve an application.D

DEFINITION Angle of measure 1

One complete revolution around a circle is 360°, and

1

360 of a complete revolution is 1°

DEFINITION

Two angles whose sum is 90°are called complementary angles.

Complementary Angles

DEFINITION

30° + 60° = 90°

Complementary Angles

90 – x

x

DEFINITION

Two angles whose sum is 180°are called supplementary angles.

Supplementary Angles

DEFINITION

50° + 130° = 180°

Supplementary Angles

180 – x

x

DEFINITION

Are formed in opposite sides of two intersecting lines andhave equal measures.

Vertical Angles

DEFINITIONVertical Angles

1

3

24


Section 2.6

a. Solve for m.

b. How many miles did you travel if the cost of the ride was $20.65?

The cost C of riding a taxi is C = 1.95 + 0.85m, where mis the number of miles (or fraction) you travel.

a. Solve for m. The cost C of riding a taxi is C = 1.95 + 0.85m, where mis the number of miles (or fraction) you travel.

C = 1.95 + 0.85m

C – 1.95 = 0.85m

C – 1.95

0.85 = m

m = C – 1.95

0.85

The cost C of riding a taxi is C = 1.95 + 0.85m, where mis the number of miles (or fraction) you travel.

b. How many miles did you travel if the cost of the ride was $20.65?

m = C – 1.95

0.85; C = 20.65

m = 20.65 – 1.95

0.85 = 18.70

0.85 = 22

22 miles were traveled.

Section 2.7

Properties of Inequalities

OBJECTIVES

Determine which of two numbers is greater.

A

OBJECTIVES

Solve and graph linear inequalities.

B

OBJECTIVES

Write, solve, and graph a compound inequality.

C

OBJECTIVES

Solve an application.D

3 > 1 or 1 < 3

DEFINITIONSOn the number line, greater numbers are always to the right of smaller ones.

0 1 2 3–1–2–3

DEFINITIONSAddition and Subtraction Properties of Inequalities.

You can add or subtract the same number on both sides of an inequality and obtain an equivalent inequality.


If x <y

then x + a<y + a

or x – b<y – b


If x >y

then x + a>y + a

or x – b>y – b

DEFINITIONSMultiplication and Division Properties of Inequalities for Positive NumbersYou can multiply or divide both sides of an inequality by any positive number and obtain an equivalent inequality.

DEFINITIONS

If x <y

Multiplication and Division Properties of Inequalities for Positive Numbers

then ax <ay

or x

a < ya

DEFINITIONS

If x >y

Multiplication and Division Properties of Inequalities for Positive Numbers

then ax >ay

or x

a > ya

Section 2.7


a. – 3 – 5

b. – 1

3 3

Fill in the blank with < or > to make the resultingstatement true.

0 – 5 – 3

0 –

13

3

a. – 3 – 5

b. – 1

3 3 <

>

0 – 5 – 3

0 –

13

3

Fill in the blank with < or > to make the resultingstatement true.

+ 2 – + 4

a 2 4

. x x xSolve and graph the inequality.

+ 24 – + 4 42 4 4

x x x

LCD = 4

–2 + + 2x x x

– + 2x x

–2 2x

– 1x

2

+ 2 – + 4

a 2 4

. x x xSolve and graph the inequality.

– 1x

0 – 1

b. x + 1 Š 3 and – 2x < 6

Solve and graph the inequality.

x Š 2 and – 2x < 6

x Š 2 and x > – 3

–3 < x Š 2

0 – 3 2

section 2.1

Documents

linear equationsprocedure

linear equations2

linear equations4

linear equations5

linear equations3

original equation

inequalitiesall real

objectivessolve applications