Acknowledgements:

We would like to dedicate this book to our math teacher, Mrs.Urooj. We would like to thank her for all her effort with us this year and all the time she spent making sure we got the best of education. By writing this book we would like to show her that we are very thankful and grateful for all her effort with us this year.




Merai: chapter 1-2,1-3,1-4 and 5-2,5-3,5-5

Siba: chapter 2 and introductory 2.4, 2.5, 2.6

Zayd: chapter 3 and PUBLISH

3.2 3.4 3.5

Adam: chapter 4 and COVER

4.2 4.3 4.4

MATH WITH SAMZ

Objective : solve one-step equations in one variable by using addition or subtraction.

Vocabulary :

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

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

Example with explanation:

Solving Equation by Using Addition

A x-2=12           

x=12+2

x=14

Since 2 is subtracted from x, add 2 on the other side

B m-½=½  

m=½+½

m=1

Since ½ is subtracted from m, add ½ on the other side

Solving Equation by Using Subtraction

A x+2=7           

x=7-2

x=5

Since 2 is added to x, subtract 2 on the other side

 

B y+0.5=0.10

y=0.10-0.5

y=0.5

Since 0.5 is added to y, subtract 0.5 on the other side  



Real World Application

A person’s maximum heart rate is the highest rate, in beats per minute, that the person’s heart should reach. One method to estimate maximum heart rate states that your age added to your maximum heart rate is 220. Using this method, write and solve an equation to find the maximum heart rate of 15 years old.

Ageadded tomaximum heart rateis220

a         +                  r                           =220

a+r=220

15+r=220

r=220-15

r=205 beats per minute

 

Chapter 1 Equations

1-2 Solving Equations by Adding or Subtracting

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

Example with explanation:

Solving Equations by Using Multiplication

A -2=m/-3

(-2)(-3)=m

6=m

Since m is divided by -3, multiply -3 on the other sideby -2

B x/5=1.5

x=(1.5)(5)

x=7.5

Since x is divided by 5, multiply 5 on the other side by 1.5

Solving Equations by Using Division

A 2x=18

x=18/2

x=9

Since x is multiplying by 2, divide 2 on the other side by 18

B 5=-2y

5/-2=y

-2.5=y

Since y is multiplying by -2, divide -2 on the other side by 5



Solving Equations That Contain Fractions

A 5/2x=25

x=(2/5) 25

x=10

The reciprocal of 5/2 is 2/5. Since x is multiplied by 5/2, multiply on the other side by 2/5

B 1/2=4y/2

(2/4)1/2 =y

¼=y

The reciprocal of 4/2 is 2/4. Since y is multiplied by 4/2, multiply on the other side by 2/4

Real World Application

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. If a plane is 10,000 feet above the ground, write and solve an equation to find the distance at which the pilot should begin descending.

Distancedivided by 3equalsheight in thousands of feet

D           /        3              =                      h

d/3=10

d=10(3)

d=30 miles from the airport

 

1-3 Solving Equations by Multiplying or Dividing

Objective: Solve equations in one variable that contain more than one step

Examples with Explanation

Solving Two-Step Equations

4-2x=10

-2x=10-4

-2x=6

x=-2/6

x=-1/3

First x multiplied by -2. Then 4 is added.subtract 4 from 10.

Since x is multiplied by -2,divide it on the other side by 6.

Solving Two-Step Equations That Contain Fractions

a/14-1/4=2/4

Method 1 use fraction operation

a/14-1/4=2/4

a/14=2/4+1/4

a/14=3/4

a=(14)¾

a=10.5

Since ¼ is subtracted from a/14, add 1/4 on the other side.

Since a is divided by 14, multiply the other side by 14.

Simplify

Simplifying Before Solving Equations

4x+2-8x=12

4x+2-8x=12

4x-8x+2=12

-4x+2=12

-4x=12-2

-4x=10

x=10/-4

x=-5/2

Combine like terms.

Since 2 is added to -4x, subtract 2 from 12 on the other side.

Since x is multiplied by -4, divide -4 on the other side

Solving Equations to Find an Indicated Value

If 2a+12=14,find the value of a+4

Step 1 find the value of a

2a+12=14

2a=14-12

2a=2

a=2/2

a=1

Since 12 is added to 2a, subtract 12 from the other side.

Since a is multiplied by 2, divide 2 on the other side.

Step 2 find the value of a+4

a+4

1+4=5

 

Chapter 1 Exercises

Lesson 2:

  1. 17=w-4

  2. s-5=3

  3. x-3.9=12.4

  4. t+5=-25

  5. 9=s+9

  6. b+⅔=2

  7. -⅚+p=2

  8. 1.75=k-0.75

  9. -⅙+h=⅙

  10. -5.2+a=-8

Lesson 3:

  1. k/4=8

  2. z/3=-9

  3. -2=w/-7

  4. 84=-12a

  5. 4m=10

  6. 9=-3/8r

  7. x/2=12

  8. 49=7c

  9. 5t=-15

  10. -52=-4c

Lesson 4:

  1. 4a+3=11

  2. 8=3r-1

  3. 42=-2d+6

  4. 1/3y+¼=5/12

  5. x/6+4=15

  6. 5=2g+1

  7. 3x+3=18

  8. 2(x+3)=10

  9. 5(h-4)=8

  10. 6=-2(7-c)

Answers:

Lesson 2:

1) 21

3) 16.3

5)0

7)17/6

9)⅓

Lesson 3:

1)32

3) 14

5)2.5

7)24

9)-3

Lesson 4:

1)2

3)-18

5)66

7)5

9)28/5

 

1-4 Solving Two-Step and Multi-Step Equations

Objective: The objective of this lesson is to learn how to solve inequalities with more than one step.

Explanation with example:

Solve each inequality and graph the solutions.

250+50y<500
  250+50y<500   Since 250 is adding to 50y, subtract 250 on the two sides.
-250         -250

 50y    <    250

50y/50  < 250/50  Divide both. sides by 50

y<5




To pass the quarter Elen needs to get 46 points. She already got 15 points on her first test. What is the least amount of points she can get on the second test to pass?

b= points on the second test

(15 plusb)divided by2must be greater than46

 

(15+b)/2 >46

 

15+b/2 >48         Multiply both sides by 2 to undo the division.

2(15+b/2) >2(46)

15+b > 92           Subtract 15 on both sides to undo the addition.

-15      -15   

b>77

 

Chapter 2

Lesson 4

Solving Two-Step and Multistep Inequalities

 

 

 

 

Solve each inequality and graph the solutions:

 

4x> 36

 

4x>36

4x/4 > 36/4 Divide both sides by 4 to undo the multiplication.

x>9




4/6x< 2

 

4/6x<2

6/4(4/6x)<6/4(2) Multiply both sides by the reciprocal of (4/6)

x<3

 



Chapter 2 Review:                                             




Lesson 3:

1- 5x> 25

2- 3x<24

3- 6/12> 2

4- ⅗<3

Lesson 4:

1- 10+2v<50

2-40-6v>180

3-50+6v<290

4-320-5v>700

 

Lesson 5:

1-x<8x+28

2-x>10x-36

3-x<16x+105

4-x>5x-40



Answers: (odd numbers)

Lesson 1: x>5,x>12

Lesson 2: v<20,v<40

Lesson 3: x< 4, x<7












Chapter 2 lesson 3

Solving inequalities by multiplying or dividing

Objective: The objective of this lesson is to learn how to solve inequalities that have variables on both sides.

 

Solve each inequality and graph the solutions:

x<6x+10

 

   x<6x+10  Subtract x from both sides.

 -x   -x

0< 5x+10    Subtract 10 from both sides to undo the addition.

-10     -10

-10<5x

-10/5 < 5x/5  Divide both side by 2 to undo the multiplication.

-2<x

 

Game A charges a fee of $120 plus $40 per week to work. Game B charges $100 per week. For how many weeks will the total cost at game A be less expensive than the cost of game B?

 

W= number of weeks

Game A feeplus $40 per week timesnumber of weeksis less expensive thanGame B charge per week timesnumber of weeks

 

120 + 40 x w < 100 x w

 

120+40w < 100w      Subtract 40 from both sides.

     -40w        -40w

120         <      60      Divide both sides by 60 to undo the multiplication.

120/60 < 60w/60

2<w

The total cost of game A will be less expensive for 2 weeks.



2.5

Solving Inequalities with Variables on Both Sides

            

Objectives:

The objective of this lesson is to identify domain and range from the relations and functions.

Vocabulary :

Relation – can also be renamed by ordered pairs.            

Functions –is a type of relation that pairs each input value with an exactly one output range .

Domain –is the input value of the first coordinates.

Range –is the output value of the second coordinates

Showing Multiple Representations of Relations.

Table

Track scoring

Place

Points

1

6

3

2

   

Graph

Tracking scoring

 

                             Mapping diagram

               
 
               
     
             
         
           
       
           

Finding the Domain and Range of a Relation

graph-2-points-c.gif

 
             
         
           
       
           

D:1<=x<=6                                                   R:3<=y<=8

 

Identifying functions

Field trip

Students X

Bus Y

50

2

100

4

125

5

 

D(50,100,125)

R(2,4,5)

  3.2

Relations and Functions

3.4

Graphing Functions

Objective:

Graphing functions given a limited input value and given an input value of a number that is real.

Graphing solution given a domain.



Step 1 :

-x+3y=6D{-4, -2, 2, 4}

-x+3y=6

+x       +x

------------

3y=x+6

3y/2=x+6/2

y=x/2+6/2

y=1/2x+3

 

x

y=1/2x+3

(x,y)

-4

y=½(-4)+3=1

(-4,1)

-2

y=½(-2)+3=2

(-2, 2)

0

y=½(0)+3=3

(0,3)

2

y=½(2)+3=4

(2, 4)

 

 






Graphing functions using a domain of all real numbers.

Step 1: use the function to generate ordered pairs by choosing several values for x.

Step 2: plot enough points to see a pattern for the graph.

Step 3: connect the points with a line or smooth curve.




3.4

Graphing Functions

 

Objectives :In this lesson you will if the graph has a positive correlation two sets lation, negative correlation, no correlation.

Vocabulary

A scatter plot is a graph that shows a possible relationship between two sets of data.

A correlation shows a relationship between two data sets

A positive correlation is both sets of data values increase

A negative correlation is one set of data values increases as the other set decreases

No correlation is when there is no relationship between the data sets.

A trend line is when you can graph a line on a scatter plot to help show a relationship between the data

Calendar year

1996

1997

1998

1999

2000

2001

2002

Species

91

79

62

11

39

10

9

 










positive correlation





negative_correlation.gif graph with negative correlation











nocorelation.gif

3.5

Scatter plots and trend lines

The rate of change is mostly used to find the amount of change in a dependent variable to the amount of change in an independent variable.

 

RATEOF CHANGE= change in dependent variable/change in independent variable

 

                                                             1st Example

This table shows the cost of a pencil in different years. Find the rate of change in cost for each time interval. During which time interval did the cost increase at the greatest rate?

 

Year

1975

1980

1985

1990

1995

Cost (cents)

6

10

12

13

22

 

STEP 1: Find the independent, and the dependent variables

DEPENDENT:cost                          INDEPENDENT: year

STEP 2: find the rates of change

 

1975 to 1980        change in cost/change in years=10-6/1980-1975=⅘=0.8=0.8cents/ year

   

   1980 to 1985        change in cost/change in years=12-10/1985-1980=⅖=0.4=0.4cents/ year

 

   1985 to 1990        change in cost/change in years=13-12/1990-1985=⅕=0.2=0.2 cents/year

    

   1990 to 1995        change in cost/change in years=22-13/1995-1990=9/5=1.8=1.8cents/year



RISE: How far a line goes up for a given distance

RUN: How far a line goes along for a given distance

SLOPE: Is the comparison of the rise, and the run for any two points on the line.

SLOPE=rise/run=change in y/change in x

Untitled 5.png     

 

Tell whether the slope of each line is positive, negative, zero, or undefined.

Untitled 8.png








This table shows the the cost of the same kind of television, but in different years. Find the rate of change in cost for each year.

 

Year

1995

2000

2005

2010

cost ($)

500$

490$

310$

250$



 



Find the slope of each line.


























Tell whether the slope of each line is positive, negative, zero, or undefined.

Untitled 9.png






This table shows a runner’s heart rate over the time. Find the rate of change of the heart rate for each minute.

 

Time min/h

3 min

7 min

15 min

19 min

25 min

Heart rate (beats per min)

78

 86

99

101

120






















CHAPTER 4                                  RATE OF CHANGE AND SLOPE

      4-3

 

SLOPE: Is the comparison of the rise, and the run for any two points on the line.

 

Slope=y2-y1/x2-x1

 

Finding Slope by Using the Slope Formula.

 

(4,6) and (9,3)



M=y2-y1/x2-x1 use the slope formula

 

M=3-6/9-4 substitute (4,6) for (x1,y1) and (x2,y2) for (9,3)

-⅗ simplify

 

M= -⅗



x

4

7

2

9

y

0

5

7

9

 

Choose any 2 points from the table Let (x1,y1) be (7,5) and (x2, y2) be (9,9)

m=y2-y1/x2-x1 use the slope formula

 

m= 5-9/7-9 substitute them

 

m=-4/-2 simplify

 

m= -4/-2

The slope of the line that contains (7, 5) and (9, 9) is -4/-2

 

1a- Find the slope of the line that contains (-6, 1) and (9, -3)

 

1b- Find the slope of the line that contains (2, 7) and (-4, -8)

 

If sometimes they don’t give you two points to find the slope you need to pick any two points from the graph.

 

Let (2, 2) and (0, -0.5) be (x1,y1) and (x2, y2)

m=y2-y1/x2-x1 use the slope formula

=-0.5-2/0-2 Substitute them

=-2.5/-2 Simplify

m=-2.5/-2

 

x

5

8

3

7

y

4

6

1

5

Choose any two points from the table and substitute them for (x1, y1) and (x2, y2)

(8, 6) and (3, 1)

m=y2-y1/x2-x1  Use the slope formula

=1-6/3-8 substitute them

=-5/-5 Simplify

m=1

A slope is undefined when its denominator is 0

Finding the slope from an equation.

 

The slope of a line is described by 4x-9y= 36

 

Step 1   Finding the x intercept

4x-9y= 36

4x-9(0)=36 substitute y for 0

4x=36

4x/4=36/4

x=9

 

Step 2   Finding the y intercept

4x-9y=36

4(0)-9y=36 substitute y for 0

-9y=36

-9y/-9=36/-9

y=-4

The line contains (9,0) and (0, 4). Use the slope formula.

m=y2-y1/x2-x1=4-0/0-9=4/-9

m=4/-9






EXERCISES

Find the slope of the line that contains each pair of points.

1- (3,9)and (2, 7)                   2- (-7, 3) and (5-9)

3- (1,-8) and (0, 4)



Find the slope of both the graph and the table





The Slope Formula

4-4

Objective : Solve systems of linear equations in two variables by substitution.

Step 1 Solve for one variable in at least one equation,if necessary.

Step 2 Substitute the resulting expression into the other equation.

Step 3 Solve that equation to get the value of the first variable.

Step 4 Substitute that value into one of the original equations and solve.

Step 5 Write the values from step 3 and 4 as an ordered pair (x,y).

 

Example with Explanation

Solving a System of Linear Equations by Substitution

1) y=2x

y=x+4

Step 1:Both equations are solved for y.

y=2x

y=x+5

Step 2: Substitute 2x for y in the second equation.

y=x+5

2x=x+5

Step 3: Solve for x

x=5

Step 4: write one of the original equations.substitute 5 for x

y=2x

y=2(5)

y=10

Step 5: Write the solution as an ordered pair (5,10)

Using the Distributive Property

x+2y=-1

x=-1-2y

4x-4y=20

Find y

4x-4y=20

4(-1-2y)-4y=20

-4-8y-4y=20

-4-12y=20

-12y=20+4

-12y=24

y=24/-12

y=-2

Find x

x+2y=-1

x+2(-2)=-1

x-4=-1

x=-1+4

x=3

(3,-2)

Real World Example

1) One high-speed internet provider has a 50 dollars setup fee and costs 30 dollars per month.Another provider has no setup fee and costs 40 dollars per month.

  1. In how many months will both providers cost the same ? What will that cost be ?

Total paidissetup feepluscost per monthtimesmonths.

Option 1

T              =50              +          30                  x        m  

Option 2

T               =0               +          40                  x        m

Step 1

t=50+30m

t=40m

Step 2

50+30m=40m

Step 3

50=20m

50/10=m

5=m

Step 4

t=40m

t=40(5)

t=200

Step 5

(5,200)

 

Chapter 5 System of Equations and Inequalities

5-2 Solving System by Substitution

 

Objective : Solve systems of linear equations in two variables by elimination.

Compare and choose an appropriate method for solving systems of linear equations.

 

Method for solving systems of equations is elimination. Like substitution, the goal of elimination is to get one equation that has only one variable.

 

Solving Systems of Equations by Elimination

Step 1: Write the system so that like terms are aligned.

Step 2: Eliminate one of the variables and solve for the other variable.

Step 3: Substitute the value of the variable into one of the original equations and solve for the other variable.

Step 4: Write the answers from steps 2 and 3 as an ordered pair, (x,y).

Examples with Explanation

Elimination Using Addition

x-2y=-19

5x+2y=1

Step 1

(x-2y=-19)+(5x+2y=1)

Step 2

6x+0=-18

6x=-18

x=-18/6

x=-3

Step 3

x-2y=-19

-3-2y=-19

-2y=-19+3

-2y=-16

y=-16/-2

y=8

Step 4 (-3,8)

Elimination Using Subtraction

3x+4y=18

-2y+4y=8

Step 1

(3x+4y=18)-(-2y+4y=8)

(3x+4y=18)+(2y-4y=-8)

Step 2

5x+0=10

5x=10

x=2

Step 3

-2x+4y=8

-2(2)+4y=8

-4+4y=8

4y=8+4

4y=12

y=12/4

y=3

Step 4 (2,3)

 

Elimination Using Multiplication First

2x+y=3

-x+3y=-12

Step 1 multiply each term in the second equation by 2 to get opposite x coefficients.

(2x+y=3)+2(-x+3y=-12)

(2x+y=3)+(-2x+6y=-24)

Step 2

7y=-21

y=-3

Step 3

2x+y=3

2x+(-3)=3

2x=6

x=3

Step 4 (3,-3)

 

Real World Application

Sam spent 24.75 dollars to buy 12 flowers for his mother. The bouquet contained roses and daisies. How many of each type of flower did sam buy?  

Write a system.Use r for the number of roses and d for the number of daisies.

2.50r+1.75d=24.75 cost of roses and daisies  

r+d=12  total number of roses and daisies   

Step 1

2.50r+1.75d=24.75+(-2.50)(r+d=12)  

2.50r+1.75d=24.75+(-2.50r-2.50d=-30.00)

Step 2

-0.75d=-5.25

d=7

Step 3

r+d=12

r+7=12

r=5

Step 4  Sam can buy 5 roses and 7 daisies.

(5,7)

Chapter 5 Exercise

Lesson 2:

1)

y=5x-10

y=3x+8

2)

3x+y=2

4x+y=20

3)

y=x+5

4x+y=20

4)

y=x+3

y=2x+4

5)

y-4x=3

2x-3y=2

Lesson 3:

1)

-x+y=5

x-5y=-9

2)

-5x+7y=11

-5x+3y=19

3)

5x+y=0

5x+2y=30

4)

-2x+y=-20
2x+y=48

5)

-x+y=-1

2x-y=0

Chapter 5 answers:

Lesson 2:

1) (9,35)

3) (3,8)

5) (-3,-9)

Lesson 3:

1) (-4,1)

3) (-6,30)

5) (-1,-2)










5-3 Solving System by Elimination