The Book of EXTREME Algebra




Authors of this book: Leanne Alhabib, Ali

 

 

Radwan, Raneem Ramadan, and Fajar

 

Adeel.



 

Published By: Almighty Green Mountain

 

Publishing Company



 

Copyright  © 2016 by Almighty Green

 

Mountain Publishing Company



 

 

h

 

 

 

 

 

This book (The Book of Extreme

 

 

Algebra), is best believed to fit in

 

 

for the age group of 12-15. It

 

 

 could be helpful for students from

 

 

grades 7-9.

 

 

Table of Contents:

 

Unit 1: Equations

 

 

Topics:

 

➢ 1-1 Variables and Expressions …...pg no:

 

➢ 1-2 Solving Equations with Variables on Both Sides…...pg no:

 

➢ 1-3 Solving for a Variable…...pg no:

 

 

Unit 2: Inequalities

 

Topics:

 

➢ 2-1 Solving two- Step and Multistep Inequalities…...pg no:

 

➢ 2-2 Solving Inequalities with Variable on Both Sides…...pg no:

 

➢ 2-3 Solving Compound Inequalities…...pg no:

 

 

 

Unit 3: Functions

 

Topics:

 

➢ 3-1 Relations and Functions…...pg no:

 

➢ 3-2 Writing Functions…...pg no:

 

➢ 3-5 Arithmetic Sequence…...pg no:

 

 

 

 

 

 

 

 

Unit 4: Linear Functions

 

 

 

 

 

Topics:

 

➢ 4-2 The Slope Formula…...pg no:

 

➢ 4-3 Slope Intercept Form…...pg no:

➢ 4-4 Point- Slope Form…...pg no:

 

Unit 5: System of Equations and Inequalities

 

 

 

Topics:

 

➢ 5-1 Solving systems by Graphing…...pg no:

 

➢ 5-2 Solving systems by Substitution…...pg no:

 

➢ 5-3 Solving systems by Elimination…...pg no:



Unit 1:

 

Equations

 

Objective:

To translate between words, Algebraic expressions, and Numerical Expressions.

 

And to evaluate Algebraic/Numerical Expressions.

 

 

Vocabulary:

Variable: a letter or symbol able to be changed

 

Constant: a value which doesn’t change

 

Numerical Expression: an expression which may contain constants or operations

 

Algebraic Expression: an expression which may contain variables, constants, or operations



+ Increased By, Sum, Plus

 

- Decreased By, Difference,  Minus

 

x Times, Product, Equal Groups of

 

/ Divided By, Quotient

 

1-1 Variables and Expressions

Examples with explanation:

 

Give two ways to write each algebraic expression in words.

 

A   x+6                                      B  m-9

The sum of x and 6                   The difference of m and 9

X increased by 6                       9 less than m

C   7xy                                     D   s/4

7 times y                                  divided by 4

The product of 7 and y           The quotient of s and 4

 

Translate from words to algebraic expressions

 

A  Ali runs one km in 19 minutes. Write an expression for the number of miles that Ali can run in m minutes.

m represents the total time Ali runs.

m/19

 

 

B  John is 4 years younger than Mike, who is y years old. Write an expression for John’s age.

y represents Mike’s age.

y-4




Real world application:

 

Ahmed drives at 95 km/h. Write an expression for the number of kilometers that Ahmed drives in t hours.

t represents the number of hours that Ahmed drives.

95 x t or 95t

1-1 Variables and Expressions

 

Try it:

 

Write each as an algebraic expression.



1) the difference of 10 and x

 

 

 

2) the quotient of j and 7

 

 

 

3) x increased by 6





Write each as a verbal expression.

 

 

8) j − 3

 

 

 

9) 5n

 

 

 

10) n ⋅ 6

 

1-1 Variables and Expressions

1-2 Solving Equations with Variables on Both Sides

Objective: To be able to solve equations in one variable which contain variable terms on both sides.

 

Vocabulary:

 

Identity: An equation that is true no matter what values are chosen.

Example: a/2 = a × 0.5 is true no matter what value is chosen for a.

 

Examples with explanation:

 

Solve each equation.

A   9x+7= 4x-8                                                    Combine like terms. Reverse the signs.

         9x-4x= -8-7                                                 Calculate.

         5x= -15                                                        Divide -15 by 5

         x= -15/5

         x= -3

 

Check: Substitute -3 for x in the original equation.

9(-3)+7=4(-3)-8

-27+7=-12-8

-20=-20 ✓

 

B  9k-7=6k+5                                                 Combine like terms. Reverse the signs.

   9k-6k=5+7                                                   Calculate.

   3k=12                                                           Divide 12 by 3

   k=12/3

   k=4

 

Check: Substitute 4 for k in the original equation.

9(4)-7=6(4)+5

36-7=24+5

29=29 ✓

 

 

1-2 Solving Equations with Variables on Both Sides

To solve more complicated equations, you need to simplify by using the Distributive Property and combining like terms.

 

Solve each equation.

 

A 7(z-8)=8z                 Distribute 7 to the expression in parentheses.

   7z-7(8)=8z              

   7z-56=8z                                      Combine like terms.

   -56=8z-7z

   -56=z

 

Check: Substitute -56 for z in the original equation.

 

7(-56-8)=8(-56)

-392-56=-448

-448=-448 ✓

 

B 9-8b+6=4(6-b)        Distribute 4 to the expression in parentheses.

  9-8b+6=4(6)-4b

   9-8b+6=24-4b                         Combine like terms.

   9+6-24=-4b+8b

   -9=4b

   -2.25=b

 

Check: Substitute -2.25 for b in the original equation.

 

9-8(-2.25)+6=4(6+2.25)

9+18+6=24+9

33=33 ✓

 

 

 

1-2 Solving Equations with Variables on Both Sides

Real world application: Michael has job offers at two companies. One company offers a starting salary at $28,000 with a raise of $3,000 each year. The second company offers a starting salary of $36,000 with a raise of $2,000 each year.

  1. After how many years would Michael’s salary be the same with both companies?

28,000+3,000x=36,000+2,000x

3,000x-2,000x=36,000-28,000

1,000x=8,000

x=8,000/1,000

x=8

Answer: 8 years.

  1. What would that salary be?

36,000+2,000(8)

36,000+16,000=

52,000

Answer: $52,000

Try It:

Solve each equation.

 

1) −20 = −4x − 6x

 

2) 6 = 1 − 2n + 5

 

3) 8x − 2 = −9 + 7x

 

4) a + 5 = −5a + 5


5) 4m − 4 = 4m

 

Objective: To solve the formulas for the variables given.

 

Vocabulary:

 

Formula: A special type of equation that shows the relationship between different variables.

Literal Equation: An equation where variables represent known values.

 

Examples with Explanation:

 

A The formula for the area of a triangle is A=1/2bh, where b represents the length of the base and h represents the height.

 

  1. Solve the formula A=1/2bh for b.

A/h x 2/1=b

Answer: b=2a/h

 

B If a triangle has an area of 192 mm2 , and the height measures 12 mm, what is the measure of the base?

b=2(192)/12

 

b=384/12

b=32mm

Answer: 32mm

1-3 Solving for A Variable

Real World Application:

 

The formula c=5p+215 relates to c, the total cost in dollars of hosting a birthday party at a skating rink, to p, the number of people attending.

  1. Solve the formula c=5p+215 for p.

c-215=5p

c-215/5=p

Answer: c-215/5

 

Try it:

 

Answer each of the following:

1. The formula y=mx+b is slope-intercept form of a line. Solve this formula for m.

2.The formula K=C+273 is used to convert temperatures from degrees celsius to kelvin. Solve this formula for C.

 

3. -10=xy+z for x



4. h-4/j=k for j



5.2p+5r=q for p

1-3 Solving for A Variable

Write each as an algebraic expression.

1) the sum of q and 8                                          2) the product of t and 12

 

Write each as a verbal expression.

 

3) x/2                                                                  4) a + 9

 

5) j − 3

 

Solve each equation.

 

6) −8 = −(x + 4)                                                 7) 12 = −4(−6x − 3)

 

8) 14 = −( p − 8)

 

9) −18 − 6k = 6(1 + 3k)

 

10) 5n + 34 = −2(1 − 7n)



11) The formula y=mx+b is slope-intercept form of a line. Solve this formula for m.



12) The formula K=C+273 is used to convert temperatures from degrees celsius to kelvin. Solve this formula for C



13)-10=xy+z for x                                    14) h-4/j=k for j



15) 2p+5r=q for p

 

Unit 1 Review

 

   Unit 2:

 

   Inequalities

 

Objective: Solve inequalities that contains more than one operation.

 

Vocabulary:

Inequality: is a sign that could show whether two numbers are greater, less, equal, or nor equal to.

2-1: Solving two- Step and Multistep Inequalities

Examples with explanation: 

A) Solving if it were positive:

129+2d≤ 570

129+2d< 570    Since 129 is adding to 2d, subtract 129 from both sides.

     -129    -129

         2d< 441             Simplify

         2d< 441

         2       2               Divide both sides by 2.

         d<220.5             Solve

Graph:                    
 
Since it is only less than, point isn’t filled. It isn’texact.

 

 

B) Solving if it were negative:

 

8-3s≤2

       8-3s≤22      Since 8 is subtracting to 3s, subtract 8 from both sides.

      -8      -8      

      -3s≤14                  Simplify

       -3s14                 Flip the symbol since it is a negative.

       -3    -3                 Divide both sides by -3.

        s≥-4.6                Solve

Graph:

 

        Since it is greater than or equal to, the point is filled.

 

 

 

C) 4(2-r)>122      

       4(2-r)>122   Start by solving the parentheses. Distribute 4 on both (constant and variable) inside the parentheses.

            4*2-4r>122         Multiply.

             8-4r>122

            8-4r>144            Solve the exponent.

           -8         -8            Subtract both sides by 8.

            -4r>136

            -4 r<136             Flip the symbol since it is a negative.

            -4      -4              Divide both sides by -4.

              r<-34                Solve

Graph:

Since it is greater than or equal to, the point is filled.

Real world application:

The competition is on, and everyone wants to show off their heavy, large watermelons in order to win the medal. The average weight of Marcus’s two watermelons must be greater than 48 kgs. One of his watermelons weighs 21kgs. What is the least number of kgs could his second watermelon weigh in order to win the medal.

 Let m represent the weight of his second watermelon. The average weight of the w  atermelon is the sum of each weight divided by 2.

 

 

(21 plus m)divided by2must be greater than48.

(21   +   m)        /        2                >               48

 

 

 

 

Since 21+m is dividing by 2, multiply both sides by 2 to undo the division.

21+m>96

-21      -21Since 21 is adding to m, subtract 21 from both sides to

                   undo the addition

     m> 75

 

Try it:

Solve and show graphing.

1) 18-2h>26                    2)   6(7+y)≤18                        3) 4+8v< 17  

                    4) 15-6i>28                        5)43+(2-e)>22

 

 2-2: Solving Inequalities with Variable on Both Sides

 2-2: Solving Inequalities with Variable on Both Sides

Objective: Solve inequalities that contain variable terms on both sides.

Vocabulary:

Like terms:two or more variables which are found on both sides.

Examples with explanation:

  1. x<2x+6

     x<2x+6   To collect the variable terms on one side, subtract x from both sides.

       -x   -x

     0< x+6  Since 6 is added to 1x, subtract 6 from both sides to undo the addition.

     -6       -6

     -6 <1x   

      1       1Since x is multiplying by 1, divide both sides by 2 to undo the multiplication.

 -6<x (or x>-6)

Graph:

 

 

B) 7(2-x)<4x

 7(2)-7(x)<4x                Distribute 7 on the left side of the inequality.

        7-7x<4x

          +7x  +7x             Add 7x to both sides do that the coefficient of x is positive.

         7     < 11x

         7<11x

        11    11      Since x is multiplied by 11, divide both sides by 11 to undo the multiplication.

         .63<x (or x>.63)

Graph:

 

 

Real world application:

The Highlights charges a fee of $600 plus $55 per week to run an ad. The Kids charges $167 per week. For how many weeks will the total cost at Highlights be less expensive than the cost at Kids?

   Let w be the number of weeks the ad runs in the paper.     

      

 

 

 

Highlights   

 

$55

 

number

  is less

  Kids  

 

number

    fee

+

 per

   *

   of

expensive

charge

   *

   of

   

week

 

weeks

   than

per week

 

weeks

   $600

  +

$55

  *

   w  

     <  

  $167

  *

   w

600+55w<  167w

       -55w   -55w             Subtract 55w from both sides.

600         <  112w

600 < 112w Since w is multiplying by 112, divide both sides by 112 to undo the multiplication.112      112

5.4<w (or w>5.4)

The total cost at Highlights is less than the cost at Kids if the ad runs for more than 5.4 weeks.

 

Try it:

1) 6(7+x)< 4(3+x)          2) -5(2-y)> 4(p+1)              3) 4k+6> 2(y+18)   

                     4) 6e-2<8(r+12)               5) 7(2-z)>2(4+x)

 

 

2-3:Solving Compound Inequalities

 

Objective:Solve compound inequalities in one variable. Graph solution sets of compound inequalities in one variable.

Vocabulary:

Compound Inequalities: When 2 simple inequalities are combined into one statement by the words AND or OR.

Intersection: An overlapping region.

Union: Combined regions. Shows the numbers that are solutions of either inequalities.

 

Examples with explanation:

A) 5<x+4<7

   5< x+4 AND x+4<7               Write the compound inequality using AND

  -4     -4              -4  -4              Solve each simple inequality.

  1 < x      AND  x < 3

 

     

B) 4x<8 OR 2x>10

   4x<8 OR 2x>10

    4x<8     2x > 10              Solve each simple inequality.

     4    4       2      2

     x< 2  OR  x > 5

   

 

 

 

 

 

 

Graph x<2

Graph x>5

 

Graph the union by combining the regions.

 

 Graph 1< x 

Graph x < 3

 

Graph the intersection by finding where the two graphs overlap.

 

Real world application:

The candy analyst recommends that a lollipop should have between 15g and 20g of sugar. Write a compound inequality to show the amount of sugar grams between the recommended range. Graph solution.

 

 

 

 

 

 

 

15   is less than or equal to   amount of sugar   is less than or equal to   20

15                   <                                s                               <                   20

15< s < 20




Try it:

1) 4<x +3<8                     2) 14<3x+8<22              3) r-2< 4 OR r-6>2     

               4) f+3<5 OR f-3>2                   5) 5<x-5<4

 

 

Chapter 2 Exercises:Questions

 

Lesson 2.1:

Solve and show graphing.

1) 167+5t<600                            2) 10-3t<14

3) 5+10r>4                                  4) 15>22+3(q-6)

5) 73<5(x+5)

 

Lesson 2.2:

Solve and show graphing.

1) 10(5+x)<4(3+x)                          2) 2b-6>b+7

3) 4(x-6)>4x                                   4) 4x-5<14x

5) 5(x-6)<13x  

 

Lesson 2.3:

Solve and show graphing.

1) -5<x+3<8                                  2) 6< 5x+7<22

3) 15<4x+5<19                             4) r-6<10 OR r-2>4

5) t+7<15 OR t+4>14

 

 

Unit 3:

 

Functions

 

3-1: Relations and Functions

 

 

Objective:

  • You will be able to identify functions.

  • Find the domain and range of relations and functions.



Vocabulary:

Relation:Relationships can also be represented by set of ordered pairs.

Domain:The domain of a relation is the first coordinates or X-values or the ordered pair.

Range:The range of a relation are the second coordinates or Y-values of the ordered pairs.

Function:A special type of relation that pairs each domain value with exactly one range value.

Examples with explanation:

 

Showing Multiple Representations Of Relation

 

Explanation about the relation  of track meet and scoring system,using this equation {(2,3),(3,2),(1,4),(4,1)},in three ways: as a graph,as a table,and as a mapping diagram.

 

 

Table

 

Track Scoring

Place

Points

2

3

3

2

1

4

4

1

Under “Place’’  write all X-Values and all Y-Value under“Points.’’





Mapping Diagram

Place Points

2     3

3      2

1      4

4      1

Under “Place’’  write all X-Values and all Y-Value under“Points.’’

Draw an arrow from each X-Value to its corresponding Y-Values.

 

Graph

Track Scoring

 

Use X- and Y-Values to plot the ordered pairs.

 

 

Finding the Domain and Range of the Relation

 

Give the domain and range of the relation.

 

 

 

 

 

  

 

All the x-values from 4 through 6 is The domain, inclusive.

All the y-values from 5 through 7 is the range,inclusive.

 

 

D:4 ≤ x ≤ 6    R:5 ≤ y≤ 7

Identifying Functions

Give the domain and range of the each relation.Tell whether the relation is a function.Explain.

A

Traveling

People X

Transportations Y

50

4

43

4

82

2



D:{50,43,82}       As you can see 4 is written twice in the table,

R:{4,2}                it is written only once  when writing the range.

This relation is a function.Each domain value is paired with exactly one range value.

 

 

B

Use the arrows to decide which domain values to each range value.

 

D:{2,6,15,13}

R:{-5,-6,0}

 

This isn’t a function.Here, each domain value does not have exactly one range value. The domain value 2 is paired with the range values -6 and 0.

 

Give the domain and range of the each relation.Tell whether the relation is a function.Explain.

C                                     
   

Like in the pictures above,draw lines to see the domain

 and range values.                                                                   

X

6

0

0

-6

Y

0

6

-6

0



 

 

 

To Compare domain and range values, make a table

 

using points from the graph.

 

This relation is not a function because the are several

 

domain values that have more than one range value.

 

For example, the domain value 0 is paired with both

 

6 and -6.

 

 

 

 

Try it:

 

1)Express the relation{(3,2),(2,5),(4,6)}as a table,as a graph,and as a mapping diagram.

 

2)

X

Y

1

3

2

4

3

5

 

3){(-2,3)(-7,5),(5,9)(-4,6)

 

4)Express the relation{(6,7),(8,9),(10,4)}as a table,as a graph,and as a mapping diagram.

 

5)

X

Y

5

7

3

6

1

8



 

3-2:Writing Functions

 

 

Objective:

 

Will be able to write an equation using a table.

     

Vocabulary:

 

Independent Variable:Is an input of a function.

 

Dependent variable:Is an output of a function.

 

Function Rule:Is an algebraic expression that defines a function.

 

Function Notation:If x is the independent variable and y is the

 

 

dependent variable,then Function Notation for y is f(x),read “f of

 

x,”where f names the function.

 

 

Examples:

 

Using a Table to Write an Equation

 

Determine a relationship between the x - and y - values.Write an

 

equation.

 

X

2

3

4

5

Y

-2

-1

0

1

 

Step 1: List the possible relationships between the first x – and y –

 

value.

 

2 – 4= -2 or 2(-2) = -4

 

Step 2: Determine if one relationship works for the remaining values.

3 – 4 = - 1  ✔     3(-4 )-1  X symbol

4 – 4 = 0     ✔    4 (-4)0   X symbol

5 – 4 = 1     ✔   5 (-4)1   X symbol



Step 3: Write an equation.

 

Y = x - 4          The value of y is 4 less than x.

 

 

Try It:

 

 

1)Determine a relationship between the x- and y-values  in the relation{(3,2),(4,1),(7,8),(10,15).Write an equation.

 

 

2)Determine a relationship between the x- and y-values  in the relation{(3,7),(5,9),(3,4),(1,2).Write an equation.

 

 

3)Determine a relationship between the x- and y-values  in the relation{(9,2),(10,11),(5,12),(3,5).Write an equation.

 

 

4)Determine a relationship between the x- and y-values  in the relation{(12,13),(14,12),(17,8),(11,15).Write an equation.

 

 

5)Determine a relationship between the x- and y-values  in the relation{(13,19),(4,23),(2,5),(12,20).Write an equation.

 

3-5 Arithmetic Sequences

 

 

Objective:

Will be able to identify Arithmetic Sequence.

 

Vocabulary:

Sequence:List of numbers that may form a pattern.

Term:Is a sequence in each number.

Arithmetic Sequence:Is when terms in a sequence by a same non zero number.

Common Difference:Is that non zero number.

 

Examples:

Tell if the sequence is an arithmetic sequence.  If it is an arithmetic sequence , find a common difference with the next 3 numbers.

20,18,16,14…

Step 1:Find the difference between each number.

20,18,16,14…

-2   -2  -2  -2

Step 2:Use the common difference you found to find the next 3 numbers.

20,18,16,14…12,10,8

From this we have found out that for each number you have substract 2.

 

 

 

 B

 

Non Arithmetic Sequence

 

  

Find the difference between these terms.

 

 

20,26,34,38

 

 

+6 +8 +4

 

 

Its random,so its not an arithmetic sequence.

 

 

 

Try it:

 

 

1)-1,-3,-5,-7…

 

 

2)-6,-8,-10,-12…

 

3)-3,-6,-9,-12…

 

4)16,20,24,28...

 

5)10,15,20,25...

 

 

 Chapter 3 Review:

 

1)3,6,9,12

 

2)Express the relation{(12,2),(3,4),(2,9)}as a table,as a graph,and as a mapping diagram.

 

3)2,4,6,8….

 

4) 

 

 

 

 

 

 

 

 

 

 

 

 

5)5,10,15,20...

 

 

6)Determine a relationship between the x- and y-values  in the relation{(3,2),(4,1),(7,8),(10,15).Write an equation.

 

 

 

7)10,20,30,40...

 

 

 

 

 

 

 

 

 

 

11) 9,18,27,36..

 

12)

 

 

 

 

 

 

 

 

 

 

 

 

13)7,14,21,28..



14)

   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

15)1,2,3,4..

 

 

Unit 4:

 

 

Linear

 

 

Functions

 

4-2:The Slope Formula:

 

Objective: The objective of this lesson is to learn how to find the slope of a line.

 

Vocabulary:

1) Slope- The slope is a number that shows the steepness of any line.

 

Examples with explanation:

 

  1. This is the slope formula:

            m =  y2-y1

                    x2-x1

 

Example number 1: The line has the coordinates (2,4) and (4,6) so to find the slope, we should substitute these values in the formula.

                                                                                                                  x1 y1       x2 y2

  m =  y2-y1                                                                                           (2,4) and (4,6)

           x2-x1

 

 m =  y2-y1                     Substitute the values

          X2-x1

 

 m = 6-4                          Subtract the values

        4-4

 

m = 2

       0

 

m = 0     

So the slope in this example is equal to zero

Example number 2: The coordinates for this line are (9,7) and (10,5)

 

 m =  y2-y1                                                                                            x1 y1     x2 y2

         X2-x1                                                                                           (9,7) and (10,5)

 

m= 5-7                              Substitute the values

     10-9                            

                                                             Subtract the values

 

 

 m= -2   Or -2 only

        1   

 

m= -2   Or -2 only           

 

Real world application:

 

The Graph shows how much food is at a restaurant at different times. Find the slope of the line using the slope formula. Then tell what the slope represents.

 

 

              

 

                

                            You can choose any two points 

 

 

 

(10,2) and (20,4)

                                       



m= y2-y1

      x2-x1

m= 20-10

      4-2

m= 10

       2

m= 5           

 

 

 

Try it:

 

 

 

1)Find the slope-

 

  (9,0) and (5,7)



 

2)  Find the slope-

 

    (8,6) and (4,8)

 

 

 

3) Find the slope and graph the line.

 

   (8,6) and (7,3)

 

 

4) Find the slope and graph the line.

 

    (-2,7) and (12,9)

 

 

5) Find the slope-

 

   (25,31) and (10,12)

 

 

 

4.3 Slope Intercept Form

 

Objective: To know how to write linear equations in Slope Intercept Form.

 

Vocabulary:

 

Y-intercept: This is the y-coordinate of the point where the line intersects the y-axis. The y-intercept always has the x-coordinate as zero.

 

X-intercept: This is the x-coordinate of a point where the line intersects the x-axis. The x-intercept always has the y-coordinate as zero.

 

Example: Writing equations in slope intercept form-

 

  1. slope= ¼ ; y-intercept= 5

       

          y=mx+b           This is the original equation

          y=¼ x+5          Substitute the slope with m and the y-intercept with b

  



How to graph in slope-intercept form:

 

 

       

   The slope is 3/2 and the y-intercept is -1.

   y=3/2 x-1

   Step One: Plot (0,1)

   Step Two: Count three units up and 2 units right.

   Step Three: Draw the line connecting the two points.

 

 

    Real Life Application:

 

To rent a car, a company charges $20.00 plus $1.00 per mile. The cost as a function of the number of miles driven is shown in the graph.

  1. Write an equation that represents the cost as a function of the number of miles.



Cost is  $1.00 per mile  times miles  plus  $20.00

  Y   =     $ 1.00              *          x      +      20





This is the equation: y=1x+20

 

Find for 100 miles:

y=1x+20

y=1(100)+20

y=100+20

y=120

The cost of the car for 100 miles is $120.

 

 

Try It:

 

 

 Graph the given slope and y-intercept.

 

 1)slope=¼; y-intercept=-2

 2)slope=7; y-intercept= 12

 3)slope=19; y-intercept= 5

 

 

 

 

Graph these equations in slope intercept form:

     

  4) y=⅗ x-17

     

  5)y=1/4x-1

 

 

 

 

4-4: Point Slope Form:

 

 Objective: To learn how to write equations in point slope form and to learn how to graph them.

 

Example with Explanation:  Writing an equation in point slope form-

 

Slope- 5/2; (-9.5)      Write the point slope form

             

y-y1=m(x-x1)           Substitute 5 for y1, 5/2 for m, and -9 for x1

                 

y-5=5/2(x-(-9)          Rewrite subtraction of negative numbers as addition

              

y-5=5/2(x+9)




Graphing equations in point slope form-

 

Graph the line described by each equation.

 

y-3=2(x-7)

 

Slope: 2 or 2/1   m is the slope (2 is in place of m)

 

 

Point: (7,3)

 

Step one: plot (1,1)

 

Step two: count 2 units up and one to the right (that is the slope-2/1)

 

Step three: Draw the line connecting the two points

 

 

 

Real Life Application:

 

   The cost to place an ad on a TV show for one straight week is a linear function of the number of lines in the ad. The costs for 2, 4 and 10 are shown. Write an equation in slope-intercept form that represents the function. Then find the cost of an ad that is 16 lines long.



Lines:

2,

  4,

  10,

   16

Cost ($):

7,

 14,

  35,

   ???

 

One: Understand the Problem-

There are two parts- an equation in slope-intercept form, and the cost of an ad that is 16 lines long.

 

Two: Make a Plan-

First, find the slope. Then, use point-slope form to write the equation. Finally, write the equation in slope-intercept form.

 

Three: Solve-

Step one: Choose any two ordered pairs from the table to find the slope.

 

m=y2-y1  = 35-14  =  21=  3.5     Here I used (4,14) and (10,35)

      x2-x1     10-4        6

 

Step two: Substitute the slope and any ordered pair from the table into the point-slope form.

y-y1=m(x-x1)

y-7=3.5(x-2)     



Step three: Write the equation in slope-intercept form by solving for y.

y-7=3.5(x-2)                    Multiply 3.5 by 2

y-7=3.5x-7

y=3.5x-7+7                       Add 7 to both sides               

y=3.5x

 

Step four: Find the cost of an ad containing 16 lines by substituting 16 for x.

y= 3.5x

y=3.5(16)

y= 56

The cost of an ad containing 16 lines is $56.

 

 

Try It:

 

Write an equation in point-slope form for the line with the given slope

 

that contains the given point.

 

 1)slope=7; (9,-2)


 2)slope=12; (5,8)


 3)slope=3; (7,9)


 4)slope=13; (-6,-7)

 

Graph the line described by the given equation:

 

 5) y-6=-(x-12)

 

 

Chapter 4 Review:

 

4-2: The Slope Formula:

 

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

 1.(5,10) and (8,-9)  

                                                 

 2.(5,12) and (-4,7)

 

Find the slope of the given lines-

    3.                                                                                  



            

 

    

     4. 




            

  Find the slope of the line described by each equation-

     

     5. 6x-5y=30                                 

 

4-3: Slope Intercept Form:

 

Graph each line given the slope and y-intercept-

   6) Slope=7/1; y-intercept= 12

 

 

 

Write the equation that describes each line in slope-intercept form.

   7) Slope=5; y-intercept=4                 

   8) Slope=-12; y-intercept= 19

   9) Slope= -18 1/4 ; y-intercept=-3    

 

 

 

Graph each line given the slope and y-intercept.

   10) Slope-6 2/5 ; y-intercept- 4

 

4-4: Point Slope Form :

 

 

Write an equation in point-slope form for the line with the given slope that contains the given point.

 

   11) Slope= -9 ⅖; (-7,5)                       

     

 

 

 

Graph the line described by each equation.

 

      

   12) y-3=-(x-7)

 

 

 

 

Write the equation that describes each line in slope-intercept form.

      

    13) Slope= -1/4 , (3,6) is on the line.  

 

    14) Slope= 22, (9,18) is on the line

 

 

 

Each pair of points is on a line . Find the intercepts.

       

    15) (5,2) and (7,4)                     

 

 

Unit 5:

 

 

System

 

 

 

of Equations

 

 

 

 

and

 

 

Inequalities

 

5-1 Solving Systems by Graphing

 

 

Objective:

 

You will be able to identify solutions of systems of linear equations

 

in 2 variables.

 

 

 

 

 

Vocabulary:

 

System of linear equations:Is 2 or more linear equations which

 

contain 2 or more variables.

 

Solution of a system of linear equations:Is 2 variables, an ordered

 

pair that satisfies every equation in the system.

 

 

B:

 

(3,2){2X –Y = 4, X –Y =5

 

 

2X –Y= 4

 

 

2(3) – 2 = 4

 

 

6 – 2 = 4

 

 

4 = 4✓

 

 

 

X –Y =5

 

 

1 = 5

 

 

 

(3, 2) makes only one equations true, so it’s not

 

a solution.

 

 

 

 

3 – 2 =5

 

 

 

 

 

 

Try It:

 

 

1) (10,6){x – y = 4, x + 1y = 16

 

 

2)(12,4){x + 2y =20, x –y = 8

 

 

3)(7,8){x – y = -1, x + y =10

 

 

4)(14,5){x + y = 20, x -2y =10

 

 

5)(20,10){x + y =30, x – y =15

 

 

5-2:Solving System of Substitution  

 

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

 

Vocabulary: substitute: it is when you change the variables into constants to solve an equation.

 

Notes:

 

Step 1: First, solve for one variable in at least one equation if necessary.

 

Step 2: Then 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), and check.

Examples with explanation:

 

A)

3x+y=14     -1st

y=x-5          -2nd

Step 1: y=x-5                         The second equation is solved for y.

Step 2: 3x+y=14

3x+ (x-5)=14                       Substitute x-5 for y in the first equation.

Step 3: solve for x:

            4x-8=14                            Simplify. Then solve for x.

               +8   +8                            Add 8 to both sides.

            4x   =  22

              4         4                          Divide both sides by 4.

                 x=5.5

Step 4: solve for y:

             y=x-5                             Write the first equation.

             y=(5.5)-5                        Substitute 5.5 for x.

             y=0.5                              

Step 5: (5.5, 0.5)                         Write solution as ordered pair.

 

B)

 

 

8y-3x=17      -1st

x-6y=12        -2nd

Step 1: x-6y=12  Solve the second equation for x by adding 6y to each side.

              +6    +6                    

                  x=6y+12

Step 2: 8y-3x=17                   

             8y-3(6y+12)=9      Substitute 6y+12 for x in the first equation.

Step 3:  Solve for y:

          8y-3(6y)-3(12)=9    Distribute -3 to the expression in parentheses.

          8y-18y-36=9                     Simplify

             -10y-36=9

                    +36 =+36                   Add 36 to both sides.

                  -10y= 45

                  -10   -10                 Divide both sides by -10.

                       y=-4.5

Step 4: Solve for x:

            x -6y=12                         Write one of the original equations.  

            x-6(-4.5)=12                   Substitute -4.5 for y.

               x+27=12

                 -27 =-27                     Subtract 27 from both sides.

                   x=-15

Step 5: (-15,-4.5)                          Write solution as ordered pair.

 

 

 

Real world application:

One sports channel costs $75 at first, then costs $23 per month. Another sports channel has no payment at first, then you must pay $50 per month.                                                                                     

A: In how many months will both sports channels cost the same? What will be the cost?

Write an equation for each option. Let t represent the total amount paid and m represents the number of months

.

Total paid   is   setup fee   plus   cost per month   times   months.

Option 1      t           =        75           +             23                  *           m                    

Option 2      t           =         0            +             50                  *           m

Step 1:    t=75+23m               Both equations are solved for t.

                t=50m

Step 2:     75+23m=50m        Substitute 75+23m for t in the second equation.

Step 3:         -23m    -23m     Solve for m. Subtract 23m from both sides.

                     75    = 27m

                     75    = 27m        Divide both sides by 27.

                     27       27

                     2.7   = m

Step 4:    t=50m                     Write one of the original equations.

                =50(2.7)                 Substitute 2.7 for m.

                = 135

Step 5:     (2.7, 135)               Write the solution as an ordered pair

Try it:

 

Solve each system by substitution.

1) y=4x-12              2) y-7x=2               3) 8x=y-3

     y=6x+14             3x-4y=25                 9x-3y=-8      

4)         4x-2y=7              5)          3x-6y=4

            y=5x-2                             4x-8=y

 

 

Objective:

 

To be able to solve systems of linear equations in two variables by elimination.

 

Vocabulary: None

 

Examples with explanation:

 

Elimination Using Subtraction

    1. 2x-3y=14

        2x+y=-10

 

2x-3y=14                                             Subtract the equations.

-(2x+y=-10)                                         Solve for y.

 

0-2y=24

-2y=24

y=24/-2

y=-12

                                                     Solve for x. Use one of the original equations.

2x-3y=14

2x-3(-12)=14

2x-36=14

2x=14+36

2x=50

x=50/2

x=25

Final Answer: (25,14)

5-3 Solving Systems by Elimination

 

Elimination using Addition

         1. 3x+y=17

             4x+2y=20

 

3x-2y=17

+(4x+2y=20)                Add the equations. Solve for x.

7x+0=37                       

7x=37

x=37/7                         Solve for y. Use one of the original equations.

 

4(37/7)+2y=20

21.14+2y=20

2y=20-21.14

2y=-1.14

y=-57/100

 

Final Answer: (37/7, -0.57)

 

Try It:

Solve each system by elimination.

1) −4x − 2y = −12

4x + 8y = −24

 

2) 4x + 8y = 20

−4x + 2y = −30

 

3) x − y = 11

2x + y = 19

 

4) −4x − 2y = 14

−10x + 7y = −25

 

5) 2x + 8y = 6

−5x − 20y = −15

5-3 Solving Systems by Elimination

 

Unit 5 Review:

  

lesson 5-1 

1)(9,1){x –y = 8, x + y = 10

 

 

2)(15,20){x + y =35, x - y = 12

 

 

3)(16,13){x – y =3, x + y = 12

 

 

4)(19,12)x – y= 7, x + y= 21

 

5) (18,2){x-y =16, x - 2y=14

  

Lesson 5.2:

1) y=4x+10             2) 11-13y=7x               3) 5x-7y=15

     y=5x-6                 10x+15y=21                 y=2x+1      

 

4) 71-2y=3x              5)  4x-15=y

    -4x+22=y                 3y-x=5

  

lesson 5-3

 

1) 5x + 4y = −30                                   2) 3x − 2y = 2

 

3x − 9y = −18                                           5x − 5y = 10

 

 

3) 3 + 2x − y = 0                                  4) −14 = −20y − 7x

 

−3 − 7y = 10x                                          10y + 4 = 2x

  

5) −7x − 8y = 9

 

−4x + 9y = −22

 

 

 

 

 

 

 

Chapter 1 review Answers

 

1. q+8

 

3. the quotient of x and 2

 

5. 3 subtracted from j

 

7. 0=x

 

9. -1=k

 

11. y-b/x=m

 

13. -10-z/y=x

 

15. q-5r/2=p

 

Chapter 2 Review:

Answer key:

 Lesson 2.1:

1) t<86.6                                         3) r>-.1                                   5) x<63.6

                                                                                            

                            

 Lesson 2.2:

1) x<-6.3                                    3) -6>0 no solution                    5) x>-3.75

                                                                         

                                                           

Lesson 2.3:

1)-8<x<5                                        3) 2.5<x<3.5                       5) t<8 OR t>10   

                           

Chapter 3 Review:

Answer key:

1)3

3)2

5)5

7)10

9)8

11)9

13)7

15)1

 

Chapter 4 Review:

Answer Key:

1)  -6 ⅓

3) ½

5) 6/5

7) y=5x+4

9)y=-18 ¼x+3

11)Answer: y-5=-9 2/5 (x+7)

13) y-6=-¼ (x-3)

15)x-intercept=3;y-intercept=-3

 

 Chapter 5 Review Answer Key:

Lesson 5-1

Answers:

1)It’s a solution.

3)It’s not a solution.

5)It’s a solution.

 

Lesson 5-2

1)  (-1.7,3.2)                                        

3)  (-2,-3)        

5)  (3.18,4.6)

 

Lessons 5-3

 

     1) (-6,0)

    3) (-1,1)

    5) (1,-2)