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 12,13,14 and 52,53,55
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
Objective : solve onestep 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 x2=12
x=12+2
x=14
Since 2 is subtracted from x, add 2 on the other side
B m½=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=72
x=5
Since 2 is added to x, subtract 2 on the other side
B y+0.5=0.10
y=0.100.5
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=22015
r=205 beats per minute
Chapter 1 Equations
Objective: Solve onestep 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
Objective: Solve equations in one variable that contain more than one step
Examples with Explanation
Solving TwoStep Equations
42x=10
2x=104
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 TwoStep Equations That Contain Fractions
a/141/4=2/4
Method 1 use fraction operation
a/141/4=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+28x=12
4x+28x=12
4x8x+2=12
4x+2=12
4x=122
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=1412
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:
17=w4
s5=3
x3.9=12.4
t+5=25
9=s+9
b+⅔=2
⅚+p=2
1.75=k0.75
⅙+h=⅙
5.2+a=8
Lesson 3:
k/4=8
z/3=9
2=w/7
84=12a
4m=10
9=3/8r
x/2=12
49=7c
5t=15
52=4c
Lesson 4:
4a+3=11
8=3r1
42=2d+6
1/3y+¼=5/12
x/6+4=15
5=2g+1
3x+3=18
2(x+3)=10
5(h4)=8
6=2(7c)
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
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 TwoStep 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
2406v>180
350+6v<290
43205v>700
Lesson 5:
1x<8x+28
2x>10x36
3x<16x+105
4x>5x40
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


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
graph with negative correlation
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=106/19801975=⅘=0.8=0.8cents/ year
1980 to 1985 change in cost/change in years=1210/19851980=⅖=0.4=0.4cents/ year
1985 to 1990 change in cost/change in years=1312/19901985=⅕=0.2=0.2 cents/year
1990 to 1995 change in cost/change in years=2213/19951990=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
Tell whether the slope of each line is positive, negative, zero, or undefined.
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.
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
43
SLOPE: Is the comparison of the rise, and the run for any two points on the line.
Slope=y2y1/x2x1
Finding Slope by Using the Slope Formula.
(4,6) and (9,3)
M=y2y1/x2x1 use the slope formula
M=36/94 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=y2y1/x2x1 use the slope formula
m= 59/79 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=y2y1/x2x1 use the slope formula
=0.52/02 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=y2y1/x2x1 Use the slope formula
=16/38 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 4x9y= 36
Step 1 Finding the x intercept
4x9y= 36
4x9(0)=36 substitute y for 0
4x=36
4x/4=36/4
x=9
Step 2 Finding the y intercept
4x9y=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=y2y1/x2x1=40/09=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 (59)
3 (1,8) and (0, 4)
Find the slope of both the graph and the table
The Slope Formula
44
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=12y
4x4y=20
Find y
4x4y=20
4(12y)4y=20
48y4y=20
412y=20
12y=20+4
12y=24
y=24/12
y=2
Find x
x+2y=1
x+2(2)=1
x4=1
x=1+4
x=3
(3,2)
Real World Example
1) One highspeed 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.
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
52 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
x2y=19
5x+2y=1
Step 1
(x2y=19)+(5x+2y=1)
Step 2
6x+0=18
6x=18
x=18/6
x=3
Step 3
x2y=19
32y=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)+(2y4y=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+(2.50r2.50d=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=5x10
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)
y4x=3
2x3y=2
Lesson 3:
1)
x+y=5
x5y=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
2xy=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)
53 Solving System by Elimination