solving one-step equationsbowmansmath.weebly.com/uploads/6/0/3/7/60377015/unit_2.pdf · subtraction...

Warm-Up #1

1. What is 14% of 0.4?

2. What is the value of

when x = 1

3. 9m – 5m + 3 simplify

4. Evaluate for a = 4, b = 13 and c = 2

a.) 2a + cb

b.) cb - 𝑎2

𝑥 + 22

x + 3

Solving One-Step Equations

Section 2-1

Goals

Goal

• To solve one-step equations

in one variable.

Rubric

Level 1 – Know the goals.

Level 2 – Fully understand the

goals.

Level 3 – Use the goals to

solve simple problems.


solve more advanced problems.

Level 5 – Adapts and applies

the goals to different and more

complex problems.

Vocabulary

• Equivalent Equations

• Addition Property of Equality

• Subtraction Property of Equality

• Isolate

• Inverse Operations

• Multiplication Property of Equality

• Division Property of Equality

A linear equation in one variable is an equation that

can be written in the form ax + b = c where a, b, and c

are real numbers and a 0.

3x + 5 = 25 4

87

a

The expressions are called the sides of the equation.

Linear Equations in One

Variable

The solution of a linear equation is the value or values

of the variable that make the equation a true statement.

The set of all solutions of an equation is called the

solution set. The solution satisfies the equation. Example:

Determine if x = – 1 is a solution to the equation.

– 3(x – 3) = – 4x + 3 – 5x

– 3[(– 1) – 3] = – 4(– 1) + 3 – 5(– 1)

– 3(– 4) = 4 + 3 + 5

12 = 12 True. x = – 1 is a solution

Solutions

Linear equations are solved by writing a

series of steps that result in the equation

x = a number

One method for solving equations is to write a series

of equivalent equations.

Two or more equations that have precisely the same

solutions are called equivalent equations.

3 + 5 = 8

1 + 7 = 2 + 6

Equivalent Equations

Two equations are said to be equivalent if every solution of either one is also a solution of the other.

( ) 6 9a x ( ) 10 13b x

The two equations shown above are equivalent, because the number 3 will satisfy both equations and 3 is the only number that will satisfy either equation.

Equivalent Equations

Inverse Operations

Add x. Subtract x.

Multiply by x. Divide by x.

An equation is like a balanced scale. To

keep the balance, you must perform the

same inverse operation on both sides.

To find solutions, perform inverse operations until

you have isolated 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.

x = a number Isolated

Variable

Inverse Operations

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.

Inverse Operations

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.

Inverse Operations

The Addition Property of Equality states that for

real numbers a, b, and c,

if a = b, then a + c = b + c.

y - 6 = 11

We need to find

the value of y.

y - 6 (+6) = 11 (+6) Adding (+6) to both sides of the equation

will maintain the balance of the equation.

y = 17

Solution to

the equation.

Left side Right side

+6 +6

Addition Property Of Equality

Example:

Solve the linear equation x 9 = 22.

Step 1: Isolate the variable x on the left side of the equation.

Step 2: Simplify the left and right sides of the equation.

Step 3: Check to verify the solution.

x 9 + 9 = 22 + 9 Add 9 to both sides of the equation.

x = 31 Apply the Additive Inverse Property.

x 9 = 22

31 9 = 22

22 = 22

Using the Addition Property

of Equality

WORDS

Addition Property of Equality

You 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

WORDS

Subtraction Property of Equality

You 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


Solution sets are written in set notation using

braces, { }. Solutions may be given in set

notation, or they may be given in the form x = 14.

Writing Math

Solve the equation.

Since 8 is subtracted from y,

add 8 to both sides to undo

the subtraction.

y – 8 = 24 + 8 + 8

y = 32

Check y – 8 = 24

32 – 8 24

24 24

To check your solution, substitute

32 for y in the original

equation.

The solution set is {32}.

Example

Solve the equation.


2.4 for t in the original equation.

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

Check 4.2 = t + 1.8

4.2 2.4 + 1.8

4.2 4.2

The solution set is {2.4}.

Example

Solve the equation. Check your answer.

Since 3.2 is subtracted from n, add

3.2 to both sides to undo the

subtraction.

n – 3.2 = 5.6

+ 3.2 + 3.2

n = 8.8

Check n – 3.2 = 5.6

8.8 – 3.2 5.6

5.6 5.6


8.8 for n in the original

equation.


Your Turn:


Since 6 is subtracted from k,

add 6 to both sides to undo

the subtraction.

–6 = k – 6

+ 6 + 6

0 = k

Check –6 = k – 6

–6 0 – 6

–6 –6

To check your solution, substitute 0

for k in the original equation.


Your Turn:


Since 6 is added to t, subtract 6 from

both sides to undo the addition.

6 + t = 14

– 6 – 6

t = 8

Check 6 + t = 14

6 + 8 14

14 14


8 for t in the original equation.


Your Turn:

The Multiplication Property of Equality states that

for real numbers a, b, and c, where c 0,

if a = b, then ac = bc.

We need to find

the value of x.

x = 28

Solution to

the equation.

1 4

7x

1

71

47

71

x Multiplying both sides of the equation by

will maintain the balance of the equation.

71

Right side Left side

× 7 × 7

Multiplication Property of

Equality

Example:

Solve the linear equation 3x = 81

Step 1: Get the coefficient of the variable x to be 1.

Step 2: Simplify the left and right sides of the equation.

Step 3: Check to verify the solution.

x = 27 Apply the Multiplicative Inverse Property.

3x = 81

81 = 81

Multiply each side of the equation by 1

.3(3x) = (81)

13

13

3(27) = 81

Using the Multiplication

Property of Equality

WORDS

Multiplication Property of Equality

You 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



Division Property of Equality

You 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 8

4

8

4 =

a c

a c

=


Since j is divided by 3, multiply from

both sides by 3 to undo the division.

–8 –8

To check your solution, substitute –24 for

j in the original equation.

–24 = j

Check

The solution set is {–24}.

Example:


Since v is multiplied by –6, divide

both sides by –6 to undo the

multiplication.

–4.8 = –6v

0.8 = v

Check –4.8 = –6v

–4.8 –6(0.8)

–4.8 –4.8


0.8 for v in the original

equation.


Example:

Solve each equation. Check your answer.

Since p is divided by 5, multiply both

sides by 5 to undo the division.


50 for p in the original

equation.

p = 50

Check

10 10


Your Turn:


Since y is multiplied by 0.5, divide both

sides by 0.5 to undo the

multiplication.

0.5y = –10

Check 0.5y = –10

0.5(–20) –10

–10 –10


–20 for y in the original

equation.

y = –20 The solution set is {–20}.

Your Turn:


Since c is divided by 8, multiply both


c = 56


56 for c in the original

equation.

7 7


Check

Your Turn:

When solving equations, you will sometimes find it

easier to add an opposite to both sides instead of

subtracting or to multiply by a reciprocal instead of

dividing. This is often true when an equation contains

negative numbers or fractions.

Solving Equations

Solve each equation.

The reciprocal of is . Since

w is multiplied by multiply

both sides by .


Example:

Solve each equation.

Since p is added to , add

to both sides to undo the

subtraction.

The solution set is . { }

Example:


Since –2.3 is added to m, add 2.3

to both sides.

–2.3 + m = 7

+2.3 + 2.3

m = 9.3

–2.3 + m = 7


To check your solution, substitute 9.3

for m in the original equation.

Check –2.3 + m = 7

–2.3 + 9.3 7

7 7

Your Turn:

5

4 + z =

Since is added to z add

to both sides.



To check your solution,

substitute 2 for z in the

original equation.

Check

Your Turn:

w = 612


The reciprocal of is . Since

w is multiplied by multiply

both sides by .

Check

102 102


substitute 612 for w in the

original equation.


Your Turn:

Ciro deposits 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 out

how much money Ciro earned mowing lawns this

year.

1

4

Example: Application

e = $1140

The original earnings were $1140 .

Write an equation to represent the

relationship.

earnings is times $285 1

4

1 4

4 1 e = 285

4 1

The reciprocal of is . Since e

is multiplied by ,

multiply both sides by

1 4 1 4

4 1

4 1

.

e = $285

Example: Continued

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 is 10,000 feet above the

ground. Write and solve an equation to find the

distance from the airport at which this plane

should begin descending.

Your Turn:

Check It Out! Example 4 Continued

d ÷ 3 = h

Write an equation to represent the

relationship.

d = 30

At 10,000 feet altitude the decent should start 30,000 feet from

the airport.

distance divided by height is 3

3 1

d

3 = 10

3 1

Substitute 10 for h. The reciprocal of

is . Since d is multiplied by

multiply both sides by .

1 3

3 1

1 3

3 1

Assignment

• 2.1 Exercises Pg. 77-79 2-68 even #70

• Pg. 80 #74

Warm-up #2

1.) If a and b are integers, then is an example of which property?

A identity property of addition

B distributive property

C commutative property of addition

D associative property of addition

2.) If x, y, and z are integers then is an example of which property?

A identity property of addition

B distributive property

C commutative property of addition

D associative property of addition

3.) 3 + p = 8

4.) 2(n + 5) = −2

Solving Two-Step Equations

Section 2-2

Goals

Goal

• To solve two-step equations

in one variable.

Rubric



goals.







complex problems.

This equation contains multiplication and addition. Equations that

contain two operations require two steps to solve. Identify the

operations in the equation and the order in which they are applied

to the variable. Then use inverse operations to undo them in

reverse over one at a time.

Many equations contain more than one operation,

such as 2x + 5 = 11.

Operations in the equation

First x is multiplied by 2.

Then 5 is added.

To solve

Subtract 5 from both

sides of the equation. Then divide both sides

by 2.

Solving Two-Step Equations

2x + 5 =11

–5 –5

2x = 6

x = 3

Subtract 5 from both sides of

the equation.

Divide both sides of the equation by 2.


Each time you perform an inverse operation, you create an

equation that is equivalent to the original equation. Equivalent

equations have the same solutions, or the same solution set. In

the example above, 2x + 5 = 11, 2x = 6, and x = 3 are all

equivalent equations.

To Solve: Inverse Operations

in the Inverse Order

Ex: x + 9 = 6 5 Ask yourself: • What is the first thing

we are doing to x? • The second thing?

Recall the order of operations as you answer these questions.

• dividing by 5

• adding 9

To undo these steps, do the inverse operations in inverse order.

The DO-UNDO Chart

Use a chart as a shortcut to

answering the questions.

DO UNDO

÷5 -9

+9 ·5

Follow the steps in the

‘undo’ column to isolate

the variable.

Ex: x + 9 = 6 5 • First subtract 9. x + 9 - 9 = 6 - 9 5 x = -3 5 • Then multiply by 5. (5) x = -3(5) 5 x = -15

Example:

Complete the do-undo chart.

DO UNDO

-2 ·3

÷ 3 +2

To solve for d:

First multiply by 3.

Then add 2.

Ex: d - 2 = 7 3

(3) d - 2 = 7(3) 3 d - 2 = 21

d - 2 = 21 +2 +2 d = 23

Example:

Remember to always use the

sign in front of the number.

DO UNDO

÷ -7 - 3

+3 · -7

To solve for a:

First subtract 3.

Then multiply by -7.

Ex: 3 - a = -2 7 • 3 - a = -2 7 -3 -3 - a = -5 7 • (-7)(- a) = (-5)(-7) 7 a = 35

Your Turn:

1) 5z + 16 = 51

2) 14n - 8 = 34

3) 4b + 8 = 10

-2

The answers:

1) DO UNDO

· 5 - 16

+16 ÷ 5

1) z = 7

2) DO UNDO

· 14 +8

-8 ÷ 14

2) n = 3

3) DO UNDO

· 4 · -2

+8 - 8

÷ -2 ÷ 4

3) b = -7


7x = 7

–4 + 7x = 3

Add 4 to both sides.

7x = 7 is equivalent to –4 + 7x = 3.

Since x is multiplied by 7, divide both sides

by 7 to undo the multiplication.


First x is multiplied by 7. Then –4 is added. –4 + 7x = 3

+ 4 +4

x = 1

Your Turn:

Check

–4 + 7(1) 3

3 3


1 for x in the original equation.

Check your answer.

–4 + 7x = 3

–4 + 7 3

Your Turn: Continued

Solve the equation.

7.2 = 1.2y

1.5 = 1.2y – 5.7

Add 5.7 to both sides.

7.2 = 1.2y is equivalent to

1.5 = 1.2y – 5.7.

Since y is multiplied by 1.2, divide both

sides by 1.2 to undo the multiplication.


First y is multiplied by 1.2. Then 5.7 is

subtracted.

+ 5.7 +5.7

6 = y

1.5 = 1.2y – 5.7

Your Turn:

Solve the equation.

n = 0

Subtract 2 from each side.

Since n is divided by 7, multiply both



First n is divided by 7. Then 2 is added.

–2 –2

= 0 = 0 is equivalent to + 2 = 2.

Your Turn:

Solve the equation.

Method 1 Use fraction operations.

Since is subtracted from , add to

both sides to undo the subtraction.

Example: Two-Step

Equations with Fractions

y = 16

Since y is divided by 8 multiply both

sides by 8.

Simplify.


Example: Continued

Method 2 Multiply by the least common

denominator (LCD) to clear fractions.

y – 6 = 10

+6 +6

y = 16

Multiply both sides by 8, the LCD of the

fractions.

Distribute 8 on the left side.

Simplify. Since 6 is subtracted from y, add 6 to



Example: Continued

Solve the equation.


Since is added to , subtract

from both sides to undo the addition.

Example: Two-Step

Equations with Fractions

Since r is multiplied by multiply both

sides by , the reciprocal.

Simplify.

The solution set is .

Example: Continued


denominator (LCD) to clear the fractions.

Multiply both sides by 12, the LCD of

the fractions.


Example: Continued

8r + 9 = 7

– 9 –9

8r =–2

Simplify. Since 9 is added 8r, subtract 9 from


Since r is multiplied by 8, divide both sides 8

to undo the multiplication.


Example: Continued

You can multiply both sides of the equation by any

common denominator of the fractions. Using the

LCD is the most efficient.

Helpful Hint



Since is subtracted from , add

to both sides to undo the subtraction.

Your Turn:

Simplify.


Since x is multiplied by multiply both

sides by , the reciprocal.



denominator (LCD) to clear the fractions.


the fractions.


4x – 5 = 50


+ 5 +5

4x = 55

Simplify. Since 5 is subtracted from 4x add 5


Simplify. Since x is multiplied by 4, divide both

sides 4 to undo the multiplication.

4x – 5 = 50

The solution set is

.


Solve the equation.


Since is added to , subtract


Your Turn:

Simplify.

Since u is multiplied by multiply both

sides by the reciprocal, .




denominator (LCD) to clear fractions.


the fractions.


6u + 4 = 7


– 4 – 4

6u = 3

Simplify. Since 4 is added to 6u subtract 4


Simplify. Since u is multiplied by 6, divide both

sides 6 to undo the multiplication.

6u + 4 = 7



Solve the equation.


Since is subtracted from , add


Your Turn:

Simplify.

The solution set is {15}. n = 15

Since n is multiplied by multiply both

sides by the reciprocal, .


Assignment

• 2.2 Exercises Pg. 84-85 1-51 odd #56,59

Warm-up #3

Warm-Up #3

Solving Multi-Step Equations

Section 2-3 Part 1

Goals

Goal

• To solve multi-step

equations in one variable.

Rubric



goals.







complex problems.

Equations that are more complicated may have to be

simplified before they can be solved. You may have

to combine like terms or use the Distributive

Property before you begin using inverse operations.


Regular price of enrollment

Application fee

A martial arts school is offering a special where new students can

enroll for half price, after a $12.50 application fee.

Ten students enrolled and paid a total of $325. To find the regular

price of enrollment, you can solve an equation.

Number of students Total

cost


Notice that this equation contains multiplication,

division, and addition. An equation that contains

multiple operations will require multiple steps to solve.

You will create an equivalent equation at each step.



2x + 1 = 21 –1 –1

2x = 20

Since 2x + 1 is divided by 3, multiply both


Since 1 is added to 2x, subtract 1 from


x = 10

Since x is multiplied by 2, divide both

sides by 2 to undo the multiplication.


Example: Solving Multi-Step

Equations

Check


substitute 10 for x in the

original equation.

7 7

Example: Continued

Solve the equation.

6 = x or x = 6

+4 +4

18 = 3x

Since 3x – 4 is divided by 2, multiply both


Since 4 is subtracted from 3x, add 4 to





Example: Solving Multi-Step

Equations

Solve the equation.

5m + 13 = 2

–13 –13

5m = –11

Since 5m + 13 is divided by 2, multiply

both sides by 2 to undo the division.

Since 13 is added to 5m, subtract 13


Since m is multiplied by 5, divide both



Your Turn:

Solve the equation.

–4 –4

–2x = –12

4 – 2x = –8

x = 6

Since 4 – 2x is divided by 4, multiply both


Since 4 is added to – 2x, subtract 4 from


Since x is multiplied by –2, divide both

sides by –2 to undo the

multiplication.


Your Turn:

You may have to combine like terms or use the

Distributive Property before you begin solving.


4x – 3x + 2

Like terms Constant

Solve 8x – 21 – 5x = –15

8x – 21 – 5x = –15

8x – 5x – 21 = –15

3x – 21 = –15

+21 = +21

3x = 6

x = 2

Use the Commutative Property of

Addition. Combine like terms.






Example: Combining Like

Terms and Solving Equations

Solve 4 = 2x + 5 – 6x

4 = 2x + 5 – 6x

4 = 2x – 6x + 5

4 = –4x + 5 –5 –5

–1 = –4x



Since 5 is added to –4x, subtract 5 from


Since x is multiplied by –4, divide both


multiplication.

The solution set is

Example: Combining Like

Terms and Solving Equations

Solve the equation.

2a + 3 – 8a = 8

2a – 8a +3 = 8

–6a + 3 = 8 –3 –3

–6a = 5



Since 3 is added to –6a, subtract 3 from


Since a is multiplied by –6, divide both


multiplication.


Your Turn:

Solve the equation.

–8 – 2d + 2 = 4

–2d –6 = 4

+6 +6

–2d = 10



Since 6 is subtracted from –2d, add 6 to


–2d + 2 – 8 = 4

d = –5

Since d is multiplied by –2, divide both


multiplication.


–8 – 2d + 2 = 4

Your Turn:

Solve the equation.

6x – 8 = 40

+8 +8

6x = 48





x = 8




4x – 8 + 2x = 40

4x + 2x – 8 = 40

4x – 8 + 2x = 40

Your Turn:

Solve the equation.

5(p – 2) = –15

5(p – 2) = –15 Distribute 5.

5(p) + 5(–2) = –15

5p – 10 = –15

+10 +10

5p = –5

p = –1

Simplify.

Since 10 is subtracted from 5p, add 10 to

both sides.

Since p is multiplied by 5, divide both

sides by 5.


Example: Distributive Property

and Solving Equations

You can think of a negative sign as a coefficient of –1.

–(x + 2) = –1(x + 2) and –x = –1x.

Helpful Hint

Solve the equation.

10y – (4y + 8) = –20

10y +(–1)(4y + 8) = –20

10y + (–1)(4y) + (–1)(8) = –20

6y – 8 = –20

+8 +8

6y = –12

Distribute –1.

Write subtraction as the addition

of the opposite.

Combine like terms.

10y – 4y – 8 = –20 Simplify.

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

Example: Distributive Property

and Solving Equations

6y = –12

y = –2

Since y is multiplied by 6, divide both


Example: Continued

Solve the equation.

3(a + 1) – 4 = 5

3(a + 1) – 4 = 5

(3)(a) + (3)(1) – 4 = 5

+ 1 +1

3a = 6

Distribute 3.

3a + 3 – 4 = 5 Simplify. Combine like terms.

Since 1 is subtracted from 3a, add 1 to


3a – 1 = 5

Since a is multiplied by 3, divide both


a = 2

Your Turn;

Solve the equation.

–4(2 – y) = 8

–4(2 – y) = 8

(–4)(2) + (–4)(–y) = 8

+8 +8

4y = 16

–8 + 4y = 8

Simplify.

Since –8 is added to 4y, add 8 to both

sides.



y = 4

Distribute –4 .

Your Turn:

Solve the equation.

d + 3(d – 4) = 20

Distribute 3.

d + 3(d – 4) = 20

d + 3(d) + 3(–4) = 20

Simplify. d + 3d – 12 = 20

4d – 12 = 20 Combine like terms.

+12 +12

4d = 32

d = 8

Since 12 is subtracted from 4d, add 12 to


Since d is multiplied by 4, divide both


Your Turn:

Solve 66 = (x + 3) 6 5

–

Write original equation. 66 = (x + 3) 6 5

–

–55 = x + 3 Simplify.

Subtract 3 from each side. – 5 8 = x

Multiplying by a Reciprocal First

Solve 66 = (x + 3) 6 5

–

Multiply by reciprocal of . 6 5

–

It is easier to solve this equation

if you don’t distribute first. 6 5

–

66 = (x + 3) 5 6

– 5 6

– 6 5

–

Solving equations systematically

is an example of deductive

reasoning. Notice how each

solution step is based on number

properties or properties of

equality.

Lin sold 4 more shirts than Greg. Fran sold 3

times as many shirts as Lin. In total, the three

sold 51 shirts. How many shirts did Greg sell?

To determine the number of shirts sold write an

equation: G + L + F = 51.

G = L – 4

F = 3L

L = L

Since the information is given in relation to Lin, set an

equation for each individual in terms of Lin.

Example: Application

G + L + F = 51

(L – 4) + (L) + (3L) = 51

5L – 4 = 51

+4 +4

5L = 55

L = 11

Since 4 is subtracted from 5L add 4 to both sides to undo the subtraction.

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

Substitute.

Combine like terms.




Example: Continued

Greg sold 7 shirts.

G = L – 4

= 11 – 4

= 7




Example: Continued

At a local gym, there is a joining fee of $59.95 and a monthly

membership fee. Sara and Martin both joined this gym.

Their combined cost for 12 months was $1319.90. How much

is the monthly fee?

Let m represent the monthly fee paid by each.

Monthly fee

for 2 is total

cost. initial fee for

2 plus

2 = 1319.90 119.90) +

12 months

(12m

Your Turn:

Since 119.90 is added to 24m, subtract 119.90 from both sides to undo the addition.

Since m is multiplied by 24, divide both sides by 24 to undo the multiplication.

–119.90 –119.90

24m = 1200.00

m = 50

Sara and Martin each paid $50 per month.

2(12m + 59.95) = 1319.90

2(12m) + 2(59.95) = 1319.90 Distribute 2.

24m + 119.90 = 1319.90


Lily and 4 of her friends want to enroll in a yoga class.

After enrollment, the studio requires a $7 processing fee.

The 5 girls pay a total of $125.75. How much does the class

cost?

number

enrolled is total cost processing fee plus

5 = 125.75 7) +

class cost

(c

Let c represent the cost of the class.

Your Turn:

5(c + 7) = 125.75

5(c) + 5(7) = 125.75

5c + 35 = 125.75

– 35 – 35

5c = 90.75

c = 18.15

The cost per person is $18.15 a month.

Since 35 is added to 5c, subtract 35 from both sides to undo the addition.

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

Distribute 5.


Assignment

• 2.3 Pt 1 Exercises Pg. 91 1-20 all

• Pg. 92 #51, 52, 57

Warm-Up #4


Section 2-3 Part 2

Two Special Techniques

Two Special Techniques for Solving Multi-Step

Equations.

1. Clear the equation of fractions by multiplying both

sides of the equation by the LCD of all denominators

in the equation.

2. Simplify equations with decimal coefficients by

multiplying both sides of the equation by a factor 10

(usually 10 or 100) to make all the coefficients whole

numbers.

Solve .

Multiply both sides by 24, the

LCD of the fractions.


Multiply by the LCD to clear the fractions.

3y – 18 = 14

+18 +18

Simplify.

Since 18 is subtracted from 3y, add

18 to both sides to undo the

subtraction. 3y = 32

Example: Clear Fractions

Using the LCD

3 3

Since y is multiplied by 3, divide

both sides by 3 to undo the

multiplication.

3y = 32

Example: Continued

3y = 32

Solve: 1 1 3

5 2 4

m

(20)M 1 1 3

20 20 205 2 4

m

Simplify

25

4m

4 10 5

4 10 15m

(10)A 4 25m

(4)D

Example: Clear Fractions

Using the LCD

Solve .

Multiply by the LCD to clear the fractions.

Multiply both sides by 12, the LCD

of the fractions.

8r + 9 = 7 –9 –9

8r = –2


Simplify. Since 9 is added to 8r,

subtract 9 from both sides to

undo the addition.

Your Turn:

8r = –2 8 8

Since r is multiplied by 8, divide

both sides by 8 to undo the

multiplication.


8r = –2

Your Turn:

1.

2.

2 5 5

3 6 4x

5

8x

1 1 1

2 3 5x

21

10x

Your Turn:

3.

4.

35

4 8

x x

8x

3 1

5 10 5

n n

2n

Solve 1.5 = 1.2y – 5.7.

First y is multiplied by 12. Then 57 is

subtracted. Work backward: Add 57

to both sides.

15 = 12y – 57

+ 57 + 57

72 = 12y


sides by 12 to undo the

multiplication.

72 = 12y

12 12

6 = y

Multiply by a factor of 10 to clear the decimals.

(10)1.5 = (10)1.2y – (10)5.7 Multiply both sides by 10, to make

the decimals whole numbers.

Example: Simplifying

Decimal Coefficients

Solve: 0.4 0.2 0.16x

(100)M 40 20 16x

(20)A 40 4x

(40)D

40 4 . . 0

01

40

0.1x

Example: Simplifying

Decimal Coefficients

Solve: 0.002 0.02 4.02k

(1000)M 2 20 4020k

( 20)A 2 4000k

(2)D 2000k

Your Turn:

Your Turn:

1.

2.

2.5 0.1 9.8n n

28n

0.25 12 4d

28d

Assignment

• 2.3 Pt 2 Exercises Pg. 91 #21-50 all

Warm-Up #5

Solving Equations With

Variables on Both Sides

Section 2-4

Goals

Goal

• To solve equations with

variables on both sides.

• To identify equations that

are identities or have no

solution.

Rubric



goals.







complex problems.

Vocabulary

• Identity

To solve an equation with variables on both sides, use

inverse operations to "collect" variable terms on one

side of the equation.

Helpful Hint

Equations 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.

Equations With Variables on

Both Sides

Solve 7n – 2 = 5n + 6.

To collect the variable terms on one

side, subtract 5n from both sides.

7n – 2 = 5n + 6

–5n –5n

2n – 2 = 6

Since n is multiplied by 2, divide both


2n = 8

+ 2 + 2

n = 4

Example: Solving Equations

with Variables on Both Sides

Solve 4b + 2 = 3b.

To collect the variable terms on one

side, subtract 3b from both sides.

4b + 2 = 3b

–3b –3b

b + 2 = 0

b = –2

– 2 – 2

Your Turn:

Solve 0.5 + 0.3y = 0.7y – 0.3.

To collect the variable terms

on one side, subtract 0.3y

from both sides.

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

Since 0.3 is subtracted from

0.4y, add 0.3 to both sides

to undo the subtraction.

Since y is multiplied by 0.4,

divide both sides by 0.4 to

undo the multiplication.

Your Turn:

To solve more complicated equations, you may

need to first simplify by using the Distributive

Property or combining like terms.

Using the Distributive

Property

Solve 4 – 6a + 4a = –1 – 5(7 – 2a).

Combine like terms.

Distribute –5 to the

expression in parentheses. 4 – 6a + 4a = –1 –5(7 – 2a)

4 – 6a + 4a = –1 –5(7) –5(–2a)

4 – 6a + 4a = –1 – 35 + 10a

4 – 2a = –36 + 10a

+36 +36

40 – 2a = 10a + 2a +2a

40 = 12a

Since –36 is added to 10a,

add 36 to both sides.

To collect the variable

terms on one side, add

2a to both sides.

Example: Simplifying Both

Sides

40 = 12a

Since a is multiplied by 12,

divide both sides by 12.

Example: Continued

Solve .

Since 1 is subtracted from

b, add 1 to both sides.

Distribute to the expression in

parentheses.

1 2

+ 1 + 1 3 = b – 1

To collect the variable terms on

one side, subtract b from

both sides.

1 2

4 = b

Your Turn:

Solve 3x + 15 – 9 = 2(x + 2).

Combine like terms.

Distribute 2 to the expression

in parentheses. 3x + 15 – 9 = 2(x + 2)

3x + 15 – 9 = 2(x) + 2(2)

3x + 15 – 9 = 2x + 4

3x + 6 = 2x + 4

–2x –2x

x + 6 = 4 – 6 – 6

x = –2

To collect the variable terms

on one side, subtract 2x

from both sides.

Since 6 is added to x, subtract

6 from both sides to undo

the addition.

Your Turn:

Summary

Solving linear equations in one variable

1) Clear the equation of fractions by multiplying both sides

of the equation by the LCD of all denominators in the

equation.

2) Use the distributive property to remove grouping symbols

such as parentheses.

3) Combine like terms on each side of the equation.

4) Use the addition property of equality to rewrite the

equation as an equivalent equation with variable terms on

one side and numbers on the other side.

5) Use the multiplication property of equality to isolate the

variable.

6) Check the proposed solution in the original equation.

A conditional equation is an equation that is true for some values

of the variable and false for other values of the variable. Normal

equations with a finite number of solutions.

A contradiction is an equation that is false for every replacement

value of the variable.

Example:

Solve the equation 4x – 8 = 4x – 1.

4x – 8 = 4x – 1

– 8 = – 1 Subtract 4x from both sides.

This is a false statement so the equation is a contradiction.

The solution set is the empty set, written { } or .

Conditional Equations &

Contradictions

Contradiction

When 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

Contradictions

An identity is an equation that is satisfied for all values of the

variable for which both sides of the equation are defined.

Example:

Solve the equation 5x + 3 = 2x + 3(x + 1).

Distribute to remove parentheses.

This is a true statement for all real numbers x.

The solution set is all real numbers.

5x + 3 = 2x + 3(x + 1).

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

Combine like terms. 5x + 3 = 5x + 3

Subtract 5x from both sides. 3 = 3

Identities

WORDS

Identity

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

NUMBERS 2 + 1 = 2 + 1

3 = 3

ALGEBRA

2 + x = 2 + x

–x –x

2 = 2

Identities

Summary

1) A conditional equation is an equation in one variable that

has a finite (normally one solution) number of solutions.

2) An identity is an equation that is true for all values of the

variable (ie. the variable is eliminated and results in a true

statement). An equation that is an identity has infinitely

many solutions.

3) A contradiction is an equation that is false for any value

of the variable (ie. the variable is eliminated and results in

a false statement). It has no solutions.

Solve 10 – 5x + 1 = 7x + 11 – 12x.

Add 5x to both sides.

Identify like terms. 10 – 5x + 1 = 7x + 11 – 12x

11 – 5x = 11 – 5x

11 = 11

+ 5x + 5x True statement.

Combine like terms on the left and the right.

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.

Example: Identity

Solve 12x – 3 + x = 5x – 4 + 8x.

Subtract 13x from both sides.

Identify like terms. 12x – 3 + x = 5x – 4 + 8x

13x – 3 = 13x – 4

–3 = –4

–13x –13x False statement.


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.

Example: Contradiction

Solve 4y + 7 – y = 10 + 3y.

Subtract 3y from both sides.

Identify like terms. 4y + 7 – y = 10 + 3y

3y + 7 = 3y + 10

7 = 10

–3y –3y False statement.


4y + 7 – y = 10 + 3y

The equation 4y + 7 – y = 10 + 3y is a contradiction. There is

no value of y that will make the equation true. There are no

solutions.

Your Turn:

Solve 2c + 7 + c = –14 + 3c + 21.

Subtract 3c both sides.

Identify like terms. 2c + 7 + c = –14 + 3c + 21

3c + 7 = 3c + 7

7 = 7

–3c –3c True statement.


2c + 7 + c = –14 + 3c + 21

The equation 2c + 7 + c = –14 + 3c + 21 is an identity. All

values of c will make the equation true. All real numbers are

solutions (infinite number of solutions).

Your Turn:

Strategy for Problem Solving

1) UNDERSTAND the problem. During this step, become

comfortable with the problem. Some way of doing this

are:

• Read and reread the problem

• Propose a solution and check.

• Construct a drawing.

• Choose a variable to represent the unknown

2) TRANSLATE the problem into an equation.

3) SOLVE the equation.

4) INTERPRET the result. Check the proposed solution in

stated problem and state your conclusion.

The product of twice a number and three is the same as the

difference of five times the number and ¾. Find the number.

1.) Understand

Read and reread the problem. If we let

x = the unknown number, then “twice a number” translates to 2x,

“the product of twice a number and three” translates to 2x · 3,

“five times the number” translates to 5x, and

“the difference of five times the number and ¾” translates to 5x – ¾.

Example:

Continued

Finding an Unknown Number

The product of

·

twice a

number

2x

and 3

3

is the same as

=

5 times the

number

5x

and ¾

¾

the difference of

–

Example continued:

2.) Translate

Continued

3.) Solve

2x · 3 = 5x – ¾

6x = 5x – ¾ (Simplify left side)

x = – ¾ (Simplify both sides)

6x + (– 5x) = 5x + (– 5x) – ¾ (Add –5x to both sides)

4.) Interpret

Check: Replace “number” in the original statement of the problem

with – ¾. The product of twice – ¾ and 3 is 2(– ¾)(3) = – 4.5. The

difference of five times – ¾ and ¾ is 5(– ¾) – ¾ = – 4.5. We get the

same results for both portions.

State: The number is – ¾.

Example continued:

A car rental agency advertised renting a Buick Century for $24.95

per day and $0.29 per mile. If you rent this car for 2 days, how

many whole miles can you drive on a $100 budget?

1.) Understand

Read and reread the problem. Let’s propose that we drive a total of

100 miles over the 2 days. Then we need to take twice the daily rate

and add the fee for mileage to get 2(24.95) + 0.29(100) = 49.90 + 29

= 78.90. This gives us an idea of how the cost is calculated, and also

know that the number of miles will be greater than 100. If we let

x = the number of whole miles driven, then

0.29x = the cost for mileage driven Continued

Example:

Your Turn:

2.) Translate

Continued

Daily costs

2(24.95)

mileage costs

0.29x

plus

+

is equal to

= 100

maximum budget

Continued:

3.) Solve

Continued

2(24.95) + 0.29x = 100

49.90 + 0.29x = 100 (Simplify left side)

0.29x = 50.10 (Simplify both sides)

29.0

10.50

29.0

29.0

x(Divide both sides by 0.29)

x 172.75 (Simplify both sides)

(Subtract 49.90 from both sides) 49.90 – 49.90 + 0.29x = 100 – 49.90

Continued:

4.) Interpret

Check: Recall that the original statement of the problem asked

for a “whole number” of miles. If we replace “number of miles”

in the problem with 173, then 49.90 + 0.29(173) = 100.07,

which is over our budget. However, 49.90 + 0.29(172) = 99.78,

which is within the budget.

State: The maximum number of whole number miles is 172.

Continued:

Assignment

• 2.4 Exercises Pg. 98-99 #1-37 odd

Equations and Problem Solving

Lesson Preview

What You'll Learn:

• To define a variable in terms of another

variable

• To model distance-rate-time problems

… And Why

• To solve real-world problems involving

distance, rate, and time, as in Examples 3–5

Some problems contain two or more unknown

quantities. To solve such problems, first decide

which unknown quantity the variable will

represent. Then express the other unknown

quantity or quantities in terms of that variable.

Warm-Up #6

Defining One Variable in Terms of Another

Geometry: The length of a rectangle is 6 in.

more than its width. The perimeter of the

rectangle is 24 in. What is the length of the

rectangle?

Consecutive integers differ by 1. The integers

50 and 51 are consecutive integers, and so are

– 10, – 9, and – 8. For consecutive integer

problems, it may help to define a variable

before describing the problem in words. Let a

variable represent one of the unknown integers.

Then define the other unknown integers in

terms of the first one.

An object that moves at a constant rate is said

to be in uniform motion. The formula d = rt

gives the relationship between distance d, rate

r, and time t.

Uniform motion problems may involve objects

going the same direction, opposite directions,

or round trips.

In the diagram below, the two vehicles are

traveling the same direction at different rates.

The distances the vehicles travel are the same.

Since the distances are equal, the products of

rate and time for the two cars are equal. For the

vehicles shown, 40 ∙ 5 = 50 ∙ 4.

A table can also help you understand

relationships in distance-rate-time problems.

Same-Direction Travel

Engineering: A train leaves a train station at 1

p.m. It travels at an average rate of 60 mi/h. A

high-speed train leaves the same station an hour

later. It travels at an average rate of 96 mi/h.

The second train follows the same route as the

first train on a track parallel to the first. In how

many hours will the second train catch up with

the first train?

Noya drives into the city to buy a software

program at a computer store. Because of traffic

conditions, she averages only 15 mi/h. On her

drive home she averages 35 mi/h. If the total

travel time is 2 hours, how long does it take her

to drive to the computer store?

Round Trip Travel

Opposite-Direction Travel

Jane and Peter leave their home traveling in

opposite directions on a straight road. Peter

drives 15 mi/h faster than Jane. After 3 hours,

they are 225 miles apart. Find Peter's rate and

Jane's rate.

Practice: Pg. 107-109

# 1-20 All

Warm-up #7

C A

A C

Section 2-6

Formulas

A Literal Equation is an equation involving two or more variables.

Formulas are special types of literal equations.

Practice: Pgs. 113-144

# 1-39 odd

Warm-Up #8

39

8

2

40

Section 2-7 Measures of Central Tendency

Objectives:

• Find mean, median, mode and range given a data set

• Collect data & analyze using measures of Central Tendency

Measures of Central Tendency

• Used to analyze, organize and summarize a set of data

• The three types are mean, median, and mode

Outlier

• data value that is much higher or much lower than the other

data values in a set

Range

• the difference between the greatest and least values in a data set


Mean

• used to describe the middle of a set of data that does NOT

contain an outlier

• the mean of a list of numbers is also called the average

• It is found by adding all the numbers in the list and dividing

by the number of numbers in the list.

Find the mean of 3, 6, 11, and 8.

74

28

4

81163

items of

items data of

number

Sum


Median

• used to describe the middle of a set of data that HAS an outlier

• middle value in a data set when the numbers are arranged in order

• if data set has an odd number of data items, the median is the

middle value

• if data set has an even number of data items, the median is the mean

of the two middle data values

The students in John's class have the following ages:

4, 29, 4, 3, 4, 11. Find the median of their ages

Arrange in order: 3, 4, 4, 11, 29

The median is 4, since it is the middle of the data set.


Mode

• used when data is non-numeric or when choosing the most popular

item

• data item that occurs most often

• a data set can have no mode (is not the same as zero mode)

• a data set can have more than one mode

The students in John's class have the following ages:

4, 29, 4, 3, 4, 11. Find the mode of their ages

The mode is 4, since it is the data item repeated most often.

Find the mean, median, and mode of the data below.

14 10 2 13 16 3 12 11

Suppose your grades on three science exams are 82, 94,

and 89. What grade do you need on your next exam to have

an average of 90?


Mean = 10.125

Median = 2, 3, 10, 11, 12, 13, 14, 16 = (11+12)/2 = 11.5

Mode = none

95

360265

4904

2654

x

x

x

904

899482

x

Write an equation

Solve the equation

Analyze Data: Line Plot

Find the mean, median and mode of the data in the line plot. Which

measure of central tendency best describes the data?

6.610

8.257.003(6.50)5(6.25)mean

median 6.25 6.25 6.25 6.25 6.25, 6.50 6.50 6.50 7.00 8.25

25.6mode

= 6.375

Analyze Data: Stem and Leaf Plot

• Use to organize data

• Separate each number into a stem and a leaf

• last digit is the leaf

• remaining digits are the stem

STEM LEAF

Gas Prices

(cost/gallon)

1.77

1.55

1.58

1.73

1.54

1.83

1.63

1.67

1.5

1.6

1.7

1.8

4 5 8

3 7 3 7

3

1.8 | 3 means 1.83

Analyze Data: Stem and Leaf Plot

Find the mean of the city mileage versus the highway mileage.

Mean City Mileage = 26.4 mi/gal

Mean Highway Mileage = 33.6 mi/gal

Practice: Pgs. 121-122

# 1-19 odd

# 23,25,28

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