1
huge thank you for all the people
who went through this book ,and
espacailly to miss urooj for helping
us in this project,and for
all the publishers that helped us.
chapter 1:-
1-1 variables and expressions
1-2 solving equations by adding or subtracting
1-3 solving equations by multiplying or dividing
chapter 2:-
2-1 graphing and writing inequalities
2-2 solving inequalities by adding or subtracting
2-3 solving inequalities by multiplying or dviding
chapter 3:-
3-1 graphing realationships
3-2 relations and functions
3-3 writing functions
chapter 4:-
4-1 identifying linear functions
4-2 using intercepts
4-3 rate of change and slope
chapter 5:-
5-1 solving systems by graphing
5-2 solving systems by substitution
5-3 solving systems by elimination 2
1-1
Variables and expressions
A variable is a letter or a symbol used to represent a value that can change.
A constant is a value that does not change
A numerical expression may contain only constants and/ or operations.
A algebraic expression may contain variables,constants and/or operations
In order to translate between algebraic expressions and words,you need to know how to write the algebraic expression in words.
+ is also known as plus, increased by, added to, sum of , more than ,etc
- is also known as minus, difference, decreased by, less than etc.
* is also known as times product, multiplied by, etc.
÷ is also known as divided by, quotient, etc
Example:
M+5 in words is m plus 5 or the sum of m and 5 or m increased by 5 or m more than 5
M-5 is m minus 5 or the difference of m and 5 or m decreased by 5 or m less than 5
M * 5 is m times 5 or the product of m and 5 or m multiplied by 5
m÷ 5 is m divided by 5 or the quotient of m and 5
To translate words to algebraic expressions look for words than indicate the action
+ = put together, with ,and, combined, etc
- = separated into,then, find how much more or less.
* = put together, per, equal groups
÷= separated into equal groups
Example
Ahmed ran 5 miles per day.write an expression for the number of miles ahmed ran in d days
5 * h or 5h
Guy was 2 inches taller than dude, who is y years old.
y-2 = dudes age
Brad bought 5 apples,and then bought a apples more. Write an expression for the amount of apples brad bought.
5+a
Jack read 5 pages every minute. Write an expression for the amount of pages jack wrote.
To evaluate algebraic expression all you need to do is substitute the numbers to the variable
Example
Evaluate x=4y=5v=6
1+X+4+y+v
1+4+5+6
16
1-1
Variables and expressions
A variable is a letter or a symbol used to represent a value that can change.
A constant is a value that does not change
A numerical expression may contain only constants and/ or operations.
A algebraic expression may contain variables,constants and/or operations
In order to translate between algebraic expressions and words,you need to know how to write the algebraic expression in words.
+ is also known as plus, increased by, added to, sum of , more than ,etc
- is also known as minus, difference, decreased by, less than etc.
* is also known as times product, multiplied by, etc.
÷ is also known as divided by, quotient, etc
Example:
M+5 in words is m plus 5 or the sum of m and 5 or m increased by 5 or m more than 5
M-5 is m minus 5 or the difference of m and 5 or m decreased by 5 or m less than 5
M * 5 is m times 5 or the product of m and 5 or m multiplied by 5
m÷ 5 is m divided by 5 or the quotient of m and 5
To translate words to algebraic expressions look for words than indicate the action
+ = put together, with ,and, combined, etc
- = separated into,then, find how much more or less.
* = put together, per, equal groups
÷= separated into equal groups
Example
Ahmed ran 5 miles per day.write an expression for the number of miles ahmed ran in d days
5 * h or 5h
Guy was 2 inches taller than dude, who is y years old.
y-2 = dudes age
Brad bought 5 apples,and then bought a apples more. Write an expression for the amount of apples brad bought.
5+a
Jack read 5 pages every minute. Write an expression for the amount of pages jack wrote.
To evaluate algebraic expression all you need to do is substitute the numbers to the variable
Example
Evaluate x=4y=5v=6
1+X+4+y+v
1+4+5+6
16
1-2
Solving equations by adding or subtracting
To solve equations with variables you need to inverse operations
Example
x-15=50
+15 +15
x =65
x=65
Check
65-15=50 correct
x-5=8
+5 +5
x =13
x=13
Check
13+5=8 correct
This also works with word problem solving.
Example
In 1673, the hope diamond was reduced from its original weight by about 45 carats. Resulting in the diamond to be about 67 carats.write and solve an equation to show how many carats the original diamond weighted.
Original weight reduced from the cut-out weight in new weight
a - 45 = 67
a-45=67
+45 +45
a =112
The original weight is 112 carats
1.3
Solving equations by multiplying or dividing
Just like the previous lesson, you can solve an equation by finding the value of the variable. You can find the variable by inversing operations
Example
x*8=5
÷8 ÷8
x = 0.625
Check
0.625*8=5 correct
It can also work on division.
Example
x÷8=5
*8*8
x =40
Check 40÷8=5 correct
Also works with fractions
3
4 x= 10
( 4 ) 3 ( 4 )
( 3 ) 4 x=( 3 ) 10
x= 13.33
1-2
Solving equations by adding or subtracting
To solve equations with variables you need to inverse operations
Example
x-15=50
+15 +15
x =65
x=65
Check
65-15=50 correct
x-5=8
+5 +5
x =13
x=13
Check
13+5=8 correct
This also works with word problem solving.
Example
In 1673, the hope diamond was reduced from its original weight by about 45 carats. Resulting in the diamond to be about 67 carats.write and solve an equation to show how many carats the original diamond weighted.
Original weight reduced from the cut-out weight in new weight
a - 45 = 67
a-45=67
+45 +45
a =112
The original weight is 112 carats
1.3
Solving equations by multiplying or dividing
Just like the previous lesson, you can solve an equation by finding the value of the variable. You can find the variable by inversing operations
Example
x*8=5
÷8 ÷8
x = 0.625
Check
0.625*8=5 correct
It can also work on division.
Example
x÷8=5
*8*8
x =40
Check 40÷8=5 correct
Also works with fractions
3
4 x= 10
( 4 ) 3 ( 4 )
( 3 ) 4 x=( 3 ) 10
x= 13.33
2-1 writing inequalities
Inequality;is a statement that two things aren't equal,to make a solution is to find any value that make the inequality true.
Objective:get to know the signs of inequality.
> greater than
<less than
≥is greater than or equal to
≤is less than or equal to
3
4
2-2Solving inequalities by adding or subtracting
Objectives:how to solve inequalities using addition
and subtraction.
Example 1:-
Using addition to solve inequalities:-
x + 6<9 since x is added to 6 , we have to subtract 9 , to remove the addition.
9-6<x
3<x
Example 2:-
Using subtraction to solve inequalities:-
D-5 >-3 since d is subtracted to the -5 , we have to add the -3,to remove the subtraction
5+3>d
8>d
4
2-3Solving inequalities by multiplying or dividing
Objectives:how to solve inequalities by
using multiplication and division
Multiplying or dividing by a positive number:-
Example 1:-
2/4r<5 since r is multiplied by 2/4 , multiply 2/4 with 5 ,then simplyfy to get the answer
10/20
r<2
Multiplying or dividing by a negative number:-
Example2:-
-45x>5 since x is multiplyed by -45 ,divide -45 from 5 ,and flipthe sign because its negitive
-45x<5
x<-9
4
Fact: Graphs can be used to illustrate many different situations
Key Words |
Segment Description |
Was Constant |
Horizontal |
Rose steadily |
Slanting Upward |
Stayed the same |
Horizontal |
Dropped Sharply |
Slanting downward |
3-2 Relations and Functions
A Relation is a set of ordered pairs. Every Relation has a domain. A domain is the set of first coordinates of the ordered pairs. Every Relation also has something called, range. The range is the set of second coordinates of the order pairs. In that case, the domain can be the x-values and the range would be the y-values. A Function is a special type of relation; that pairs each domain value with exactly one range value.
Example (1,3), (2,5), (3,6), (4,4)
Track |
Score |
Place |
Point |
1 |
3 |
2 |
5 |
3 |
6 |
4 |
4 |
MAPPING DIAGRAM
3-2 Writing Functions
Any algebraic expression that defines a function is called a Function Rule.
EXAMPLE:
X |
1 |
2 |
3 |
4 |
Y |
6 |
7 |
8 |
9 |
First list the possible relationships between the x and y values.
Then, determine if one relationship works for the remaining value.
Write an equation.
1+5=6
2+5=7
3+5=8
4.2
Using Intercepts
Always remember that there are the y-intercepts and the x-intercepts. The y-intercept being the y- coordinate of the point where the graph intersects the y-axis, and the x-intercept being the x-coordinate of the point where the graph intersects the x-axis.
Example 1: Finding Intercepts
5x - 4y = 25
To find the x-intercept, replace y with 0 and solve for x.
5x - 4(0) = 25
5x - 0 = 25
5x = 25
5x = 25
5 5
X = 5 The x-intercept is 5.
5x - 4y = 25
5(0) - 4y = 25
0 - 4y = 25
-4y =25
-4y = 25
4 4 Y = 6.25. The y-intercept is 6.25.
Example 2: Travel Application
The New York subway runs 1600 meters from Time Square to the Empire State Building. It’s travels at a speed of 200 meters per minute. The function f(x) = 1600 - 200x gives the subway’s distance in meters of the Empire State Building after x minutes. Graph this function and find the intercepts. What does each intercept represent.
Neither time nor distance can be negative, so choose several non negative values of x. Use the function to generate ordered pairs.
X |
0 |
2 |
4 |
6 |
8 |
f(x) = 1600-200x |
1600 |
1200 |
800 |
400 |
0 |
Graph the ordered pairs. Connect the point with a line.
Y- intercept: 1600.
This is the starting distance from Time Square. (time = 0)
X-intercept: 8.
This is the time when the subway reaches empire state.
(distance = 0)
Example 3: Graphing Linear Equations by Using Intercepts
Use intercepts to graph the line described by each equation.
4x - 8y = 16
Step 1) Find the intercepts. x-intercept y-intercept
4x - 8y = 16 4x - 8y = 16
4x - 8(0) = 16 4(0) - 8y = 16
4x = 16 -8y = 16
4x = 16
4 4 -8y = 16
X = 4 8 8
Y = -2
Step 2: Graph the Line.
Plot (4, 0) and (0, -2)
Connect with a straight line.
Example 3: Graphing Linear Equations by Using Intercepts
Use intercepts to graph the line described by each equation.
4x - 8y = 16
Step 1) Find the intercepts. x-intercept y-intercept
4x - 8y = 16 4x - 8y = 16
4x - 8(0) = 16 4(0) - 8y = 16
4x = 16 -8y = 16
4x = 16
4 4 -8y = 16
X = 4 8 8
Y = -2
Step 2: Graph the Line.
Plot (4, 0) and (0, -2)
Connect with a straight line.
The slope formula
The slope formula is m= y1 - y2
X1 - x2
So if the points are (1,2) and (2,6), y1 will be 2 and y2 will be 6.(1,2)(2,6)
x1,y1 x2,y2
So solve m= 2-6
1-2
So the slope will be -4
Remember, a slope is rise
run
So the slope of (1,2) and (2,6) is -4
1
You can also find the slope on graphs and tables.
let (-1,-2) be x1,y1 and (2,2) be x2,y2
So the equation will be m -2-2
-1-2 so the slope is 1.33
Also in graphs
x |
1 |
3 |
5 |
y |
2 |
5 |
6 |
Pick any 2 points like (1,2) and (3,5)
And do the same equation m=2-5 1-3 so the slope is 1.5
A direct variation is a special type of linear relationship that can be written in a form y= kx, where k is a nonzero constant called the constant of variation.
Identifying Direct Variation from Equation
Example 1;
y= 8x The constant of variation is 8.
b) -9x +15y = 0
-9x + 15y= 0
+9x + 9x
15y = 9x y= 9 x
15 15 15
What happens if you have to solve for k?
y= kx
y = kx
X x
Y = k
X
Example 2: Identifying Direct Variations from Ordered Pairs
x |
1 |
2 |
3 |
y |
12 |
24 |
36 |
Method 1: Write an equation.
y= 12x Each y-value is 12 times the corresponding x-value.
This is a direct variation because it can be written as y= kx, where k = 12.
Method 2: Find y for each ordered pair.
X
12 = 12 24 = 12 36 = 12
1 2 3
Example 3: Writing and Solving Direct Variation Equations
The value of y varies directly with x, and y= 12 when x = 24. Find y when x =37.
Method 1: Find the value of k and then write the equation.
Y = kx
12 = k(24)
1 = k The equation is y= 1 x. When x = 37, y = 1(37) = 18.5
2 2 2
4.5
5.1
Solving systems by graphing
To tell whether the ordered pair is a solution, you need to substitute the x and y with the variables.
Example
(4,2);x+4y=12
5y+7x=34
So you need to substitute the x and y from (4,2) to the 2 equations. So your equations will be like:
4+4(2)=12
5(2)+7(4)=34
Now simplify
4+8=12 correct
10+24=34 correct
(4,2) makes both equation true, (4,2) is a solution of the system
Another example:
(4,2);x+4y=15
5y+7x=40
Simplify
4+4(2)=15
5(2)+7(4)=40
4+8=15 incorrect
10+24=40 incorrect
The point(4,2) didn't solve both equations correct. (4,2)is not a solution of the system
also , both equations must be correct, it not then the point isn't a solution.
You can also find the solution by a graph
To find the solution for 2 linear equations, you can find the point by plotting them
Example
The 2 linear equations are
{y=x-3 }
{y=-x-1}
So the graph should be like this:
The point where they intersect is the solution.
Checking
(1,-2)
-2=-x-3 y=-x-1
-2=-(1)-3 -2=-(1)-1
-2=-2 correct -2=-2 correct
The points (1,-2) solved the equations, the points(1,-2) are the solutions to the system.
5.2
5.2
Solving Systems by Substitution
Simpler Solving Systems of Equations 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 1st variable. |
Step 4 Substitute that value into one of the original equations and solve. |
Step 5 Write the values from Steps 3 and 4 as an ordered pair, (x,y), and check. |
____________________________________________________________
Solving a System of Linear Equations by Substitution
Solve each system by substitution.
A
{y = 20x
{y = x + 5
Step 1 Write the equation given to you without the brackets.
y = 20x
y = x + 5
Step 2
y = x + 5
20x = x + 5 Substitute 2x for y in the second equation.
Step 3 Solve for x.
20x = x + 5 Solve for x.
-x -x
x= 5
Step 4
y= 20x Write one of the original equations.
y= 20(5) Simplify
y= 100
Step 5
(5, 100) Write the solution as an ordered pair.
____________________________________________________________
Using the Distributive Property
Solve by Substitution
{2y- 5x =3
{x- 4y = 7
Step 1
x- 4y = 7 Solve the 2nd equation for x by moving 4y to the other side.
x = 7 + 4y
x= 7 + 4y
Step 2
2y- 5x =3 Substitute 7 + 4y for x in the 1st equation.
2y- 5(4y + 7) =3
Step 3
2y- 5(4y) - (7) =3 Distribute 5 to the expressions in the parentheses.
2y - 20y - 35 =3 Simplify
2y - 20y = 3 + 35 Move like terms to each other's own sides.
-18y = 38 Simplify further.
y= 2.1
Step 4
x- 4y = 7 Write one of the original equations.
x -4(2.1) = 7 Substitute 2.1 for y.
x-8.4 =7 Simplify
x= 7 + 8.4
x= 15.4
Step 5 Write the solution as an ordered pair.
(15.4, 2.1)
____________________________________________________________
5.3
Solving systems by elimination
Solving systems by eliminations by following these steps
Solving systems of equations by elimination |
1. Write the systems so that the terms are aligned |
2. Eliminate one of the variables and solve for the other variable |
3. Substitute the value of the variable into one of theof the original equations and solve for the other variable |
4. Write the answers from step 2 and 3 as an ordered pair(x,y) and check |
Use these steps to solve the system
with addition
Example
{x-3y=-19
{6x+3y=1
x- 3y = -19←Step 1
+6x+3y= 1
7x +0= -18← step 2
7x= -18
7x= 7x
x= -2.57
x-3y=-19←step 3
-2.57-3y=-19
+2.57 -19
-3y=-19s
-3 -3
y= 6.33 (2.57,6.33)← step 4
Solving using subtraction
{x-3y=25
{-5x+3y=1
x-3y=-19← step 1
-(-5x+3y=1)
x-3y=25
+5x-3y=-1
6x+0=24← step 2
6x =24
6 6
x=4
x-3y=25← step 3
4-3y=25
-4 =-4
3y=21
y=7
(4,7)← step 4