holt algebra 1 2-1 solving equations by adding or subtracting warm up evaluate. 1. + 4 2. 0.51 +...

Holt Algebra 1

2-1Solving Equations by Adding or Subtracting

Warm UpEvaluate.

1. + 4

2. 0.51 + (0.29)

Give the opposite of each number.

3. 8 4.

Evaluate each expression for a = 3 and b = 2.

5. a + 5 6. 12 b

0.8

13

23

23

3 23

–8

14 8

23

Holt Algebra 1


Solve one-step equations in one variable by using addition or subtraction.

Objective

Holt Algebra 1


An equation is a mathematical statement that two expressions are equal.

A solution of an equation is a value of the variable that makes the equation true.

To find solutions, isolate the variable. A variable is isolated when it appears by itself on one side of an equation, and not at all on the other side.

Holt Algebra 1


Inverse Operations

Operation Inverse Operation

Addition Subtraction

Subtraction Addition

Isolate a variable by using inverse operations which "undo" operations on the variable.

An equation is like a balanced scale. To keep the balance, perform the same operation on both sides.

Holt Algebra 1


Solve the equation. Check your answer.

Example 1A: Solving Equations by Using Addition

Since 8 is subtracted from y, add 8 to both sides to undo the subtraction.

y – 8 = 24 + 8 + 8

y = 32

Holt Algebra 1



Example 1B: Solving Equations by Using Addition

= z34

+ 716

+ 716

= z – 716

516 Since is subtracted from z, add to

both sides to undo the subtraction.

716

716

Holt Algebra 1



Example 2B: Solving Equations by Using Subtraction

Since 1.8 is added to t, subtract 1.8 from both sides to undo the addition.

4.2 = t + 1.8 –1.8 – 1.8

2.4 = t

Holt Algebra 1


Check It Out! Example 3b

z = 2

+ 3 4

+ 3 4

Solve – + z = . Check your answer. 5 4

3 4

3 4

Since – is added to z, add

to both sides.

3 4

– + z =

5 4

3 4

Holt Algebra 1


Over 20 years, the population of a town decreased by 275 people to a population of 850. Write and solve an equation to find the original population.

Example 4: Application

Write an equation to represent the relationship.

+ 275 + 275

p =1125

p – d = c

original population minus

current population

decrease in

populationis

p – 275 = 850 Since 275 is subtracted from p, add 275 to both sides to undo the subtraction.

p – d = c

The original population was 1125 people.

Holt Algebra 1


A person's maximum heart rate is the highest rate, in beats per minute, that the person's heart should reach. One method to estimate maximum heart rate states that your age added to your maximum heart rate is 220. Using this method, write and solve an equation to find a person's age if the person's maximum heart rate is 185 beats per minute.

Check It Out! Example 4

Holt Algebra 1


Check It Out! Example 4 Continued

a + r = 220

ageadded to

220maximum heart rate is


– 185 – 185

a = 35

a + 185 = 220 Substitute 185 for r. Since 185 is added to a, subtract 185 from both sides to undo the addition.

a + r = 220

A person whose maximum heart rate is 185 beats per minute would be 35 years old.

Holt Algebra 1


WORDS

Addition Property of EqualityYou can add the same number to both sides of an equation, and the statement will still be true.

NUMBERS

3 = 3 3 + 2 = 3 + 2 5 = 5

ALGEBRA a = b a + c = b + c

Properties of Equality

Holt Algebra 1


WORDS

Subtraction Property of EqualityYou can subtract the same number from both sides of an equation, and the statement will still be true.

NUMBERS

7 = 7 7 – 5 = 7 – 5 2 = 2

ALGEBRA a = b a – c = b – c


Holt Algebra 1

2-2Solving Equations by Multiplying or Dividing

Warm-Up

Solve each equation.

1. r – 4 = –8

2.

3. This year a high school had 578 sophomores enrolled. This is 89 less than the number enrolled last year. Write and solve an equation to find the number of sophomores enrolled last year.

Holt Algebra 1


Solve one-step equations in one variable by using multiplication or division.

Objective

Holt Algebra 1


Inverse Operations

Operation Inverse Operation

Multiplication Division

Division Multiplication

Solving an equation that contains multiplication or division is similar to solving an equation that contains addition or subtraction. Use inverse operations to undo the operations on the variable.

Holt Algebra 1


Solve the equation.

Example 1A: Solving Equations by Using Multiplication

Since j is divided by 3, multiply both sides by 3 to undo the division.–24 = j

–8 =j3

Holt Algebra 1




Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication.y = –20

0.5y = –10

Holt Algebra 1


Remember that dividing is the same as multiplying by the reciprocal. When solving equations, you will sometimes find it easier to multiply by a reciprocal instead of dividing. This is often true when an equation contains fractions.

Holt Algebra 1


Solve the equation.

Example 3A: Solving Equations That Contain Fractions

w = 24

w = 2056 The reciprocal of is . Since w is

multiplied by , multiply both sides

by .

56

65

566

5

Holt Algebra 1


Solve the equation. Check your answer.Check It Out! Example 3a

– = b14

15

The reciprocal of is 5. Since b is

multiplied by , multiply both sides

by 5.

151

5

= b54–

Holt Algebra 1




Ciro puts of the money he earns from mowing lawns into a college education fund. This year Ciro added $285 to his college education fund. Write and solve an equation to find how much money Ciro earned mowing lawns this year.

14

one-fourth times earnings equals college fund

m = $1140

Substitute 285 for c. Since m is divided by 4, multiply both sides by 4 to undo the division.

Ciro earned $1140 mowing lawns.

Holt Algebra 1


Check it Out! Example 4


The distance in miles from the airport that a plane should begin descending, divided by 3, equals the plane's height above the ground in thousands of feet. A plane began descending 45 miles from the airport. Use the equation to find how high the plane was flying when the descent began.

Distance divided by 3 equals height in thousands of feet

15 = h

Substitute 45 for d.

The plane was flying at 15,000 ft when the descent began.

Holt Algebra 1


WORDS

Multiplication Property of EqualityYou can multiply both sides of an equation by the same number, and the statement will still be true.

NUMBERS

6 = 6 6(3) = 6(3) 18 = 18

ALGEBRA a = b ac = bc


Holt Algebra 1


Properties of EqualityDivision Property of EqualityYou can divide both sides of an equation by the same nonzero number, and the statement will still be true.

WORDS

a = b (c ≠ 0)

8 = 8

2 = 2

ALGEBRA

NUMBERS 84

84

=

ac

ac=

Holt Algebra 1


Lesson Quiz: Part 1Solve each equation.

1.

2.

3. 8y = 4

4. 126 = 9q

5.

6.

21

2.8

–14

40

Holt Algebra 1


Lesson Quiz: Part 2

7. A person's weight on Venus is about his or her weight on Earth. Write and solve an equation to find how much a person weighs on Earth if he or she weighs 108 pounds on Venus.

910

Holt Algebra 1

2-3Solving Two-Step and Multi-Step Equations

Warm UpEvaluate each expression.

1. 9 –3(–2)

2. 3(–5 + 7)

3.

4. 26 – 4(7 – 5)

Simplify each expression.

5. 10c + c

6. 8.2b + 3.8b – 12b

7. 5m + 2(2m – 7)

8. 6x – (2x + 5)

15

6

18

–4

11c

0

9m – 14

4x – 5

Holt Algebra 1


Solve equations in one variable that contain more than one operation.

Objective

Holt Algebra 1


Alex belongs to a music club. In this club, students can buy a student discount card for $19.95. This card allows them to buy CDs for $3.95 each. After one year, Alex has spent $63.40.

To find the number of CDs c that Alex bought, you can solve an equation.

Cost per CD Total cost

Cost of discount card

Notice that this equation contains multiplication and addition. Equations that contain more than one operation require more than one step to solve. Identify the operations in the equation and the order in which they are applied to the variable. Then use inverse operations and work backward to undo them one at a time.

Holt Algebra 1


Solve 18 = 4a + 10.

Example 1A: Solving Two-Step Equations

18 = 4a + 10

–10 – 10

8 = 4a

8 = 4a4 4

2 = a

Holt Algebra 1


Solve –4 + 7x = 3.

Check it Out! Example 1a

–4 + 7x = 3

+ 4 + 4

7x = 7

7x = 77 7

x = 1

Holt Algebra 1


Solve 1.5 = 1.2y – 5.7.

Check it Out! Example 1b

1.5 = 1.2y – 5.7

+ 5.7 + 5.7

7.2 = 1.2y

7.2 = 1.2y1.2 1.2

6 = y

Holt Algebra 1


Solve .

Example 2A: Solving Two-Step Equations That Contain Fractions

Method 1 Use fraction operations.

Holt Algebra 1


Example 2A Continued

Simplify.

Holt Algebra 1


Solve .

Example 2B: Solving Two-Step Equations That Contain Fractions

Method 1 Use fraction operations.

Holt Algebra 1


Example 2B Continued

Holt Algebra 1


Equations that are more complicated may have to be simplified before they can be solved. You may have to use the Distributive Property or combine like terms before you begin using inverse operations.

Holt Algebra 1


Solve 8x – 21 - 5x = –15.

Example 3A: Simplifying Before Solving Equations

Use the Commutative Property of Addition.

8x – 21 – 5x = –15

8x – 5x – 21 = –15

3x – 21 = –15 Combine like terms.Since 21 is subtracted from 3x, add 21

to both sides to undo the subtraction.+ 21 +21

3x = 6

x = 2

Since x is multiplied by 3, divide both sides by 3 to undo the multiplication.

Holt Algebra 1


Solve 10y – (4y + 8) = –20

Example 3B: Simplifying Before Solving Equations

Write subtraction as addition of the opposite.10y + (–1)(4y + 8) = –20

10y + (–1)(4y) + (–1)( 8) = –20 Distribute –1 on the left side.

Combine like terms.

10y – 4y – 8 = –20

6y – 8 = –20+ 8 + 8

6y = –12

y = –2

Simplify.

Since 8 is subtracted from 6y, add 8 to both sides to undo the subtraction.

6 66y = –12 Since y is multiplied by 6,

divide both sides by 6 to undo the multiplication.

Holt Algebra 1


Sara paid $15.95 to become a member at a gym. She then paid a monthly membership fee. Her total cost for 12 months was $735.95. How much was the monthly fee?


Holt Algebra 1



Let m represent the monthly membership fee that Sara must pay. That means that Sara must pay 12m. However, Sara must also add the amount she spent to become a gym member.

Make a Plan

total cost

initial membership

monthly fee= +

735.95 = 12m + 15.95

Holt Algebra 1



735.95 = 12m + 15.95

Solve3

– 15.95 – 15.95

720 = 12m

720 = 12m

12 12

60 = m

Since 15.95 is added to 12m, subtract 15.95 from both sides to undo the addition.

Since m is multiplied by 12, divide both sides by 12 to undo the multiplication.

Holt Algebra 1


If 3d – (9 – 2d) = 51, find the value of 3d.

Example 5B: Solving Equations to Find an Indicated Value

Since 9 is subtracted from 5d, add 9 to both sides to undo the subtraction.

Step 1 Find the value of d.

3d – (9 – 2d) = 51

Since d is multiplied by 5, divide both sides by 5 to undo the multiplication.

+9 +9

5d = 60

d = 12

3d – 9 + 2d = 51

5d – 9 = 51

Holt Algebra 1


If 3d – (9 – 2d) = 51, find the value of 3d.

Example 5B Continued

d = 12

Step 2 Find the value of 3d.

3(12) To find the value of 3d, substitute 12 for d.

36 Simplify.

Holt Algebra 1

2-4Solving Equations with Variables on Both Sides

Warm UpSimplify.

1. –7(x – 3)

2.

3. 15 – (x – 2)

Solve.

4. 3x + 2 = 8

5.

Holt Algebra 1


Solve equations in one variable that contain variable terms on both sides.

Objective

Holt Algebra 1


To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation.

Helpful HintEquations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms.

Holt Algebra 1


Solve 7n – 2 = 5n + 6.

Example 1: Solving Equations with Variables on Both Sides

7n – 2 = 5n + 6

–5n –5n

2n – 2 = 6

2n = 8

+ 2 + 2

n = 4

Holt Algebra 1


Solve 0.5 + 0.3y = 0.7y – 0.3.


0.5 + 0.3y = 0.7y – 0.3

–0.3y –0.3y

0.5 = 0.4y – 0.3

0.8 = 0.4y

+0.3 + 0.3

2 = y

Holt Algebra 1


To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.

Holt Algebra 1


Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Example 2: Simplifying Each Side BeforeSolving Equations

4 – 6a + 4a = –1 –5(7 – 2a)

4 – 6a + 4a = –1 – 35 + 10a

4 – 2a = –36 + 10a

+36 +3640 – 2a = 10a

+ 2a +2a

40 = 12a

Holt Algebra 1


Solve .

Check It Out! Example 2A

+ 1 + 13 = b – 1

4 = b

Holt Algebra 1


An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions.

A contradiction is an equation that is not true for any value of the variable. It has no solutions.

Holt Algebra 1


WORDS

IdentityWhen solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions.

NUMBERS2 + 1 = 2 + 1

3 = 3

ALGEBRA

2 + x = 2 + x –x –x 2 = 2

Identities and Contradictions

Holt Algebra 1


ContradictionWhen solving an equation, if you get a false equation, the original equation is a contradiction, and it has no solutions.

WORDS

x = x + 3 –x –x 0 = 3

1 = 1 + 2 1 = 3

ALGEBRA

NUMBERS

Identities and Contradictions

Holt Algebra 1


Solve 10 – 5x + 1 = 7x + 11 – 12x.

Example 3A: Infinitely Many Solutions or No Solutions

10 – 5x + 1 = 7x + 11 – 12x

11 – 5x = 11 – 5x

11 = 11

+ 5x + 5x

10 – 5x + 1 = 7x + 11 – 12x

The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.

Holt Algebra 1


Solve 12x – 3 + x = 5x – 4 + 8x.

Example 3B: Infinitely Many Solutions or No Solutions

12x – 3 + x = 5x – 4 + 8x

13x – 3 = 13x – 4

–3 = –4

–13x –13x

12x – 3 + x = 5x – 4 + 8x

The equation 12x – 3 + x = 5x – 4 + 8x is a contradiction. There is no value of x that will make the equation true. There are no solutions.

Holt Algebra 1


Jon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs will that be?


Person Bulbs

Jon 60 bulbs plus 44 bulbs per hour

Sara 96 bulbs plus 32 bulbs per hour

Holt Algebra 1


Example 4: Application Continued

Let b represent bulbs, and write expressions for the number of bulbs planted.

60 bulbs

plus

44 bulbs each hour

the same

as

96 bulbs

plus

32 bulbs each hour

When is ?

60 + 44b = 96 + 32b

60 + 44b = 96 + 32b– 32b – 32b

To collect the variable terms on one side, subtract 32b from both sides.

60 + 12b = 96

Holt Algebra 1



Since 60 is added to 12b, subtract 60 from both sides.

60 + 12b = 96–60 – 60

12b = 36Since b is multiplied by 12,

divide both sides by 12 to undo the multiplication.

b = 3

Holt Algebra 1



After 3 hours, Jon and Sara will have planted the same number of bulbs. To find how many bulbs they will have planted in 3 hours, evaluate either expression for b = 3:

60 + 44b = 60 + 44(3) = 60 + 132 = 192

96 + 32b = 96 + 32(3) = 96 + 96 = 192

After 3 hours, Jon and Sara will each have planted 192 bulbs.

Holt Algebra 1


Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg?


Let g represent Greg's age, and write expressions for his age.

four times Greg's

age

decreased by

3is

equal to

three times Greg's

age

increased by

7 .

4g – 3 = 3g + 7

Holt Algebra 1



4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides.

g – 3 = 7

–3g –3g

Since 3 is subtracted from g, add 3 to both sides.

+ 3 + 3

g = 10

Greg is 10 years old.

Holt Algebra 1

2-5 Solving for a Variable

Warm-up


1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4

3. 6 – 7(a + 1) = –3(2 – a)

4. 4(3x + 1) – 7x = 6 + 5x – 2

5.

6. A painting company charges $250 base plus $16 per hour. Another painting company charges $210 base plus $18 per hour. How long is a job for which the two companies costs are the same?

Holt Algebra 1


Solve a formula for a given variable.

Solve an equation in two or more variables for one of the variables.

Objectives

Holt Algebra 1


A formula is an equation that states a rule for a relationship among quantities.

In the formula d = rt, d is isolated. You can "rearrange" a formula to isolate any variable by using inverse operations. This is called solving for a variable.

Holt Algebra 1


Solving for a VariableStep 1 Locate the variable you are asked to

solve for in the equation.

Step 2 Identify the operations on this variable and the order in which they are applied.

Step 3 Use inverse operations to undo operations and isolate the variable.

Holt Algebra 1



The formula C = d gives the circumference of a circle C in terms of diameter d. The circumference of a bowl is 18 inches. What is the bowl's diameter? Leave the symbol in your answer.

Now use this formula and the information given in the problem.

Holt Algebra 1



Solve the formula d = rt for t. Find the time in hours that it would take Ernst Van Dyk to travel 26.2 miles if his average speed was 18 miles per hour.

Now use this formula and the information given in the problem.

d = rt

Holt Algebra 1


Example 2A: Solving Formulas for a Variable

The formula for the area of a triangle is A = bh, where b is the length of the base, and is the height. Solve for h.

A = bh

2A = bh

Holt Algebra 1


Dividing by a fraction is the same as multiplying by the reciprocal.

Remember!

Holt Algebra 1


Example 2B: Solving Formulas for a VariableThe formula for a person’s typing speed is ,where s is speed in words per minute,

w is number of words typed, e is number of errors, and m is number of minutes typing. Solve for e.

ms = w – 10e–w –w

ms – w = –10e

Holt Algebra 1


A formula is a type of literal equation. A literal equation is an equation with two or more variables. To solve for one of the variables, use inverse operations.

Holt Algebra 1


Example 3: Solving Literal Equations

A. Solve x + y = 15 for x.

x + y = 15 Locate x in the equation.

Since y is added to x, subtract y from both sides to undo the addition.

–y –yx = –y + 15

B. Solve pq = x for q.

pq = x Locate q in the equation.Since q is multiplied by p, divide

both sides by p to undo the multiplication.

Holt McDougal Algebra 1

2-6 Solving Absolute-Value Equations

Warm Up


1. 3x + 5 = 17

2. r – 3.5 = 8.7

3. 4t – 7 = 8t + 3

4.

5. 2(y – 5) – 20 = 0



Solve equations in one variable that contain absolute-value expressions.

Objectives



Recall that the absolute-value of a number is that number’s distance from zero on a number line. For example, |–5| = 5 and |5| = 5.

5 4 3 2 0 1 2 3 4 56 1 6

5 units

For any nonzero absolute value, there are exactly two numbers with that absolute value. For example, both 5 and –5 have an absolute value of 5.

To write this statement using algebra, you would write |x| = 5. This equation asks, “What values of x have an absolute value of 5?” The solutions are 5 and –5. Notice this equation has two solutions.

5 units



To solve absolute-value equations, perform inverse operations to isolate the absolute-value expression on one side of the equation. Then you must consider two cases.



Additional Example 1A: Solving Absolute-Value Equations

Solve the equation.

|x| = 12|x| = 12

Case 1 x = 12

Case 2 x = –12

The solutions are {12, –12}.

Think: What numbers are 12 units from 0?

Rewrite the equation as two cases.

12 units 12 units

10 8 6 4 0 2 4 6 8 1012 2 12•••


2-6 Solving Absolute-Value EquationsAdditional Example 1B: Solving Absolute-Value Equations

3|x + 7| = 24

|x + 7| = 8

The solutions are {1, –15}.

Case 1 x + 7 = 8

Case 2 x + 7 = –8

– 7 –7 – 7 – 7x = 1 x = –15

Since |x + 7| is multiplied by 3, divide both sides by 3 to undo the multiplication.

Think: What numbers are 8 units from 0?

Rewrite the equations as two cases. Since 7 is added to x subtract 7 from both sides of each equation.

Solve the equation.



Solve the equation.


8 =|x 2.5| Think: What numbers are

8 units from 0?

Case 18 = x 2.5

+2.5 +2.5

10.5 = x

+2.5 +2.55.5 = x

Case 2 8 = x 2.5

Rewrite the equations as two cases.

The solutions are {10.5, –5.5}.

8 =|x 2.5|

Since 2.5 is subtracted from x add 2.5 to both sides of each equation.



The table summarizes the steps for solving absolute-value equations.

1. Use inverse operations to isolate the absolute-value expression.

2. Rewrite the resulting equation as two cases that do not involve absolute values.

3. Solve the equation in each of the two cases.

Solving an Absolute-Value Equation



Not all absolute-value equations have two solutions. If the absolute-value expression equals 0, there is one solution. If an equation states that an absolute-value is negative, there are no solutions.


2-6 Solving Absolute-Value EquationsAdditional Example 2A: Special Cases of Absolute-Value

Equations

Solve the equation. 8 = |x + 2| 8

8 = |x + 2| 8+8 + 8

0 = |x + 2|

0 = x + 22 22 = x

Since 8 is subtracted from |x + 2|, add 8 to both sides to undo the subtraction.

There is only one case. Since 2 is added to x, subtract 2 from both sides to undo the addition.

The solution is {2}.


2-6 Solving Absolute-Value EquationsAdditional Example 2B: Special Cases of Absolute-Value

Equations

Solve the equation.

3 + |x + 4| = 0

3 + |x + 4| = 03 3

|x + 4| = 3

Since 3 is added to |x + 4|, subtract 3 from both sides to undo the addition.

Absolute value cannot be negative.

This equation has no solution.



Remember!

Absolute value must be nonnegative because it represents a distance.


2-6 Solving Absolute-Value EquationsCheck It Out! Example 2a

Solve the equation.

2 |2x 5| = 7

2 |2x 5| = 7 2 2

|2x 5| = 5

Since 2 is added to –|2x – 5|, subtract 2 from both sides to undo the addition.

Absolute value cannot be negative.

|2x 5| = 5

This equation has no solution.

Since |2x – 5| is multiplied by negative 1, divide both sides by negative 1.




Solve the equation.

6 + |x 4| = 6

6 + |x 4| = 6+6 +6

|x 4| = 0

x 4 = 0+ 4 +4

x = 4

Since –6 is added to |x 4|, add 6 to both sides.

There is only one case. Since 4 is subtracted from x, add 4 to both sides to undo the addition.



First convert millimeters to meters.

3 mm = 0.003 m Move the decimal point 3 places to the left.

The length of the beam can vary by 0.003m, so find two numbers that are 0.003 units away from 3.5 on a number line.

Additional Example 3: Engineering ApplicationA support beam for a building must be 3.5 meters long. It is acceptable for the beam to differ from the ideal length by 3 millimeters. Write and solve an absolute-value equation to find the minimum and maximum acceptable lengths for the beam.


2-6 Solving Absolute-Value EquationsAdditional Example 3 Continued

Case 1x – 3.5 = 0.003

+3.5 =+3.5 x = 3.503

Case 2x – 3.5 =–0.003

+3.5 =+3.5 x = 3.497

Since 3.5 is subtracted from x, add 3.5 to both sides of each equation.

The minimum length of the beam is 3.497 meters and the maximum length is 3.503 meters.

3.501

0.003 units

3.502 3.5033.5002.4992.4982.497

0.003 units

You can find these numbers by using the absolute-value equation |x – 3.5| = 0.003. Solve the equation by rewriting it as two cases.



First convert millimeters to meters.

180 mm = 0.180 m Move the decimal point 3 places to the left.

The height of the bridge can vary by 0.18 m, so find two numbers that are 0.18 units away from 134 on a number line.

Sydney Harbour Bridge is 134 meters tall. The height of the bridge can rise or fall by 180 millimeters because of changes in temperature. Write and solve an absolute-value equation to find the minimum and maximum heights of the bridge.




Case 1x – 134 = 0.18

+134 =+134 x = 134.18

Since 134 is subtracted from x add 134 to both sides of each equation.

The minimum height of the bridge is 133.82 meters and the maximum height is 134.18 meters.

134.06

0.18 units

134

0.18 units


134.12 134.18133.94133.88133.82

Case 2x – 134 =–0.18

+134 =+134 x = 133.82

You can find these numbers by using the absolute-value equation |x – 134| = 0.18. Solve the equation by rewriting it as two cases.

Holt Algebra 1

2-6 Rates, Ratios, and ProportionsWarm-Up


1. 15 = |x| 2. 2|x – 7| = 14

3. |x + 1|– 9 = –9 4. |5 + x| – 3 = –2

5. 7 + |x – 8| = 6

–15, 15 0, 14

–1 –6, –4

no solution

6. Inline skates typically have wheels with a diameter of 74 mm. The wheels are manufactured so that the diameters vary from this value by at most 0.1 mm. Write and solve an absolute-value equation to find the minimum and maximum diameters of the wheels. |x – 74| = 0.1; 73.9 mm; 74.1 mm

Holt Algebra 1

2-6 Rates, Ratios, and Proportions

Write and use ratios, rates, and unit rates.Write and solve proportions.

Objectives

Holt Algebra 1


A ratio is a comparison of two quantities by division. The ratio of a to b can be written a:b or , where b ≠ 0. Ratios that name the same comparison are said to be equivalent.

A statement that two ratios are equivalent, such as , is called a proportion.

Holt Algebra 1


Reading Math

Read the proportion as

“1 is to 15 as x is to 675”.

Holt Algebra 1


Example 1: Using Ratios

The ratio of the number of bones in a human’s ears to the number of bones in the skull is 3:11. There are 22 bones in the skull. How many bones are in the ears?

There are 6 bones in the ears.

Holt Algebra 1


The ratio of games lost to games won for a baseball team is 2:3. The team has won 18 games. How many games did the team lose?


The team lost 12 games.

Holt Algebra 1


A rate is a ratio of two quantities with different

units, such as Rates are usually written as

unit rates. A unit rate is a rate with a second

quantity of 1 unit, such as or 17 mi/gal. You

can convert any rate to a unit rate.

Holt Algebra 1


Example 2: Finding Unit Rates

Raulf Laue of Germany flipped a pancake 416 times in 120 seconds to set the world record. Find the unit rate. Round your answer to the nearest hundredth.

The unit rate is about 3.47 flips/s.

Holt Algebra 1


A rate such as in which the two quantities

are equal but use different units, is called a

conversion factor. To convert a rate from one

set of units to another, multiply by a conversion

factor.

Holt Algebra 1


Example 3A: Converting Rates

Serena ran a race at a rate of 10 kilometers per hour. What was her speed in kilometers per minute? Round your answer to the nearest hundredth.

The rate is about 0.17 kilometer per minute.

Holt Algebra 1


Helpful Hint

In example 3A , “1 hr” appears to divide out, leaving “kilometers per minute,” which are the units asked for. Use this strategy of “dividing out” units when converting rates.

Holt Algebra 1


Example 3B: Converting Rates

A cheetah can run at a rate of 60 miles per hour in short bursts. What is this speed in feet per minute?

Step 1 Convert the speed to feet per hour.

The speed is 5280 ft per minute.

= =

Holt Algebra 1


In the proportion , the products a • d and b

• c are called cross products. You can solve a

proportion for a missing value by using the

Cross Products property.

Cross Products Property

WORDS NUMBERS ALGEBRA

In a proportion, cross products are equal.

2 • 6 = 3 • 4

If and b ≠ 0

and d ≠ 0then ad = bc.

Holt Algebra 1


Example 4: Solving Proportions

Solve each proportion.

3(m) = 5(9)

3m = 45

m = 15

Use cross products.

Divide both sides by 3.

Use cross products.

6(7) = 2(y – 3)

42 = 2y – 6+6 +648 = 2y

24 = y

A. B.

Add 6 to both sides.Divide both sides by 2.

Holt Algebra 1


A scale is a ratio between two sets of measurements, such as 1 in:5 mi. A scale drawing or scale model uses a scale to represent an object as smaller or larger than the actual object. A map is an example of a scale drawing.

Holt Algebra 1


Example 5A: Scale Drawings and Scale Models

A contractor has a blueprint for a house drawn to the scale 1 in: 3 ft.

A wall on the blueprint is 6.5 inches long. How long is the actual wall?

blueprint 1 in. actual 3 ft.

x • 1 = 3(6.5)

x = 19.5The actual length of the wall is 19.5 feet.

Holt Algebra 1


Lesson Quiz: Part 1

1. In a school, the ratio of boys to girls is 4:3. There are 216 boys. How many girls are there?162

Find each unit rate. Round to the nearest hundredth if necessary.

2. Nuts cost $10.75 for 3 pounds. $3.58/lb

3. Sue washes 25 cars in 5 hours. 5 cars/h

4. A car travels 180 miles in 4 hours. What is the car’s speed in feet per minute? 3960 ft/min

Holt Algebra 1


Lesson Quiz: Part 2


5.

6.

7. A scale model of a car is 9 inches long. The scale is 1:18. How many inches long is the car it represents?

6

16

162 in.

Holt Algebra 1

2-7 Applications of Proportions

Warm UpEvaluate each expression for a = 3, b = –2, c = 5.1. 4a – b 2. 3b2 – 5

3. ab – 2c


4. 5.

14

16

9

7

6.4

Holt Algebra 1


Use proportions to solve problems involving geometric figures.

Use proportions and similar figures to measure objects indirectly.

Objectives

Holt Algebra 1


Similar figures have exactly the same shape but not necessarily the same size.

Corresponding sides of two figures are in the same relative position, and corresponding angles are in the same relative position. Two figures are similar if and only if the lengths of corresponding sides are proportional and all pairs of corresponding angles have equal measures.

Holt Algebra 1


When stating that two figures are similar, use the symbol ~. For the triangles above, you can write ∆ABC ~ ∆DEF. Make sure corresponding vertices are in the same order. It would be incorrect to write ∆ABC ~ ∆EFD.

You can use proportions to find missing lengths in similar figures.

Holt Algebra 1

2-7 Applications of ProportionsExample 1A: Finding Missing Measures in Similar

FiguresFind the value of x the diagram.

∆MNP ~ ∆STU

M corresponds to S, N corresponds to T, and P corresponds to U.

6x = 56

The length of SU is cm.

Holt Algebra 1


Reading Math

• AB means segment AB. AB means the length of AB.

• A means angle A. mA the measure of angle A.

Holt Algebra 1



Find the value of x in the diagram if ABCD ~ WXYZ.

ABCD ~ WXYZ

x = 2.8

The length of XY is 2.8 in.

Holt Algebra 1


You can solve a proportion involving similar triangles to find a length that is not easily measured. This method of measurement is called indirect measurement. If two objects form right angles with the ground, you can apply indirect measurement using their shadows.

Holt Algebra 1


Example 2: Measurement Application

A flagpole casts a shadow that is 75 ft long at the same time a 6-foot-tall man casts a shadow that is 9 ft long. Write and solve a proportion to find the height of the flag pole.

The flagpole is 50 feet tall.

Holt Algebra 1


Helpful Hint

A height of 50 ft seems reasonable for a flag pole. If you got 500 or 5000 ft, that would not be reasonable, and you should check your work.

Holt Algebra 1



A woman who is 5.5 feet tall casts a shadow 3.5 feet long. At the same time, a building casts a shadow 28 feet long. Write and solve a proportion to find the height of the building.

The building is 44 feet tall.

3.5x = 154

x = 44

Holt Algebra 1


If every dimension of a figure is multiplied by the same number, the result is a similar figure. The multiplier is called a scale factor.

Holt Algebra 1


Example 3A: Changing Dimensions

The radius of a circle with radius 8 in. is multiplied by 1.75 to get a circle with radius 14 in. How is the ratio of the circumferences related to the ratio of the radii? How is the ratio of the areas related to the ratio of the radii?

Circle A Circle B

Radii:

The ratio of the circumference is equal to the ratio of the radii.The ratio of the areas is the square of the ratio of the radii.

Circumference: Area:

Holt Algebra 1


Example 3B: Changing DimensionsEvery dimension of a rectangular prism with length of 12, cm, and height 9 cm is multiplied by to get a similar rectangular prism. How is the ratio of the volumes related to the ratio of the corresponding dimensions?

The ratio of the volumes is the cube of the ratio of the corresponding dimensions.

Prism A Prism B

V = lwh (12)(3)(9) = 324 (4)(1)(3) = 12

Holt Algebra 1


Helpful Hint

A scale factor between 0 and 1 reduces a figure. A scale factor greater than 1 enlarges it.

Holt Algebra 1

2-8 Percents

Warm-Up

Find the value of x in each diagram.

1. ∆ABC ~ ∆MLK 2. RSTU ~ WXYZ

3. A girl that is 5 ft tall casts a shadow 4 ft long. At the same time, a tree casts a shadow 24 ft long. How tall is the tree?

Holt Algebra 1

2-8 Percents

Solve problems involving percents.

Objective

Holt Algebra 1

2-8 Percents

A percent is a ratio that compares a number to 100. For example,

To find the fraction equivalent of a percent write the percent as a ratio with a denominator of 100. Then simplify.

To find the decimal equivalent of a percent, divide by 100.

Holt Algebra 1

2-8 Percents

Some Common Equivalents

Percent

Fraction

Decimal

100%40%25%20% 50% 60% 75% 80%10%

The greatest percent shown in the table is 100% or 1.

But percents can be greater than 100%. For example,

120% = = 1.2. You can also find percents that are

less than 1%. For example, 0.5%= = 0.005. You

can use the proportion = to find unknown

values.

Holt Algebra 1

2-8 Percents

Example 1A: Finding the Part

Find 30% of 80.

Method 1 Use a proportion.

100x = 2400

x = 24

30% of 80 is 24.

is

of

%

100

Holt Algebra 1

2-8 Percents

Example 1B: Finding the Part

Find 120% of 15.

Method 2 Use an equation.

x = 120% of 15

x = 1.20(15)

x = 18

120% of 15 is 18.

Holt Algebra 1

2-8 Percents

Check It Out! Example 1a

Find 20% of 60.


100x = 1200

x = 12

20% of 60 is 12.

Holt Algebra 1

2-8 Percents


Find 210% of 8.

Method 2 Use an equation.

x = 210% of 8

x = 2.10(8)

x = 17

210% of 8 is 16.8.

Holt Algebra 1

2-8 Percents

Example 2A: Finding the PercentWhat percent of 45 is 35? Round your answer to the nearest tenth.


45x = 3500

x ≈ 77.835 is about 77.8% of 45.

Holt Algebra 1

2-8 Percents


The serving size of a popular orange drink is 12 oz. The drink is advertised as containing 5% orange juice. How many ounces of orange juice are in one serving size?

100x = 60

x = 0.6A 12 oz orange drink contains 0.6 oz of orange juice.

Holt Algebra 1

2-8 Percents

Warm-Up

1. Find 20% of 80.

2. What percent of 160 is 20?

3. 35% of what number is 40?

4. 120 is what percent of 80?

5. Find 320% of 8.

6. 65 is 0.5% of what number?

Find each value. Round to the nearest tenth if necessary.

16

12.5%

114.3

150%

25.6

13,000

Holt Algebra 1

2-9 Application of Percents

Warm Up 1. Write 0.03 as a percent. 2. Write as a decimal.

Find each value. Round to the nearest tenthif necessary.3. Find of 50. 4. ∆ABC ~ ∆MLK

5. A girl that is 5 ft tall casts a shadow 4 ft long. At the same time, a tree casts a shadow 24 ft long. How tall is the tree?

Holt Algebra 1


Use common applications of percents.Estimate with percents.

Objectives

Holt Algebra 1


A commission is money paid to a person or a company for making a sale. Usually the commission is a percent of the sale amount.

Holt Algebra 1


Caution!

You must convert a percent to a decimal or a fraction before doing any calculations with it.

Holt Algebra 1


Example 1: Business Application

Mr. Cortez earns a base salary of $26,000 plus a sales commission of 5%. His total sales for one year were $300,000. Find his total pay for the year.

Write the formula for total pay.total pay = base salary + commission Write the formula for commission.

Substitute values given in the problem.= 26,000 + 5% of 300,000= base + % of total sales

Write the percent as a decimal.= 26,000 + (0.05)(300,000)Multiply.= 26,000 + 15,000Add.= 41,000

Mr. Cortez’s total pay was $41,000.

Holt Algebra 1


Interest is the amount of money charged for borrowing money, or the amount of money earned when saving or investing money. Principal is the amount borrowed or invested. Simple interest is interest paid only on the principal.

Simple Interest Paid Annually

Time in years

Interest rate peryear as a decimal

Principal

Simple interest

Holt Algebra 1


Example 2A: Financial Application

Find the simple interest paid for 3 years on a $2500 loan at 11.5% per year.

I = Prt

I = (2500)(0.115)(3)

I = 862.50

The amount of interest is $862.50.

Holt Algebra 1


Example 2B: Financial Application

After 6 months, the simple interest earned on an investment of $5000 was $45. Find the interest rate.

I = Prt Write the formula for simple interest.

45 = 2500r

0.018 = r

The interest rate is 1.8%.

Substitute the given values.

Multiply 5000 .. Since r is

multiplied by 2500, divide

both sides by 2500 to undo

the multiplication.

Holt Algebra 1


Helpful Hint

When you are using the formula I= Prt to find

simple interest paid annually, t represents

time in years. One month is year.

Holt Algebra 1



Find the simple interest earned after 2 years on an investment of $3000 at 4.5% interest earned annually.

I = Prt Write the formula for simple interest.

I = 270

The interest earned is $270.

Substitute the given values.

Multiply.

I = (3000)(0.045)(2)

Holt Algebra 1


A tip is an amount of money added to a bill for service. it is usually a percent of the bill before sales tax is added. Sales tax is a percent of an item’s cost.

Holt Algebra 1


Sales tax and tips are sometimes estimated instead of calculated exactly. When estimating percents, use percents that you can calculate mentally.

• Find 10% of a number by moving the decimal point one place to the left.

• Find 1% of a number by moving the decimal point two places to the left.

• Find 5% of a number by finding of 10% of the number.

Holt Algebra 1


Example 3A: Estimating with Percents

Lunch at a restaurant is $27.88. Estimate a 15% tip.

Step 1 First round $27.88 to $30.

Step 2 Think: 15% = 10% + 5% 10% of $30 = $3.00

Step 3 Think 5% = 10% ÷ 2 = $3.00 ÷ 2 = $1.50 Step 4 15 = 10% + 5% =

$3.00 + $1.50 = $4.50

The tip should be about $4.50.

Move the decimal point one place left.

Holt Algebra 1

2-10 Percent Increase and Decrease

Warm Up

1. Find 30% of 40.

2. Find 28% of 60. Solve for x.

3. 22 = x(50)

4. 17.2 = x(86)



16.8

12

38

0.44

25%

120%

Holt Algebra 1


Warm-Up

1. Find the simple interest paid for 9 months on a a $500 loan at 8% per year.

2. After 2 years the simple interest earned on an investment of 4000 was $216. Find the interest rate.

3. A family’s dinner check was $38.82. Estimate a 15% tip.

4.The sales tax rate is 7.1%. Estimate the sales tax on a television set that costs $399.

5. Find the result when 70 is decreased by 20%.

Holt Algebra 1


Find percent increase and decrease.

Objective

Holt Algebra 1


A percent change is an increase or decrease given as a percent of the original amount. Percent increase describes an amount that has grown and percent decrease describes an amount that has be reduced.

Holt Algebra 1


Example 1A: Finding Percent Increase and Decrease

Find each percent change. Tell whether it is a percent increase or decrease.

From 8 to 10

= 0.25

= 25%

Simplify the numerator.

8 to 10 is an increase, so a change from 8 to 10 is a 25% increase.

Holt Algebra 1


Helpful Hint

Before solving, decide what is a reasonable answer. For Example 1A, 8 to 16 would be a 100% increase. So 8 to 10 should be much less than 100%.

Holt Algebra 1



From 200 to 110

= 0.45

= 45%

Simplify the numerator.

200 to 110 is an decrease, so a change from 200 to 110 is a 60% decrease.


Holt Algebra 1


Example 2: Finding Percent Increase and Decrease

A. Find the result when 12 is increased by 50%.

0.50(12) = 6 Find 50% of 12. This is the amount of increase.

12 + 6 =18 It is a percent increase, so add 6 to the original amount.

12 increased by 50% is 18.

B. Find the result when 55 is decreased by 60%.

0.60(55) = 33 Find 60% of 55. This is the amount of decrease.

55 – 33 = 22 It is a percent decrease so subtract 33 from the the original amount.

55 decreased by 60% is 22.

Holt Algebra 1


Common application of percent change are discounts and markups.

A discount is an amount by which an original price is reduced.

discount = % of original price

final price = original price – discount

A markup is an amount by which a wholesale price is increased.

final price = wholesale cost markup+

markup wholesale cost= % of

Holt Algebra 1


Example 3: Discounts The entrance fee at an amusement park is

$35. People over the age of 65 receive a 20% discount. What is the amount of the discount? How much do people over 65 pay?

Method 1 A discount is a percent decrease. So find $35 decreased by 20%.

0.20(35) = 7 Find 20% of 35. This is the amount of the discount.

35 – 7 = 28 Subtract 7 from 35. This is the entrance fee for people over the age of 65.

Holt Algebra 1


Example 3A: Discounts

Method 2 Subtract the percent discount from 100%.

100% – 20% = 80% People over the age of 65 pay 80% of the regular price, $35.

0.80(35) = 28 Find 80% of 35. This is the entrance fee for people over the age of 65.

35 – 28 = 7 Subtract 28 from 35. This is the amount of the discount.

By either method, the discount is $7. People over the age of 65 pay $28.00.

Holt Algebra 1


Example 3B: Discounts

A student paid $31.20 for art supplies that normally cost $52.00. Find the percent discount .

$52.00 – $31.20 = $20.80 Think: 20.80 is what percent of 52.00? Let x represent the percent.20.80 = x(52.00)

0.40 = x

40% = x

The discount is 40%

Since x is multiplied by 52.00, divide both sides by 52.00 to undo the multiplication.

Write the answer as a percent.

Holt Algebra 1



A $220 bicycle was on sale for 60% off. Find the sale price.

Method 2 Subtract the percent discount from 100%.

100% – 60% = 40% The bicycle was 60% off of 100% .

0.40(220) = 88 Find 40% of 220.

By this method, the sale price is $88.

Holt Algebra 1


Example 4: Markups The wholesale cost of a DVD is $7. The markup is 85%. What is the amount of the markup? What is the selling price?Method 1A markup is a percent increase. So find $7 increased by 85%.0.85(7) = 5.95

7 + 5.95 = 12.95

Find 85% of 7. This is the amount of the markup.

Add to 7. This is the selling price.

Subtract from 12.95. This is the amount of the markup.

12.95 7 = 5.95

By either method, the amount of the markup is $5.95. The selling price is $12.95.

Method 2Add percent markup to 100%

The selling price is 185% of the wholesale price, 7.

100% + 85% = 185%

Find 185% of 7. This is the selling price.

1.85(7) = 12.95

Holt Algebra 1



A video game has a 70% markup. The wholesale cost is $9. What is the selling price?

Method 1

A markup is a percent increase. So find $9 increased by 70%.

0.70(9) = 6.30 Find 70% of 9. This is the amount of the markup.

9 + 6.30 = 15.30 Add to 9. This is the selling price.

The amount of the markup is $6.30. The selling price is $15.30.

Holt Algebra 1


Lesson Quiz: Part 1

1. from 20 to 28.

2. from 80 to 62.

3. from 500 to 100.

4. find the result when 120 is increased by 40%.

5. find the result when 70 is decreased by 20%.


40% increase

22.5% decrease

80% decrease

168

56

Holt Algebra 1


Lesson Quiz: Part 2


6. A movie ticket costs $9. On Mondays, tickets are 20% off. What is the amount of discount? How much would a ticket cost on a Monday?

7. A bike helmet cost $24. The wholesale cost was $15. What was the percent of markup?

$1.80; $7.20

60%

holt algebra 1 2-1 solving equations by adding or subtracting warm up evaluate. 1. + 4 2. 0.51 +...

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