[PDF] - Equations and Identities Multi Step Equations Distributing PDF Document

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7th Grade

Equations & Inequalities

2015-10-16 www.njctl.org

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Click on a topic to go to that section.

Graphing and Writing Inequalities in One Variable Simple Inequalities Involving Addition and Subtraction Simple Inequalities Involving Multiplication and Division Glossary & Standards Equations and Identities Solving an Equation for a Variable One Step Equations Two Step Equations Multi Step Equations Distributing Fractions in Equations Writing and Solving Algebraic Equations

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Equations and Identities

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Slide 5 / 249 Equations and Identities

What is an equation? How is it different than an identity? Discuss in your groups.

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An equation is created when two expressions are set equal to one another such that they are equal for some values of their variables, but not for all. If they are equal for all values of their variables, then that is an identity, not an equation. So, not all mathematical statements which include an equal sign are equations...some are identities.

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Here are some identities: 2 + 3 = 5 9 - 2 = 7 5(x - 3) = 5x - 15 0.5y = y/2 These are always true...there are no values that can be assigned to the variables for which these would be untrue.

Slide 8 / 249 Equations and Identities

Here are some equations: s = 15/t v = 12 - 9.8t x = 10 + 6t - 4.9t2 In all these cases, the variables are interdependent. They are

nly true for certain sets of variables.

Changing the value of t on the right side of the equation, changes the possible values of s, v or x on the left side...and vice versa. These are equations (not identities) since knowing the value of

ne variable changes the possible value(s) of the other(s).

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s = 15/t v = 12 - 9.8t x = 10 + 6t - 4.9t2 Above are simplified physics equations in which t represents time, s represents the speed; v is for final velocity and x is for final position. In these equations, we have only two variables, but in later math courses, there will be more than two. For instance, here are more general versions of those same equations.

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s = d/t v = vo + at x = xo + vot + 1/2at2 In these cases, there are up to five variables, which all depend on each other. You'll work with these more in Algebra I, but it's important to see that equations define the relationship between variables since the equation is only true for certain sets of values.

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Let's use this table to find some solutions to the equation s = d/t; where s represents speed (in meters/second), d represents distance (in meters) and t represents time (in seconds). d (m) t (s) s (m/s) 30 2 60 4 90 6 120 2 240 4 360 6 We've entered the distance traveled and the time it took to travel that distance in two

f the columns.

Use the equation (s = d/t) to find the speeds and fill in the blank column. s = d/t

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Note that in the first three sets of answers, the object was moving at a speed of 15 m/s. The final three sets of answers are for an object traveling four times faster, at 60 m/s. d (m) t (s) s (m/s) 30 2 15 60 4 15 90 6 15 120 2 60 240 4 60 360 6 60 But, in all cases, knowing the value of two of the three variables determines the values of the third. s = d/t

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Knowing any two of the values, determines the third. Try it in this case, where the equation is still s = d/t, but the values of different variables are provided. Can you still figure out the unknown values and complete the table? d (m) t (s) s (m/s) 20 2 5 20 100 10 120 2 240 80 6 60 s = d/t

Math Practice

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This was more challenging, since the equation is solved for s...not for d or t. Filling in values will be even more challenging when we work with equations that have more variables. d (m) t (s) s (m/s) 20 2 10 100 5 20 100 10 10 120 2 60 240 3 80 360 6 60 We need a way to solve an equation for any variable, so that we can find its value, given the values of the others. Solving an equation for a variable means to get it alone

n the left side of the equation.

s = d/t

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Solving an Equation for a Variable

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The variable in this equation is d. Solving for a variable means having it alone on the left side. Right now, the equation is not solved for a variable.

Solving for a Variable

d 10 8 =

Our goal is to be able to solve any equation for any variable that appears in it. Let's look at a simple equation first.

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Like in any game there are a few rules. There are four rules which will allow you to solve any equation for any of its variables, not just in 7th grade Math, but in most any of your subsequent mathematics and science courses. So, learning these rules and how to apply them is very important.

The Rules for Solving Equations Slide 18 / 249

Here are the four rules. Let's examine them, one at a time.

1. To "undo" a mathematical operation, you must perform the

inverse operation.

2. You can do anything you want (except divide by zero) to one

side of an equation, as long as you do the same thing to the

ther.
3. If there is more than one operation going on, you must undo

them in the opposite order in which you would do them, the

pposite of the "order of operations."
4. You can always switch the left and right sides of an equation.

The Rules

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1. To "undo" a mathematical operation, you must do the opposite.

We learned earlier that for every mathematics operation, there is an inverse operation which undoes it: when you do both

perations, you get back to where you started.

When the variable for which we are solving is connected to something else by a mathematical operation, we can eliminate that connection by using the inverse of that operation.

The Rules Slide 20 / 249

2. You can do anything you want (except divide by zero) to one

side of an equation, as long as you do the same to the other side. If the two expressions on the opposite sides of the equal sign are equal to begin with, they will continue to be equal if you do the same mathematical operation to both of them. This allows you to use an inverse operation on one side, to undo an operation, as long as you also do it on the other side. You can just never divide by zero (or by something which turns out to be zero) since the result of that is always undefined.

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3. If there is more than one operation going on, you must undo

them in the opposite order in which you would do them, the

pposite of the "order of operations."

The operations which are connected to a variable must be "undone" in the reverse order from the Order of Operations. So, when solving for a variable, you: first have to do addition/subtraction, then multiplication/division, then exponents/roots, finally parentheses. The order of the steps you take to untie a knot are the reverse of the order used to tie it.

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4. You can always switch the left and right sides of an equation.

Once an equation has been solved for a variable, it'll be a lot easier to use if that variable is moved to the left side. Mathematically, this has no effect since the both sides are equal. But, it's easier to use the equation if the side for which you are solving is on the left and values are substituted on the right.

The Rules Slide 23 / 249

Let's solve this equation for "d" That means that when we're done we'll have d alone

n the left side of the equation.

The Rules

d 10 8 =

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1

First, is d already alone? If not, what is with it?

A

8

B

d

C

10

D

it is already alone d 10

8 =

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2

What mathematical operation connects d and 10?

A

d is added to 10

B

d is multiplied by 10

C

d is divided by 10

D

10 is subtracted from d d 10

8 =

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3

What is the opposite of dividing d by 10?

A

dividing 10 by d

B

dividing by s into 10

C

multiplying d by 10

D

multiplying by 10 by d

Rule 1. To "undo" a mathematical operation, you must do the

pposite.

d 10 8 =

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4 What must we also do if we multiply the right side by 10?

A

divide the left side by 10

B

multiply the left side by 10

C

divide the left side by d

D

divide the left side by d

Rule 2. You can do anything you want (except divide by zero) to

ne side of an equation, as long as you do the same thing to the
ther.

d 10 8 =

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5 Is there more than one mathematical operation acting

n "d"?

Yes No Rule 3. If there is more than one operation going on, you must undo them in the opposite order in which you would do them, the opposite

f the "order of operations."

d 10 8 =

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Are we done?

Applying Rules 1 and 2

1. To "undo" a mathematical operation, you must do the opposite.
2. You can do anything you want (except divide by zero) to one

side of an equation, as long as you do the same thing to the other. So we undo d being divided by t, by multiplying both sides by t.

d 10 (8) = (10) (10) 80 = d d 10

8 =

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We've now solved our equation for d.

Applying Rule 4

Rule 4. You can always switch the left and right sides of an equation.

d 80 = d = 80

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When solving equations, the goal is to isolate the variable on one side of the equation in order to determine its value (the value that makes the equation true). x + 7 = 32 The goal: get "x" by itself on one side of the equal sign.

Solving Equations Slide 32 / 249

For each equation, write the inverse operation needed to solve for the variable. a.) y + 7 = 14 subtract 7 b.) a - 21 = 10 add 21 c.) 5s = 25 divide by 5 d.) x = 5 multiply by 12 12

click

Inverse Operations

click click click

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Think about this... To solve c - 3 = 12 Which method is better? Why? Discuss at your tables. Kendra Added 3 to each side of the equation c - 3 = 12 +3 +3 c = 15 Ted Subtracted 12 from each side, then added 15. c - 3 = 12

12 -12

c - 15 = 0 +15 +15 c = 15

Inverse Operations Slide 34 / 249

6 Whose method is better to solve? g + 5 = 19 A Miguel B Amanda g + 5 = 19

5

g = 14

5
19

g - 14 = 0 +14 g = 14 g + 5 = 19

19

+14

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7 Whose method is better to solve? A Maria B James

(5) x = 150

= 30 x 5

(5)

(150) 30 30 = 30 x 5 = 1 x 150 (150)

x = 150

= 30 x 5

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Shown above is an identity; it is true for any value of x. These three ways of writing this fraction are identical.

Negative Sign

x 5 _ x 5 _ x 5 _ = =

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8 What is the inverse operation needed to solve this equation? 7x = 49 A Addition B Subtraction C Multiplication D Division

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9 What is the inverse operation needed to solve this equation? x - 3 = -12 A Addition B Subtraction C Multiplication D Division

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10 If h represents a number, which equation is a correct translation of: “Sixty more than 9 times a number is 375”? A 9h = 375 B 9h + 60 = 375 C 9h - 60 = 375 D 60h + 9 = 375

From the New York State Education Department. Office of Assessment Policy, Development and

Administration. Internet. Available from www.nysedregents.org/IntegratedAlgebra; accessed 17, June, 2011.

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One Step Equations

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HINTS: To solve equations, you must work backwards through the order of

perations to find the value of the variable.

Remember to use inverse operations in order to isolate the variable

n one side of the equation.

Whatever you do to one side of an equation, you MUST do to the

ther side!

One Step Equations Slide 42 / 249

Examples: y + 9 = 16

9 -9 The inverse of adding 9

y = 7 is subtracting 9 6m = 72 6 6 The inverse of multiplying by 6 m = 12 is dividing by 6 Remember - whatever you do to one side of an equation, you MUST do to the

ther!!!

One Step Equations

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Slide 43 / 249 One Step Equations

Try These! Solve each equation. x - 8 = -2 +8 +8 x = 6 2 = x - 6 +6 +6 8 = x

click click

Slide 44 / 249 One Step Equations

Try These! Solve each equation. x + 2 = -14

2 -2

x = -16 7 = x + 3

3 -3

4 = x

click click

Slide 45 / 249 One Step Equations

Try These! Solve each equation. 15 = x + 17

17 -17
2 = x

x + 5 = 3

5 -5

x = -2

click click

Slide 46 / 249 One Step Equations

Try These! Solve each equation. 3x = 15 3 3 x = 5

4x = -12
4 -4

x = 3

25 = 5x

5 5

5 = x

click click click

Slide 47 / 249 One Step Equations

Try These! Solve each equation. x 2 x = 20 = 10 (2) (2) x

6

x = -216 = 36 (-6) (-6)

click click

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11 Solve. x - 6 = -11

SLIDE 9

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12 Solve. j + 15 = -17

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13 Solve.

115 = -5x

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14 Solve. = 12 x 9

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15 Solve. 51 = 17y

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16 Solve.

3 = x

7

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17 Solve. 23 + t = 11

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18 Solve. 108 = 12r

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Two-Step Equations

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Let's solve this equation for "t"

Solving for t

15 t 5 =

That means that when we're done we'll have t alone

n the left side of the equation.

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19 Is t already alone? If not, what is with it?

A

5

B

15

C

t

D

it is already alone 15

t 5 =

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20 What mathematical operation connects d to t?

A

t is being divided by 15

B

15 is being divided by t

C

15 is being multiplied by t

D

t is being subtracted from 15 15

t 5 =

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21 What is the opposite of dividing 15 by t?

A

dividing 15 by t

B

dividing 5 by t

C

multiplying 15 by t

D

multiplying t by 15

Rule 1. To "undo" a mathematical operation, you must do the

pposite.

15 t 5 =

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22 What must we do if we multiply the right side by t?

A

divide the left side by t

B

multiply the left side by t

C

divide the left side by 15

D

divide the left side by 5

Rule 2. You can do anything you want (except divide by zero) to

ne side of an equation, as long as you do the same thing to the
ther.

15 t 5 =

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23 Is there more than one mathematical operation acting

n "t"?

Yes No Rule 3. If there is more than one operation going on, you must undo them in the opposite order in which you would do them, the opposite

f the "order of operations."

15 t 5 =

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Are we done?

Solving for t

1. To "undo" a mathematical operation, you must do the opposite.
2. You can do anything you want (except divide by zero) to one

side of an equation, as long as you do the same thing to the other. So we undo d being divided by t, by multiplying both sides by t.

15 t 5 = (t) (t) 5t = 15 15 t 5 =

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24 What mathematical operation connects 5 to t? A

t is being divided by 15

B

t is being divided into 5

C

t is being multiplied by 5

D

t is being subtracted from 5 5t = 15

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25 What is the opposite of multiplying t by 5? A

dividing t by 5

B

dividing t by t

C

multiplying t by t

D

multiplying t by 5 5t = 15

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5

1. To "undo" a mathematical operation, you must do the opposite.

5

t = 3

Solving for t

2. You can do anything you want (except divide by zero) to one

side of an equation, as long as you do the same thing to the

ther.

5t = 15 5t 15 5 =

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26 Is t alone on the left? If not, what is with it? A

5

B

15

C

t

D

it is alone

t = 3

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Sometimes it takes more than one step to solve an equation. Remember that to solve equations, you must work backwards through the order of operations to find the value of the variable. This means that you undo in the opposite order (PEMDAS): 1st: Addition & Subtraction 2nd: Multiplication & Division 3rd: Exponents 4th: Parentheses **Grouping Symbols could be said in the 4th step, instead of Parentheses, making GEMDAS in some cases**

Order of Operations Slide 69 / 249

One way to remember the reverse operations is SADMEG Subtraction Addition Division Multiplication Exponents Grouping Symbols

Order of Operations Slide 70 / 249

Example: 4x + 2 = 10

2 - 2

Undo addition first 4x = 8 4x = 8 4 4 Undo multiplication second x = 2 Remember - whatever you do to one side

f an equation, you MUST do to the other!!!

Two Step Equations Slide 71 / 249 Two Step Equations

2y - 9 = -13

+9

2y = -4

y = 2 +9

2

Undo subtraction first Undo multiplication second

2y = -4

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5b + 3 = 18

3 -3

5b = 15 5b = 15 5 5 b = 3

Two Step Equations

Solve the equation then slide the box down.

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Solve the equation then slide the box down. + 6 = 10 w = 8 w 2 w 2 = 4

6 -6

w 2 = 4 2 2

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3j - 4 = 14 +4 +4 3j = 18 3j = 18 3 3 j = 6

Two Step Equations

Solve the equation then slide the box down.

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2x + 3 = -1
3 -3
2x = -4
2 -2

x = 2

Two Step Equations

Solve the equation then slide the box down.

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2m - 4 = 22

+4 +4

2m = 26
2 -2

m = -13

Two Step Equations

Solve the equation then slide the box down.

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m 3 = 5 +5 = +5 m = 15 m 3

5 = 0

3 3

Two Step Equations

Solve the equation then slide the box down.

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27 Solve the equation. 5x - 6 = -56

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28 Solve the equation. 14 = 3c + 2

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29 Solve the equation. x 5

4 = 24

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30 Solve the equation. 5r - 2 = -12

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31 Solve the equation. 14 = -2n - 6

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32 Solve the equation. + 7 = 13 x 5

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33 Solve the equation. + 2 = -10 x 3

SLIDE 15

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34 Solve the equation.

2.5x - 4 = 3.5

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35 Solve the equation. 3.3x - 4 = -13.9

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36 Solve the equation.

x + (-5.1) = -2.3

6

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37 Solve the equation. 2.8x - 7 = -1.4

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Multi-Step Equations

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Slide 90 / 249 Steps for Solving Multiple Step Equations

As equations become more complex, you should:

1. Simplify each side of the equation.

(Combine like terms and use the distributive property.)

2. Use inverse operations to solve the equation.

Remember, whatever you do to one side of an equation, you MUST do to the other side!

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Example: 5x + 7x + 4 = 28 Combine Like Terms 12x + 4 = 28 Undo Addition

4 - 4

12x = 24 Undo Multiplication 12x = 24 12 12 x = 2

Multiple Step Equations Slide 92 / 249

Example:

1 = 2r - 7r +19

Combine Like Terms

1 = -5r + 19

Undo Subtraction

19 = - 19
20 = -5r

Undo Multiplication

20 = -5r
5 -5

4 = r r = 4

Multiple Step Equations Slide 93 / 249

Try this 12h - 10h + 7 = 25

Multiple Step Equations Slide 94 / 249

Try this 17 - 9f + 6 = 140

Multiple Step Equations Slide 95 / 249

Try this

17q + 7q -13 = 27

Multiple Step Equations Slide 96 / 249

Always check to see that both sides of the equation are simplified before you begin solving the equation. When an equation is simplified there should be at most one term for each variable and one constant term.

Hint

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In order to solve for a variable, it can't be in parentheses with other variables or numbers. If that is how the equation is given, you have to use the distributive property to get it on its own. For example: You need to use the distributive property as shown below as a first step in solving for x. 5(x - 6) - 3x = 8 Distributive Property 5x - 30 - 3x = 8 Now we can combine like terms and solve

Distributive Property Slide 98 / 249

Examples: 5(20 + 6) = 5(20) + 5(6) 9(30 - 2) = 9(30) - 9(2) 3(5 + 2x) = 3(5) + 3(2x)

2(4x - 7) = -2(4x) - (-2)(7)

Remember: The distributive property is a(b + c) = ab + ac

Distributive Property Slide 99 / 249

Example: 2(b - 8) = 28 Distribute the 2 2b - 16 = 28 Add 16 to both sides 2b = 44 Divide both sides by 2 b = 22

Multiple Step Equations Slide 100 / 249

Example: 3r + 4(r - 2) = 13 Distribute the 4 3r + 4r - 8 = 13 Combine like terms 7r - 8 = 13 Add 8 to both sides 7r = 21 Divide by 7 r = 3

Multiple Step Equations Slide 101 / 249

Try this! 3(w - 2) = 9

Distributive Property Slide 102 / 249

Try this 4(2d + 5) = 92

Distributive Property

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Try this 6m + 2(2m + 7) = 54

Distributive Property Slide 104 / 249

38 Solve. 9 + 3x + x = 25

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39 Solve

8e + 7 +3e = -13

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40 Solve

27 = 8x - 4 - 2x - 11

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41 Solve n - 2 + 4n - 5 = 13

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42 Solve 32 = f - 3f + 6f

SLIDE 19

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43 Solve 6g - 15g + 8 - 19 = -38

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44 Solve 3(a - 5) = -21

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45 Solve 4(x + 3) = 20

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46 Solve 3 = 7(k - 2) + 17

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47 Solve 2(p + 7) -7 = 5

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48 Solve 3m - 1m + 3(m - 2) = 19.75

SLIDE 20

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49 Solve. 5.8w + (-2.9w) - 18 = 5.2

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50 Solve. 2(1.75r + 3.2) - 2.25r = 11.4

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Distributing Fractions in Equations

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You always need to have the variable for which you are solving end up in the numerator. Also, it's often easier to just get rid of fractions as a first step.

Distributing Fractions Slide 119 / 249

(-3 + 3x) = 3 5 72 5 3(-3 + 3x) = 72 Distribute

9 + 9x = 72

Add 9 to both sides 9x = 81 Divide both sides by 9 x = 9

Distributing Fractions

Multiply both sides of this equation by the lowest common denominator (LCD) of both sides to eliminate the fractions. Then this just becomes like all the equations we've been solving. In this case, the LCD is 5...so just multiply both sides by 5 to get:

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Example: Multiply both sides by 5 Simplify fractions 3(x + 2) = 45 Distribute the 12 3x + 6 = 45 Subtract 6 from both sides 3x = 39 Divide both sides by 3 x = 13 3 5 (x + 2) = 9 (5)(3) 5 (x + 2) = (9)(5)

Distributing Fractions

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Example: Multiply both sides by 21 Simplify fractions 12(x + 9) = 14 Distribute the 12 12x + 108 = 14 Subtract 108 from both sides 12x = -94 Divide both sides by 12 x = 7 10/12 Simplify fraction x = 7 5/6 4 7 (x + 9) = 2 3 (21)(4) 7 (x + 9) = (2)(21) 3

Distributing Fractions Slide 122 / 249

Example: Multiply both sides by 16 Simplify fractions Distribute the 2 2x + 1 = 7 Subtract 1 from both sides 2x = 6 Divide both sides by 2 x = 3 1 8 (x + ) = 7 16 1 2 (16)(1) 8 (x + ) = 7(16) 16 1 2 (2) (x + ) = 7 1 2

Distributing Fractions Slide 123 / 249

51 Solve the equation. 2x - 3 = 9 5

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52 Solve the equation. 2x + 3 = 4 3

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53 Solve: 2 3 x + 5 6 = 7 12

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54 Solve

3

5 1 2 x + = 1 10

SLIDE 22

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Writing and Solving Algebraic Equations

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Recall from our last chapter that we translated phrases/sentences into expressions. Now, since we know how to solve equations, we can extend this understanding to writing and solving algebraic equations, and using them to solve word problems. Let's start off by reviewing key words for our operations. What are some of them?

Writing and Solving Algebraic Equations Slide 129 / 249

List words that indicate equals

Writing and Solving Algebraic Equations Slide 130 / 249

Remember that in order to write an expression or equation from a word problem, it's helpful to translate the English into a math sentence before making an equation. These next few examples review this process with equations.

Writing and Solving Algebraic Equations Slide 131 / 249 Writing Equations

Tim's age is equal to Kathy's plus 8 T = K + 8 Tim is eight years older than Kathy. Write an equation for Kathy's age. click for mathematical equation click for equation in words Write an equation in words. Then translate that into a mathematical equation.

Slide 132 / 249 Writing Equations

Bob's height equals double Fred's height less 6 B = 2F - 6 click for mathematical equation click for equation in words Bob is 6 inches less than twice the height of Fred. Write an equation for Bob's height. Write an equation in words. Then translate that into a mathematical equation.

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Write an equation in words. Then translate that into a mathematical equation.

Writing Equations

Speed equals distance divided by time s = d/t Speed is equal to the distance traveled in an amount of time. click for equation in words click for mathematical equation

Slide 134 / 249 Writing and Solving Algebraic Equations

Write the equation for each sentence. Then check your answer. Three less than five times a number is twenty-two.

Slide 135 / 249 Writing and Solving Algebraic Equations

Now, solve the equation that you wrote to represent the sentence below. Three less than five times a number is twenty-two.

Slide 136 / 249 Writing and Solving Algebraic Equations

Write the equation for each sentence. Then check your answer. Three fourths of the sum of a number and eight is the same as 39.

Slide 137 / 249 Writing and Solving Algebraic Equations

Now, solve the equation that you wrote to represent the sentence below. Three fourths of the sum of a number and eight is the same as 39.

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55 Which equation can be made from the sentence below: The difference between 15 and a number is 18. A n - 18 = 15 B 18 - n = 15 C n - 15 = 18 D 15 - n = 18

SLIDE 24

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56 What is the solution to the equation that represents the sentence below? The difference between 15 and a number is 18.

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57 Which equation can be made from the sentence below: Twelve more than the quotient of a number and four is -3. A - 12 = -3 B + 12 = -3 C - 12 = 4 D + 12 = 4 n 4 n 4 n

3

n

3

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58 What is the solution to the equation that represents the sentence below? Twelve more than the quotient of a number and four is -3.

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59 Which equation can be made from the sentence below: Two fifths of a number increased by 7 is -1. A n + (-1) = 7 B n - (-1) = 7 C n + 7 = -1 D n - 7 = -1 2 5 2 5 2 5 2 5

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60 What is the solution to the equation that represents the sentence below? Two fifths of a number increased by 7 is -1.

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We can use our algebraic translating skills to solve other problems. We can use a variable to show an unknown. A constant will be any fixed amount. If there are two separate unknowns, relate one to the other.

Writing and Solving Algebraic Equations

SLIDE 25

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The school cafeteria sold 225 chicken meals today. They sold twice the number of grilled chicken sandwiches than chicken

tenders. How many of each were sold?

2c + c = 225

chicken sandwiches chicken tenders total meals

c + 2c = 225 3c = 225 3 3 c = 75 The cafeteria sold 150 grilled chicken sandwiches and 75 tenders.

Writing and Solving Algebraic Equations Slide 146 / 249

Julie is matting a picture in a frame. Her frame is 9 inches wide and her picture is 7 inches wide. How much matting should she put on either side? 2m + 7 = 9.5

7 -7

2m = 2.5 2 2 m = 1.25 = 1 Julie needs 1 inches on each side. 1 4 1 4

both sides

f the mat

size of picture size of frame as a decimal

1 2

Writing and Solving Algebraic Equations

2m + 7 = 9.5

Slide 147 / 249

Many times with equations there will be one number that will be the same no matter what (constant) and one that can be changed based on the problem (variable and coefficient). Example: George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all?

Writing and Solving Algebraic Equations Slide 148 / 249

George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all? Notice that the video games are "per game" so that means there could be many different amounts of games and therefore many different prices. This is shown by writing the amount for one game next to a variable to indicate any number of games.

30g

cost of

ne video

game number

f games

Writing and Solving Algebraic Equations Slide 149 / 249

George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all? Notice also that there is a specific amount that is charged no matter what, the flat fee. This will not change so it is the constant and it will be added (or subtracted) from the other part of the problem.

30g + 7

cost of

ne video

game number

f games

the cost

f the

shipping

Writing and Solving Algebraic Equations Slide 150 / 249

George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all? "Total" means equal so here is how to write the rest of the equation.

30g + 7 = 127

cost of

ne video

game number

f games

the total amount the cost of shipping

Writing and Solving Algebraic Equations

SLIDE 26

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George is buying video games online. The cost of the video is $30.00 per game and shipping is a flat fee of $7.00. He spent a total of $127.00. How many games did he buy in all? Now you can solve it.

Writing and Solving Algebraic Equations Slide 152 / 249

61 Lorena has a garden and wants to put a gate to her fence directly in the middle of one side. The whole length of the fence is 24 feet. If the gate is 4 feet, how many feet should be on either side of the fence?

1 2

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62 Lewis wants to go to the amusement park with his

family. The cost is $12.00 for parking plus $27.00 per

person to enter the park. Lewis and his family spent $147. Which equation shows this problem?

A

12p + 27 = 147

B

12p + 27p = 147

C

27p + 12 = 147

D

39p = 147

Slide 154 / 249

63 Lewis wants to go to the amusement park with his

family. The cost is $12.00 for parking plus $27.00 per

person to enter the park. Lewis and his family spent $147. How many people went to the amusement park WITH Lewis?

Slide 155 / 249

64 Mary is saving up for a new bicycle that is $239. She has $68.00 already saved. If she wants to put away $9.00 per week, how many weeks will it take to save enough for her bicycle? Which equation represents the situation?

A

9 + 68 = 239

B

9d + 68 = 239

C

68d + 9 = 239

D

77d = 239

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65 Mary is saving up for a new bicycle that is $239. She has $68.00 already saved. If she wants to put away $9.00 per week, how many weeks will it take to save enough for her bicycle?

SLIDE 27

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66 You are selling t-shirts for $15 each as a fundraiser. You sold 17 less today then you did yesterday. Altogether you have raised $675. Write and solve an equation to determine the number of t-shirts you sold today. Be prepared to show your equation!

Slide 158 / 249

67 Rachel bought $12.53 worth of school supplies. She still needs to buy pens which are $2.49 per pack. She has a total of $20.00 to spend on school supplies. How many packs of pens can she buy? Write and solve an equation to determine the number of packs of pens Rachel can buy. Be prepared to show your equation!

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68 The product of -4 and the sum of 7 more than a number is -96. Write and solve an equation to determine the number. Be prepared to show your equation!

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69 A magazine company has 2,100 more subscribers this year than last year. Their magazine sells for $182 per year. Their combined income from last year and this year is $2,566,200. Write and solve an equation to determine the number of subscribers they had each year. Be prepared to show your equation! How many subscribers last year?

Slide 161 / 249

Teachers: Use this Mathematical Practice Pull Tab for the next 2 SMART Response slides.

Slide 162 / 249

70 The length of a rectangle is 9 cm greater than its width and its perimeter is 82 cm. Write and solve an equation to determine the width

f the rectangle.

Be prepared to show your equation!

SLIDE 28

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71 The perimeter of a regular hexagon is 13.2 cm. Write and solve an equation to determine the length

f a side of the hexagon.

Be prepared to show your equation!

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72 Rebecca and Megan are shopping at a store that sells jewelry, scarves, and purses. The cost of all the items at the store include tax. Part A Rebecca buys some scarves that cost $5 each and two purses that cost $12 each. The cost of Rebecca's total purchase is $39. Write an equation that can be used to find n, the number of scarves that Rebecca buys. Drag and drop the appropriate variable or number into each box.

From PARCC sample test

n 2 5 12 17 24 39

+ =

Slide 165 / 249

73 (Continued from previous slide.) Part B Megan buys 3 bracelets and 3 necklaces. Each bracelet costs $5. Megan pays the clerk $40 and gets $4 change. What is the cost, in dollars, of one necklace?

From PARCC sample test

Slide 166 / 249

74 Jessica rented 1 video game and 2 movies for a total of $11.50 · The video game cost $4.75 to rent. · The movies cost the same amount each to rent. What amount did Jessica pay to rent each movie?

From PARCC sample test

Slide 167 / 249

75 Devon exercised the same amount of time each day for 5 days last week. · His exercise included walking and swimming. · Each day he exercised, he walked for 10 minutes. · He exercised for a total of 225 minutes last week. What is the number of minutes Devon swam each of the 5 days last week?

From PARCC sample test

Slide 168 / 249

76 At the start of the month, the value of an investment was $48.45. By the end of the month, the value of the investment changed by a loss of $13.80. What was the value, in dollars, of the investment at the end of the month?

From PARCC sample test

SLIDE 29

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Graphing and Writing Inequalities with One Variable

Return to Table

f Contents

Slide 170 / 249

When you need to use an inequality to solve a word problem, you may encounter one of the phrases below.

Important Words Sample Sentence Equivalent

Translation

is more than Trenton is more than 10 miles away. t > 10 is greater than A is greater than B. A > B must exceed The speed must exceed 25 mph. The speed is greater than 25 mph. s > 25

Inequalities Slide 171 / 249

Important Words Sample Sentence Equivalent Translation

cannot exceed Time cannot exceed 60 minutes. Time must be less than or equal to 60 minutes. t < 60 is at most At most, 7 students were late for class. Seven or fewer students were late for class. n< 7 is at least Bob is at least 14 years old. Bob's age is greater than or equal to 14. B > 14 When you need to use an inequality to solve a word problem, you may encounter one of the phrases below.

Inequalities Slide 172 / 249

How are these inequalities read? 2 + 2 > 3 Two plus two is greater than 3 2 + 2 ≥ 4 Two plus two is greater than or equal to 4 2 + 2 < 5 Two plus two is less than 5 2 + 2 ≤ 5 Two plus two is less than or equal to 5 2 + 2 ≤ 4 Two plus two is less than or equal to 4 2 + 2 > 3 Two plus two is greater than or equal to 3

Inequalities Slide 173 / 249 Writing Inequalities

Let's translate each statement into an inequality. x is less than 10 20 is greater than or equal to y x < 10 words inequality statement translate to 20 > y

Slide 174 / 249

Write an inequality for each sentence below. Double a number is at most four. Three plus a number is at least six. 2x ≤ 4 3 + x ≥ 6

click click

Writing Inequalities

SLIDE 30

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Five less than a number is less than twice that number. The sum of two consecutive numbers is at least thirteen. Three times a number plus seven is at least nine. x - 5 < 2x x + (x + 1) ≥ 13 3x + 7 > 9

click click click

Write an inequality for each sentence below.

Writing Inequalities Slide 176 / 249

$10.50 10.5 at least > An employee earns e A store's employees earn at least $10.50 per hour. Define a variable and write an inequality for the amount the employees may earn per hour. Let e represent an employee's wages.

Writing Inequalities Slide 177 / 249

77 You have $200 to spend on clothes. You already spent $140 and shirts cost $12. Which equation shows this scenario?

A

200 < 12x + 140

B

200 ≤ 12x + 140

C

200 > 12x + 140

D

200 ≥ 12x + 140

Slide 178 / 249

78 A sea turtle can live up to 125 years. If one is already 37 years old, which scenario shows how many more years could it live? A 125 < 37 + x B 125 ≤ 37 + x C 125 > 37 + x D 125 ≥ 37 + x

Slide 179 / 249

79 The width of a rectangle is 3 in longer than the

length. The perimeter is no less than 25 inches.

A 4a + 6 < 25 B 4a + 6 ⁯≤ 25 C 4a + 6 > 25 D 4a + 6 ≥ 25

Slide 180 / 249

80 The absolute value of the sum of two numbers is less than or equal to the sum of the absolute values

f the same two numbers.

A B C D

SLIDE 31

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Remember: Equations have one solution. Solutions to inequalities are NOT single numbers. Instead, inequalities will have more than one value for a solution. This would be read as, "The solution set is all numbers greater than or equal to negative 5."

Solution Sets

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Let's name the numbers that are solutions of the given inequality. r > 10 Which of the following are solutions? {5, 10, 15, 20} 5 > 10 is not true So, not a solution 10 > 10 is not true So, not a solution 15 > 10 is true So, 15 is a solution 20 > 10 is true So, 20 is a solution Answer: {15, 20} are solutions of the inequality r > 10

Solution Sets Slide 183 / 249

Let's try another one. 30 ≥ 4d; {3, 4, 5, 6, 7, 8} 30 ≥ 4d 30 ≥ (4)3 30 ≥ 12 30 ≥ 4d 30 ≥ (4)4 30 ≥ 16 30 ≥ 4d 30 ≥ (4)5 30 ≥ 20 30 ≥ 4d 30 ≥ (4) 6 30 ≥ 24 30 ≥ 4d 30 ≥ (4)7 30 ≥ 28 30 ≥ 4d 30 ≥ (4)8 30 ≥ 32

click to reveal click to reveal click to reveal click to reveal click to reveal click to reveal

Solution Sets Slide 184 / 249

81 Which of the following are solutions to the inequality: x > 11 {9, 10, 11, 12} Select all that apply. A 9 B 10 C 11 D 12

Slide 185 / 249

82 Which of the following are solutions to the inequality: m < 15 {13, 14, 15, 16} Select all that apply. A 13 B 14 C 15 D 16

Slide 186 / 249

83 Which of the following are solutions to the inequality: x > 34 {32, 33, 34, 35} Select all that apply. A 32 B 33 C 34 D 35

SLIDE 32

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84 Which of the following are solutions to the inequality: 3x > 15 {4, 5, 6, 7} Select all that apply. A 4 B 5 C 6 D 7

Slide 188 / 249

85 Which of the following are solutions to the inequality: 6y < 42 {6, 7, 8, 9} Select all that apply A 6 B 7 C 8 D 9

Slide 189 / 249

Since inequalities have more than one solution, we show the solution two ways. The first is to write the inequality. The second is to graph the inequality on a number line. In order to graph an inequality, you need to do two things:

1. Draw a circle (open or closed) on the number that is your

boundary.

2. Extend the line in the proper direction.

Graphing Inequalities Slide 190 / 249 Graphing Inequalities - The Circle

An open circle on a number shows that the number is not part of the solution. It is used with "greater than" and "less than". The word equal is not included. < > A closed circle on a number shows that the number is part of the solution. It is used with "greater than

r equal to" and "less than or equal to". ≤ ≥

Determining Whether to Use an Open or Closed Circle

Slide 191 / 249 Determining Which Direction to Extend the Line

Extend Line to the Left: If the variable is smaller than the number, then you extend your line to the left (since smaller numbers are on the left). Extend the line to the left in these situations: # > variable variable < # Extend Line to the Right: If the variable is larger than the number, then you extend your line to the right (since bigger numbers are

n the right).

Extend the line to the right in these situations: # < variable variable > #

Slide 192 / 249

When graphing inequalities, ask yourselves each question below. What is the number in the inequality? What kind of circle should be used? In what direction does the line go?

Graphing Inequalities

SLIDE 33

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Step 1: Rewrite this as x < 5. Step 2: What kind of circle? Because it is less than, it does not include the number 5 and so it is an open circle.

1
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5

1 2 3 4 5

Graphing Inequalities

x is less than 5

Slide 194 / 249

Step 4: Draw a line, thicker than the horizontal line, from the dot to the

arrow. This represents all of the numbers that fulfill the inequality.

Step 3: Draw an arrow on the number line showing all possible

solutions. Numbers greater than the variable, go to the right.

Numbers less than the variable, go to the left.

1
2
3
4
5

1 2 3 4 5 x < 5

Graphing Inequalities

1
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5

1 2 3 4 5

Slide 195 / 249

Step 1: Rewrite this as x ≤ 5. Step 2: What kind of circle? Because it is less than or equal to, it does include the number 5 and so it is a closed circle.

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5

1 2 3 4 5

Graphing Inequalities

x is less than or equal to 5

Slide 196 / 249

Step 4: Draw a line, thicker than the horizontal line, from the dot to the

arrow. This represents all of the numbers that fulfill the inequality.

Step 3: Draw an arrow on the number line showing all possible

solutions. Numbers greater than the variable, go to the right. Numbers

less than the variable, go to the left.

1
2
3
4
5

1 2 3 4 5 x ≤ 5

Graphing Inequalities

1
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1 2 3 4 5

Slide 197 / 249

1 2 3 4 5 6 7 8 9 10

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You try Graph the inequality x > 3.5 Graph the inequality

2.5 > x

click 3.5 on the number line for answer click -2.5 on the number line for answer

Graphing Inequalities

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Slide 198 / 249

Try these. Graph the inequalities.

1. x > -3
2. x < 4
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5

1 2 3 4 5

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1 2 3 4 5

Graphing Inequalities

SLIDE 34

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Try these. State the inequality shown. 1. 2.

Graphing Inequalities

1
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1 2 3 4 5

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1 2 3 4 5

Slide 200 / 249

86 This solution set would be x ≥ -4. True

False

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Slide 201 / 249

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87 A x > 1.5 B x < 1.5 C x < 1.5 D x > 1.5 State the inequality shown.

Slide 202 / 249

5 6 7 8 9 10 11 12 13 14 15

88 A 11 < x B 11 > x C 11 > x D 11 < x State the inequality shown.

Slide 203 / 249

89 A x > -1.5 B x < -1.5 C x ≤ -1.5 D x ≥ -1.5 State the inequality shown.

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1 2 3 4 5

E x > -0.5 F x < -0.5 G x ≤ -0.5 H x ≥ -0.5

Slide 204 / 249

90 A

4 < x

B

4 > x

C

4 ≤ x

D

4 ≥ x

State the inequality shown.

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3
4
5

1 2 3 4 5

SLIDE 35

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91 A x > 0 B x < 0 C x ≤ 0 D x ≥ 0 State the inequality shown.

Pull Pull

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1 2 3 4 5

Slide 206 / 249

Simple Inequalities Involving Addition and Subtraction

Return to Table

f Contents

Slide 207 / 249

x + 3 = 13

3 - 3

x = 10 Who remembers how to solve an algebraic equation? Does 10 + 3 = 13 13 = 13 Be sure to check your answer! Use the inverse of addition

Inequalities Involving Addition & Subtraction Slide 208 / 249

Solving one-step inequalities is much like solving one-step equations. To solve an inequality, you need to isolate the variable using the properties of inequalities and inverse operations. Remember, whatever you do to one side, you do to the other.

Inequalities Involving Addition & Subtraction Slide 209 / 249

12 > x + 5

5 -5 Subtract to undo addition

7 > x To find the solution, isolate the variable x. Remember, it is isolated when it appears by itself on one side of the equation, or in this case, inequality.

Inequalities Involving Addition & Subtraction Slide 210 / 249

7 > x The symbol is > so it is an open circle and it is numbers less than 7 so it goes to the left. 1 2 3 4 5 6 7 8 9 10

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Inequalities Involving Addition & Subtraction

SLIDE 36

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A. j + 7 > -2

Solve and graph.

9 is not included in solution set; therefore we graph

with an open circle.

A. j + 7 > -2
7 -7

j > -9

Inequalities Involving Addition & Subtraction Slide 212 / 249

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B. r - 2 ≥ 4

Solve and graph. r - 2 ≥ 4 +2 +2 r ≥ 6

Inequalities Involving Addition & Subtraction Slide 213 / 249

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5 ≥ w

4
4

9 ≥ w + 4 w ≤ 5

C. 9 ≥ w + 4

Solve and graph.

slide to reveal click to reveal

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Inequalities Involving Addition & Subtraction Slide 214 / 249

92 Solve the inequality. 3 < s + 4 ____ < s

Slide 215 / 249

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A B C D 93 Solve the inequality and graph the solution.

4 + b < -2

Slide 216 / 249

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A B C D 94 Solve the inequality and graph the solution.

8 > b - 5

SLIDE 37

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95 Solve the inequality. m + 6.4 < 9.6 m < ______

Slide 218 / 249

Simple Inequalities Involving Multiplication and Division

Return to Table

f Contents

Slide 219 / 249

Since x is multiplied by 3, divide both sides by 3 for the inverse operation.

Multiplying or Dividing by a Positive Number

3x > -36 3x > -36 3 3 x > -12

Slide 220 / 249

Solve the inequality. 2 3 r < 4 3 2

( )

r < 6 Since r is multiplied by 2/3, multiply both sides by the reciprocal of 2/3.

2 3 r < 4 3 2

( )

Multiplying or Dividing by a Positive Number Slide 221 / 249

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96 3k > 18 A B C D

Slide 222 / 249

97 A B C

30 ≥ 3q

10 ≥ q

10 ≤ q
10 ≥ q

D 10 ≤ q

SLIDE 38

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98 x 2 A B C D < -3

Slide 224 / 249

99 A B C g > 27 g > 36 g > 108 g ≥ 36 D g ≥ 108 3 4

Slide 225 / 249

100 A B C

21 ≥ 3d

d ≥ -7 d > -7 d < -7 D d ≤ -7

Slide 226 / 249

Lab: Multiplying or Dividing by a Negative Number · Sometimes you must multiply or divide to isolate the variable. · Multiplying or dividing both sides of an inequality by a negative number gives a surprising result. · Find out what happens by completing the lab given below. Now let's see what happens when we multiply or divide by negative numbers.

Multiplying or Dividing by a Negative Number Slide 227 / 249

1. Write down two different numbers and put the appropriate

inequality (< or >) between them.

2. Apply each rule to your original two numbers from step 1 and
simplify. Write the correct inequality(< or >) between the answers.
A. Add 4
B. Subtract 4
C. Multiply by 4
D. Multiply by -5
E. Divide by 4
F. Divide by -4

Lab: Multiplying or Dividing by a Negative Number Slide 228 / 249

3. What happened with the inequality symbol in your results?
4. Compare your results with the rest of the class.
5. What pattern(s) do you notice in the inequalities?
6. How do different operations affect inequalities?
7. Write a rule for inequalities.

Lab: Multiplying or Dividing by a Negative Number

SLIDE 39

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Dividing each side by -3 changes the ≥ to ≤.

3y ≥ 18
3y ≤ 18
3 -3

y ≤ -6 Solve and graph. A.

Multiplying or Dividing by a Negative Number Slide 230 / 249

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Divide each side by -7 Flip the sign because you divided by a negative.

7m > -28
7m < -28
7 -7

m < 4 Solve and graph. B.

Multiplying or Dividing by a Negative Number Slide 231 / 249

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Divide each side by 5. The sign does NOT change because you did not divide by a negative. 5m > -25 5m > -25 5 5 m > -5 Solve and graph. C.

Multiplying or Dividing by a Negative Number Slide 232 / 249

D. -8y > 32

Solve and graph.

E. -9f > -54

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Multiplying or Dividing by a Negative Number Slide 233 / 249

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You multiplied by a negative.

r

2 < 5

2

( )

r > -10 Multiply both sides by the reciprocal of

1/2.
r

2 > 5

2

( )

Why did the inequality change?

move to reveal

Multiplying or Dividing by a Negative Number Slide 234 / 249

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1. -6h < 42

Try these. Solve and graph each inequality.

2. 4x > -20

Multiplying or Dividing by a Negative Number

SLIDE 40

Slide 235 / 249

3. 5m < 30

Try these. Solve and graph each inequality.

4. > -3

a

2

Multiplying or Dividing by a Negative Number

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101 Solve and graph. 3y < -6

If you can't put the inequality in your responder, just put the number.

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102 Solve and graph. x

4 < -2

If you can't put the inequality in your responder, just put the number.

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103 Solve and graph.

5y ≤ -25

If you can't put the inequality in your responder, just put the number.

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104 Solve and graph. n

2 > 2

If you can't put the inequality in your responder, just put the number.

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105 Consider the inequality 5x < 30. Part A Natalia says that any value of x less than 25 makes the inequality true. · Use a specific example to disprove Natalia's statement. · Explain why your example disproves her statement. Part B Describe in words all values of x that make the inequality true. Explain your answer.

Students type their answers here

From PARCC sample test

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Glossary & Standards

Return to Table

f Contents

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Closed Circle

Shows that the number is part of the solution. It is used with "greater than or equal to" and "less than or equal to".

≤ and ≥ make

makes makes x ≥ -5

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The Distributive Property

A property that allows you to multiply all the terms

n the inside of a set of parenthesis by a term on

the outside of the parenthesis.

a(b + c) = ab + ac a(b + c) = ab + ac a(b - c) = ab - ac 3(2 + 4) = (3)(2) + (3)(4) = 6 + 12 = 18 3(2 - 4) = (3)(2) - (3)(4) = 6 - 12 = -6 3(x + 4) = 48 (3)(x) + (3)(4) = 48 3x + 12 = 48 3x = 36 x = 12

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Equation

A mathematical statement, in symbols, that two things are exactly the same (or equivalent).

4x+2 = 14 3y + 2 = 11 11 - 1 = 3z + 1 7x = 21

(where x = 3) (where z = 3) (where y = 3)

a.k.a. function d = rt

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Identity

An equation that is equal for all values of their variables

3(x - 1) = 3x - 3 3x - 3 = 3x - 3

5 + 2 = 7

3x - 1 = 3x + 1

3x -3x
1 = +1

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Inverse Operation

The operation that reverses the effect of another operation.

Addition

Subtraction Multiplication Division + _ x ÷ 11 = 3y + 2

2
2

9 = 3y ÷ 3 ÷ 3 3 = y

5 + x = 5

x = 10 + 5 + 5

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Open Circle

Shows that the number is not part of the solution. It is used with "greater than" and "less than".

< and > make

makes makes x < 5

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Solution Set

The set of all the solutions of an inequality.

x ≥ -5 x is greater than or equal to -5

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Standards for Mathematical Practices

Click on each standard to bring you to an example of how to meet this standard within the unit. MP8 Look for and express regularity in repeated reasoning. MP1 Make sense of problems and persevere in solving them. MP2 Reason abstractly and quantitatively. MP3 Construct viable arguments and critique the reasoning of others. MP4 Model with mathematics. MP5 Use appropriate tools strategically. MP6 Attend to precision. MP7 Look for and make use of structure.