MATHEMATICS SUMMER REVIEW RESOURCE PACKET … · MATHEMATICS SUMMER REVIEW RESOURCE PACKET ... of -! ”

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MATHEMATICS SUMMER REVIEW

RESOURCE PACKET

TO BE COMPLETED IN PREPARATION FOR

ALGEBRA 2 5.0

This Resource Packet provides an overview, including examples, of key

problem solving skills that you learned in Algebra 1. In order to be

successful in Algebra 2 5.0, it is essential that you possess a high level of

mastery for these types of problems prior to beginning the school year. As

you will discover, the majority of our time in Algebra 2 will be spent on

learning entirely new and advanced concepts that you have never seen

before. This leaves little time for relearning old concepts.

This Resource Packet has been designed to support your demonstration of

mastery through completion of problems as presented in the Mathematics

Summer Review - Practice Problem Packet (provided separately). As noted

on the Practice Problem Packet, completion of this assignment is strongly

encouraged. Upon request on the first day of school, your teacher will

provide the solutions and will be available to answer any questions that may

arise. (See the Practice Problem Packet for specific details on grading and

follow-up testing on these review topics).

Make sure that you allocate an appropriate amount of time to complete this

packet before the beginning of the school year.

START EARLY – DO NOT PROCRASTINATE.


Mathematics Review

In preparation for ALGEBRA 2 5.0

RESOURCE PACKET

CONTENTS

SECTION TOPIC PAGE

1. Order of Operations 3

2. Solving Linear Equations 5

3. Solving Inequalities in One Variable 7

4. Slope and Equations of Lines 13

5. Solving Systems of Equations 26

6. Graphing Inequalities in Two Variables 30

7. Rules of Exponents 32

8. Operations with Polynomials 35

9. Factoring Polynomials 37

10. Radical Expressions 38


REVIEW SECTION 1 – ORDER OF OPERATIONS

Vocabulary

Numerical Expression: An expression, such as 6 + 5, that consists of

numerals combined by operations and names a particular number, called the

value of the expression.

Algebraic Expression: An expression, such as 2x + 3, that consists of

numerals and variables combined with operations.

In order to avoid misunderstandings and errors, mathematicians have agreed

on certain rules for computation called Order of Operations. (PEMDAS)

Order of Operations

P Represents grouping or inclusion symbols. The basic grouping symbols

are parentheses ( ), brackets [ ], and the fraction bar. Expressions

within grouping symbols should be treated as a single value and should

be evaluated first. The fraction bar indicates that the numerator and

the denominator should each be treated as a single value. When more

than one grouping symbol is used, evaluate within the innermost

grouping symbol first.

E Represents exponents. Perform all operations with exponents.

M Multiplication and Division are considered to be on the same level

D of operational importance. Therefore, perform all multiplication and

division in order from left to right.

A Addition and Subtraction are also on the same level of operational

S importance. Perform all addition and subtraction in order from left

to right.


Use the Order of Operations to simplify each expression.

Example 1: Evaluate 3 12 4− 3 12 4 3 3 0− = − =

Example 2: Evaluate 2 26 4 3 7− + 2 26 4 3 7 36 4 3 49− + = − +

36 12 49

24 49 73

= − +

= + =

Example 3: Evaluate ( )2

3 2 5 7+ ( ) ( )2 2

3 2 5 7 3 7 7+ =

( )3 49 7

147 7 21

=

= =

To evaluate an algebraic expression means to replace each variable by a

given value. Then apply the Order of Operations to simplify the result.

Example 4: Evaluate 2

3

2a

a b

−

+ if 5 and 3a b= =

2

3

5 2 25 2 23

5 27 325 3

− −= =

++

Example 5: Evaluate ( )22

8 33

a b b − +

if = =5 and 2a b

first


( )

( )

2

2

28 5 2 3 2

3

28 3 3 2

3

28 9 3 2

3

272 6

3

278 52

3

− +

+

+

+

=

REVIEW SECTION 2 – SOLVING LINEAR EQUATIONS

1) Multiply using the Distributive Property if necessary

2) Undo addition or subtraction

3) Undo multiplication or division

Example 1:

+ =−

=−

=−

5 8 2

5 10

2

y

y

y

Example 2:

34 16

23

122

8

x

x

x

+ =

=

=

Example 3:

Subtract 8

Divide by 5

Subtract 4

Multiply by 2

3


− =

=

=

1 54

64

24

x

x

x

Example 4:

( )10 2 5

10 20 5

10 25

25 5

10 2

z

z

z

z

− =

− =

=

= =

Example 5:

( )− = −

− = −

− =−

2 7 2 14

2 14 2 14

14 14

f f

f f

Example 6:

− =− +

=

20 3 3 5

20 5

x x

Example 7:

Use the distributive property

Add 20

Divide by 10 (Don’t forget to

simplify)

Distribute

Subtract 2f

Since -14 = -14 is always true, the original

equation is an identity and the solution set

includes all Real numbers.

Add 3x

Since 20 5, there are no solutions that

make it true. Therefore, there is no

solution for this equation.

Add 1

Multiply by 4


( ) ( )− + =− − +

− + = −

= −

− =−

=

2 5 10 3 2

2 10 10 3 2

2 3 2

2

2

x x

x x

x x

x

x

Example 8:

( )2 4 2 7 3

2 4 2 7 3

4 4 7 3

4 3 3

7 3

3

73

7

x x

x x

x x

x x

x

x

x

− − = +

− + = +

− = +

− = +

− =

=−

=−

Distribute

Combine like terms

Subtract 3x

Divide by -1

Distribute

Combine like terms

Subtract 4

Subtract 3x

Divide by -7

Rewrite in simplest terms.


REVIEW SECTION 3 – SOLVING INEQUALITIES

IN ONE VARIABLE

Properties of Inequalities

Equivalent Inequalities are inequalities that have the same solution.

Addition Property of Inequalities: Adding a constant or variable term

to both sides of an inequality does not change the inequality.

Numerical Example:

4 17

4 3 17 3

7 20

+ +

4 3 17 3+ + Add 3 to

Algebraic Example 1 (add constant to both sides):

3 7 5 4

3 7 7 5 4 7

3 5 11

x x

x x

x x

− +

− + + +

+

4 3 17 3+ + Add 3

Algebraic Example 2 (add variable term to both sides):

11 4 2 3

11 4 2 2 3 2

13 4 3

x x

x x x x

x

− − +

− + − + +

−

Subtraction Property of Inequalities: Subtracting a constant or

variable term from both sides of an inequality does not change the

inequality. (Same idea as addition property of inequality)

Numerical Example:

31 8

31 6 8 6

25 2

− −

4 3 17 3+ + Add 3 to

Algebraic Example 1 (subtract constant from both sides):

8 5 3 8

8 5 8 3 8 8

8 13 3

x x

x x

x x

− +

− − + −

−

Given (true statement)

Add 3 to both sides

New inequality (still true!)

Given

Add 7 to both sides

Inequality stays the same!


Subtract 6 from both sides


Given

Subtract 8 from both sides


Given

Add 2x to both sides



Multiplication Property of Inequalities: Multiplying both sides of an

inequality by a positive constant does not change the inequality.

However, multiplying by a negative constant switches the direction of

the inequality.

Numerical Example 1 (multiply by a positive constant):

8 13

8 2 13 2

16 26

−

− • •

−

4 3 17 3+ + Add 3 to

Numerical Example 2 (multiply by a negative constant):

15 4

15 3 4 3

45 12

•− •−

− −

4 3 17 3+ + Add 3 to

Algebraic Example 1 (multiply by a positive constant):

( ) ( )

117

21

2 17 22

34

x

x

x

4 3 17 3+ + Add 3

Algebraic Example 2 (multiply by a negative constant):

36

14

14 3 146

3 14 3

28

x

x

x

−

− − − −

.Add 3

Since division by some number is the same as multiplying by the

reciprocal of that number, it should seem reasonable that the rules

for division in inequalities are the same as those outlined above for

multiplication in an inequality. See the next page for details.


Multiply both sides by 2


Given




Multiply both sides by -3

Direction of inequality switches.

Given


14−



Division Property of Inequalities: Dividing both sides of an inequality

by a positive constant does not change the inequality however dividing

by a negative constant switches the direction of the inequality.

Numerical Example 1 (division by a positive constant):

2 30

2 30

2 21 15

4 3 17 3+ + Add 3 to

Numerical Example 2 (division by a negative constant):

44 22

44 22

11 114 2

−

−

− −

−

4 3 17 3+ + Add 3 to

Algebraic Example 1 (division by a positive constant):

5 35

5 35

5 57

x

x

x

4 3 17 3+ + Add 3

Algebraic Example 2 (division by a negative constant):

7 21

7 21

7 73

x

x

x

−

−

− −

−

4 3 17 3+ + Add 3

NOTE: The key to successful transformations of

inequalities using multiplication or division is as follows:

When you multiply or divide by a negative number (i.e.

when the number “doing the work” is negative), the

direction of the inequality sign changes.


Divide both sides by 2


Given

Divide both sides by 5



Divide both sides by -11


Given

Divide both sides by -7



Rewriting Inequalities:

Generally, inequalities are written with the variable on the left side of

the inequality sign. When you solve an inequality, sometimes the

variable ends up on the right side so you need to rewrite the inequality

into an easier form to graph properly.

Examples:

Rewrite 5 x as 5x

Rewrite 7 x− as 7x −

Rewrite 2

5x− as

2

5x −

Remember: As seen in all of these examples, the arrow point of

the inequality sign always points to the same number or variable as

in its original form. However, the arrow on your graph now points

in the same direction as the inequality.

Graphing Inequalities

When you solve an inequality, one of the key ways to represent your

solution is by graphing the solution on a number line. The graph of an

inequality in one variable is the set of points on a number line that

represent all solutions to the inequality.

Example 1: Graph 3 › x

Solution: Rewrite as x ‹ 3.

Graph with an open circle at x = 3.

Shade region of number line with values that

are less than 3 as indicated by the

inequality.

Remember: Always use an open circle for the ‹ or › inequality

symbols to show that the solution does not include a particular

value. (Notice that with the variable on the left, the inequality

sign points in the same direction as the arrow on the graph.)

4 3 2 1 0 -1 -2


Example 2: Graph x ≥ 4

Solution: Graph with a closed circle at x = 4.

Shade region of number line with values that

are greater than or equal to 4 as indicated

by the inequality.

Remember: Always use a closed circle for the ≤ or ≥ inequality

symbols to show that the solution does include a particular value.

Solving Inequalities and Graphing Solution

Example 3: Solve and Graph

- 7 6

1

x

x

−

Add 7 to each side

Graph


x+2 7

x 5

Subtract 2 from each side

Graph

More Examples – Solving Inequalities with Various Strategies


8 7 6 5 4 3 2

4 3 2 1 0 -1 -2

8 7 6 5 4 3 2


to “clear” fractions.

30 25 20 15 10 5 0

53

3 * 5 *33

15

x

x

x



Example 7: Solve and Graph 3x + 2 › 14

3x › 12 Subtract 2

x › 4 Divide by 3

Example 8: Solve and Graph 4(x - 1) ≥ 8

4x – 4 ≥ 8 Distribute

4x ≥ 12 Add 4

x ≥ 3 Divide by 4

Example 9: Solve and Graph 11 – 2x ≥ 3x + 16

-2x ≥ 3x + 5 Subtract 11

-5x ≥ 5 Subtract 3x

x ≤ -1 Divide by -5

Divide by –4.

Since dividing by a

negative number,

direction of inequality

sign must be changed.

3 2 1 0 -1 -2 -3

8 7 6 5 4 3 2

7 6 5 4 3 2 1

2 1 0 -1 -2 -3 -4

− −

-4x 8

-4x 8

4 4

x -2


REVIEW SECTION 4 – SLOPE AND EQUATIONS OF LINES

When graphing a linear equation, the line has a certain amount of “steepness”

associated with it. In order to measure the amount of steepness, we define

the “slope” of the line as the following ratio:

=Change in y direction Rise

Slope =m =Change in x direction Run

.

The terms “Rise” and “Run” are used in Algebra to describe the steepness of

a line. The vertical portion of a graph from one distinct point to another

(Rise) as compared to the dimension of the horizontal distance from one

point to another (Run). When taken together, Rise over Run tells the story

of the slope, or steepness, of the steps. The slope of a linear equation is the

same at any point along the graph of the equation.

DETERMINING SLOPE DIRECTLY FROM A GRAPH

1) Identify the coordinates of a point on the graph where the line

crosses at the intersection of grid lines on the graph paper.

2) Identify a second point which also crosses at the intersection of

grid lines.

3) Starting at the first point, draw a step using only vertical and

horizontal moves that will take you to the second point. The

amount of movement in the vertical direction is the “change in y”

and the amount of movement in the horizontal direction is the

“change in x.” Use these values to calculate the slope as shown

above.

NOTE:

Movement is a move in the positive y direction

Movement is a move in the negative y direction

Movement is a move in the positive x direction

Movement is a move in the negative x direction


NOTE: As shown in the above graph, any line that has a positive

value for slope will rise from left to right.

NOTE: As shown in the above graph, any line that has a negative

value for slope will fall from left to right.

Example 1: Find the slope of the line

shown in the graph to the right.

Pick two convenient points that cross

at the intersection of grid lines on the

graph paper. Since slope is the same

at all points along the graph, the size

of the “step” created does not matter.

Based on a movement of +3 in the y

direction and +2 in the x direction, the

slope of this equation would be 3

2.



Based on a movement of +8 in the y

direction and -4 in the x direction, the

slope of this equation would be −

8

4

which is then simplified to -2.

3 = +y

2x = +

8 = +y

4 = −x

SCALE: Each block is 1 unit wide by 1 unit high



DETERMINING SLOPE FROM TWO POINTS

Since the slope of a line (usually labeled “m”) represents the ratio of

change in y direction to change in x direction, then slope may be

calculated from two ordered pair solutions ( ) ( )2211 ,and, yxyx for the

equation by the following formula:

( )( )

− = =

−

2 1

2 1

yDifference in y coordinatesSlope = m =

Difference in x coordinates x

y y

x x

Example 1:

Find the slope of a line that goes through the points (-2,-4)

and (3, 8).

The ordered pair (-2, -4) can be labeled Point 1 and (3, 8) is

labeled Point2. The x and y coordinates for each point are

labeled using subscripts to keep track of the numbers that will

be used in the slope calculation.

1 1

Point 1

( 2, 4 )

,x y

− −

2 2

Point 2

( 3, 8 )

,x y

( )( )( )( )

8 4Difference in y coordinates 12Slope = m =

Difference in x coordinates 3 2 5

− −= =

− −

Example 2:

Find the slope of a line that goes through the points (4, -2)

and (-1, 5).

1 1

Point 1

( 4, 2 )

,x y

− −

2 2

Point 2

( 1, 5 )

,x y

( )( )( )( )

5 2 7Difference in y coordinates 7Slope = m =

Difference in x coordinates 1 4 5 5

− −= = = −

− − −


DETERMINING SLOPE FROM THE SLOPE-INTERCEPT

FORM OF AN EQUATION

If an equation is written in the slope intercept form of y= mx + b,

then the slope may be obtained directly from m, the numerical

coefficient of x.

Example:

Given the equation y x= +3 1 , the slope of this equation

will be 3

1

3

1or

−

−.

SPECIAL VALUES OF SLOPE

➢ SLOPE = 0

The slope of a horizontal line is always zero (0).

Example 2:

Find the slope of a line that goes through the points (-6, 4)

and (18, 4).



The movement from one point to the

other is +7 in the x direction but zero

in the y direction (since y values at

both points are the same). As a result

0

07

= = =

ym

x

0 =y because y

values are the same

at both points

7 = +x



1 1

Point 1

( 6, 4 )

,x y

−

2 2

Point 2

( 18, 4 )

,x y

( )

( )( )4 4 0Difference in y coordinates

Slope = m = 0Difference in x coordinates 18 6 24

−= = =

− −

➢ UNDEFINED SLOPE or NO SLOPE

The slope of a vertical line is considered to be “Undefined” and is

sometimes called “No Slope.”

Example 4:

Find the slope of a line that goes through the points (-9, -16)

and (-9, 3).

1 1

Point 1

( 9, 16 )

,x y

− −

2 2

Point 2

( 9, 3 )

,x y

−

( )( )( )( )

Difference in y coordinatesSlope = m =

Difference in x coordinates

3 16 19

9 9 0Undefined

− −= = =

− − −



The movement from one point to the

other is +6 in the y direction but zero

in the x direction (since x values at

both points are the same). As a result

6

0

= = =

ym Undefined

x

0 =x because x

values are the same

at both points 6 = +y




WRITING LINEAR EQUATIONS

While the slope intercept form of a linear equation, = +y mx b , is one of the

most convenient forms to use in order to easily graph a linear equation, the

point-slope form of a linear equation is the preferred format to use when

writing linear equations for a line based on a given value of slope along with

the coordinates of a solution point on the line.

The point-slope form of a linear equation is:

( ) ( )− = −1 1y y m x x

( )1 1

: = slope

, = coordinates of an ordered pair

on the line (and therefore, part

of solution set)

where m

x y

Upon careful examination, you will see that the point-slope form of a linear

equation is nothing more than an algebraic transformation of the slope

formula for a general value of x and y:

( )( )

( ) ( )−

= − = −−

11 1

1

ym ; Cross multiply to obtain:

x

yy y m x x

x

WRITING LINEAR EQUATIONS GIVEN A POINT AND A

SLOPE

Example: Write the equation of a line that has a slope of −3

4 and goes

through the point ( )−, )3 1 .


( ) ( )

( )( ) ( )

: ; ;= − = = −

− = −

− − = − −

+ = − +

= − + −

= − +

1 1

1 1

33 1

4

31 3

43 9

14 4

3 91

4 43 5

4 4

Given m x y

y y m x x

y x

y x

y x

y x

WRITING LINEAR EQUATIONS GIVEN TWO POINTS

Example: Write the equation of a line that goes through the points

( )− ,2 5 and ( )− −,6 9 .

The starting point for this problem is then to first find the slope of the line

and then find the equation in the same manner as demonstrated in the last

example:

=− = −

= =−

1 2

1 2

Given: x 2 x 6

5 9y y

( )( )

( )( )( ) ( )( )

− −− −= = = =

− − − − −

2 1

2 1

9 5y 14 7Slope = m =

x 6 2 4 2

yy

x x

( ) ( )

( ) ( )( )

− = −

− = − −

− = +

= + +

= +

1 1

75 2

27

5 72

77 5

27

122

y y m x x

y x

y x

y x

y x

Note that no numbers are

substituted into the

equation for the x and y

variables that do not have a

subscript.


GRAPHING LINEAR EQUATIONS

The graph of a linear equation is a straight line. When you first started to

graph linear equations, you probably picked several values of x, calculated

the corresponding values of y using the given equation, plotted the points

and then connected the points with a line. You have also learned to graph

linear equations by three other methods: a) using a given point on the line

and slope, b) using slope and y-intercept and c) using x and y-intercepts. In

Algebra 2, you will be expected to be able to use these three “shortcut”

methods to graph linear equations. You will also be expected to develop a

linear equation from only a graph of the equation.

GRAPHING LINEAR EQUATIONS USING A POINT AND SLOPE

1) If given a point on a line and the slope of the line through that point, the

location of the given point represents a starting point for drawing the

graph. Plot the given point.

2) The slope, commonly represented by the variable m, indicates the

direction to “move away” from this given point in a stepwise manner.

Using the location of the given point as a starting point, find a second

point on the graph by moving away from the given point using the vertical

and horizontal components given by the slope. Find one point on either

side of the given point to confirm the trend indicated by slope.

3) Draw a straight line through given point and the points constructed

through use of the slope.

A few notes on interpreting “direction of movement” from slope (m)

➢ If m = an integer value like 2, rewrite it as 2

1 to represent a move of

2 in the positive y direction ( ) and 1 in the positive x direction (→).

Remember that this value of slope could also be represented by −

−

2

1

which would be 2 in the negative y direction () and 1 in the negative x

direction ().

➢ If m is a negative value like −2

3, remember that you can associate the

negative sign with either the numerator −

2

3 or the denominator

−

2

3 to obtain the correct direction of movement from a given point.


EXAMPLE: Graph the equation of a line that goes through the

point (2, -1) and has a slope of −3

4.

Step 1: Graph the given point in Quadrant IV as shown on the

graph below.

Step 2: Plot additional points based on the stepwise movement

given by the slope. Since m = −3

4, then move away from

the given point by moving 3 units in the negative y

direction and then 4 units in the positive x direction

(i.e. =− =,y x3 4 ). Another option is to move 3 units in

the positive y direction followed by 4 units in the

negative x direction (i.e. = =−,y x3 4 ).

Step 3: Draw a straight line through the given point and the

constructed points.

3y = +

4x = +

4x = −

3y = −

x Starting point

at (2, -1)

y



A reminder about “Special Cases” where Slope = 0 or is “Undefined”

➢ SLOPE = 0: The graph of an equation such as y= (some value) is

always represented by a horizontal line and is commonly referred to as

a Constant Function. Since all points along a horizontal line have the

same value for the y-coordinate, the value of y is always equal to

zero and as a result, the slope of this line is always zero (0).

EXAMPLE: Graph the equation of a line that goes through the point (4, -3)

and has a slope of 0. What is the equation of this line?

➢ SLOPE = UNDEFINED: The graph of an equation such as

x= (some value) is always represented by a vertical line. Since all

points along a vertical line have the same x-coordinate, the value of

x is always equal to zero and as a result, the slope of this line is

always “Undefined.” An undefined slope is sometimes also described

as “No Slope.”

EXAMPLE: Graph the equation of a line that goes through the point (3, 2)

and has an undefined slope. What is the equation of this line?

x

y

=3x

x

y

= −3y




GRAPHING LINEAR EQUATIONS BY SLOPE AND Y-INTERCEPT

SLOPE-INTERCEPT FORM:

= +

= slopewhere

= y-intercept

my mx b

b

1) If the equation is not in the “slope-intercept” form of y = mx+b,

complete the necessary transformations to get it into this form.

2) The y-intercept is given by the value of b in the equation and is the

starting point for completing this graph. This point would be graphed

on the y-axis as the ordered pair (0, y-intercept).

3) The slope is given by the value of m and indicates the direction to

“move away” from the y-intercept in a stepwise manner. Using the

y-intercept as a starting point, find a second point on the graph by

moving away from the y-intercept. Find one point on either side of

the y-intercept to confirm the trend indicated by slope.

4) Draw a straight line through y-intercept and constructed points.

Example 1: Graph: 5

13

y x= −

Identify the key graphing parameters and graph:

y-intercept:

Since b = -1. the y-intercept

is located at (0, -1).

Slope:

Since m = 5

3, then “move away”

from the y-intercept starting point

by moving 5 units in the positive y

direction and then 3 units in the

positive x direction. It could also

mean a move of 5 units in the

negative y direction and then 3 units in the negative x direction

since −

−

5

3 = 5

3.

5 = +y

3x = +

3x = −

5 = −y

x

Starting point is at

y intercept (0, -1)



Example 2: Graph: 2

35

y x= − +

Identify the key graphing parameters and graph:

y-intercept:

Since b = 3. the y-intercept

is located at (0, 3).

Slope:

The slope m = 2

5− which can be

rewritten as 2

5

− or 2

5−. Using the

value of 2

5

− , “move away” from the

y-intercept by moving 2 units in

the negative y direction and then 5

units in the positive x direction. Another option is to use the

value of 2

5− which would mean to “move away” from the y-

intercept by moving 2 units in the positive y direction and then

5 units in the negative x direction.

Example 3: Graph: − + =2 3 6x y

Step 1: First convert equation to slope intercept form:

( )

Addition Property of Equality

Multiplication Property of Equality

− + =

+ +

= +

= +

2 3 3

2 2

1 13 2 3

3 3

21

3

x y

x x

y x

y x

.

y

2 = +y

5 = −x

5 =x

2 = −y

x


y intercept (0, 3)



Step 2: Identify the key graphing parameters and graph:

y-intercept: Since b = 1. the y-intercept is located at

(0, 1).

Slope: Since m = 2

3, then “move away” from the y-

intercept starting point by moving 2 units in

the positive y direction and then 3 units in

the positive x direction. It could also mean

a move of 2 units in the negative y direction

and then 3 units in the negative x direction

since −

−

2

3 = 2

3.

y

x


y intercept (0, 1)



REVIEW SECTION 5 – SOLVING SYSTEMS OF EQUATIONS

Vocabulary: Two or more linear equations with the same variables represent

a system of linear equations or a linear system.

Three possible cases exist when solving a system of equations:

1) If one ordered pair satisfies all of the equations in the

system, then the solution is that ordered pair, (x, y).

2) If an infinite number of ordered pairs satisfy all of the

equations in the system, then the graphs of both equations

are the same line.

3) If no ordered pairs satisfy all of the equations, then the

graphs of these equations are parallel, and therefore do not

intersect. As a result, the system has no solution, and is

denoted by the Greek symbol .

When the solution to a system of linear equations in two variables is

an ordered pair of numbers (x, y), you can check your answer by

substituting this ordered pair back into each original equation of the

system. The result should be a true statement. The point (x, y)

lies on the graph of each equation, and is the point of intersection of

the two lines.

Method #1: SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY

Example: Solve the system by “Graphing”. -3x + y = -7

2x + 2y = 10

Solution: Rewrite each equation in slope-intercept form.

1) -3x + y = -7 y = 3x –7 Slope: 3 y int: (0. –7)

2) 2x + 2y = 10 y = -x + 5 Slope: -1 y int: (0, 5)


Method #2: SOLVING SYSTEMS

OF EQUATIONS BY LINEAR

COMBINATION (also known as Elimination)

The objective in this method is to get the coefficients of one variable

to be opposite numbers.

Example: Solve the system by “Elimination.” 4x - 3y = 11

3x + 2y = -13

To eliminate the y

4x - 3y = 11 Multiply by 2 8x - 6y = 22

3x + 2y = -13 Multiply by 3 9x + 6y = -39

Add the equations 17x = -17

Solve for x x = -1

Now substitute -1 for x in either of the original equations, and solve for y.

3x + 2y = -13

3(-1) + 2y = -13 Substitute –1 for x

– 3 + 2y = -13 Simplify

2y = -10 Add 3 to each side

y = -5 Solve for y

The solution is the ordered pair ( -1, -5 ).

•

•

• Solution to system is given by

coordinates of intersection of

the two equations. In this case,

the solution is (3, 2)

Graph of Equation # 1 Graph of Equation # 2


Method #3: SOLVING SYSTEMS OF EQUATIONS BY SUBSTITUTION

This method of solving systems by substitution should be considered

when at least one of the variables in the system of equations has a

coefficient of one (1) as demonstrated in the following examples:

Example 1: Solve the system by “Substitution.” -3x + y = -11

5x + 3y = 9

Since the variable y in the first equation has a coefficient of 1, solve

that equation for y in terms of x.

y = 3x - 11

Next substitute 3x - 11 for y in the other equation and solve for x.

5x + 3y = 9

5x + 3(3x – 11) = 9 Substitute 3x - 11 for y

5x + 9x –33 = 9 Distribute the 3

14x – 33 = 9 Combine like terms

14x = 42 Add 33 to each side

x = 3 Divide each side by 14

Find the value of y by substituting 3 for x in one of the original

equations (your choice).

-3(3) + y = -11 or 5(3) + 3y = 9

-9 + y = -11 15 + 3y = 9

y = -2 3y = -6


y = -2

The solution is the ordered pair ( 3, -2 ).


Example 2: Solve the system by “Substitution”. -3x + 4y = 16

x - 3y = -7

Since the variable x in the second equation has a coefficient of 1,

solve that equation for x in terms of y:

x = 3y - 7

Substitute 3y - 7 for x in the first equation and solve for y.

-3x + 4y = 16

-3(3y – 7) + 4y = 16 Substitute 3y - 7 for x

-9y + 21 + 4y = 16 Distribute the -3

-5y +21 = 16 Combine like terms

-5y = -5 Subtract 21 from both sides

y = 1 Divide each side by -5

Find the value of x by substituting 1 for y in one of the original

equations (your choice).

-3x + 4(1) = 16 or x – 3(1) = -7

-3x = 12 x = -4

x= -4

The solution is the ordered pair ( -4, 1 ).


REVIEW SECTION 6 – INEQUALITIES IN TWO VARIABLES

GRAPHING A LINEAR INEQUALITY IN TWO VARIABLES

Example 1: Graph x – y < 2

a) The corresponding equation is x – y = 2. In

slope-intercept form, this equation is y = x – 2.

Graph the boundary line using the y-intercept of -2

and the slope of 1. This line should be a dashed

line, or a broken line, to show that the points on

the line are not solutions since the original

inequality is only “ ”, not “ ”.

b) Pick a point that is not on the line, and use its

coordinates to determine the region to shade. For

ease of calculation, always consider using the origin (0,0) as the test

point unless it is on the line. Substituting the values of the

coordinates into the original inequality will result in either a

true or false statement.

c) Using (0, 0) as a test point in the given inequality x – y < 2,

results in 0 - 0 < 2, which is a true statement. Since the test

point generated a true statement, shade the same region as the test

point. In this case, shade above the line. This shaded region

represents the solution set of the given inequality, and consists of all

points above the graph of y = x – 2.

OR

If the coordinates had generated a false statement, then the region

on the other side of the line would have been shaded instead. The

actual solution set would have been the region that did not contain the

test point.

NOTE: The type of line used for the boundary between solution

points and non-solution points always depends on the type of inequality

given in the original problem.

Inequality Symbol “Boundary” line type

or Dashed or broken line

y

Test Point (0, 0)

Any point in shaded

region is a member of the solution set.

Points on line are excluded from solution

set in this example

(dashed line).


or Solid line

Example 2: Graph 2x + 3y -6

a) The corresponding equation is 2x + 3y -6= . In

slope-intercept form, this equation is 2

y = - x - 23

.

Graph this line using a y-intercept of -2 and a slope

of 2

-3

. Draw a solid line to show that the points on

the line are part of the solution set since the

original inequality is “ ”, and not just “ ”.

b) Pick a point that is not on the line, and use its

coordinates to determine the region to shade.

Again, for ease of calculation, always consider using the origin (0,0)

as the test point unless it is on the line. If (0,0) is on the boundary

line, you must try a different test point. Substituting the values of

the coordinates into the original inequality results in either a true or

false statement.

c) Using (0, 0) as a test point in the given inequality 2x + 3y -6 ,

results in ( ) ( )2 0 + 3 0 -6 , which is false! Since the test point

generated a false statement, shade the region on the other side of

the line. In this case, shade below the line. This shaded region

represents the solution set of the given inequality, and consists of all

points below the graph of 2

y = - x - 23

.

Testing another point on the other side of the boundary line confirms

that the lower area of this graph contains all of the solution points

for this inequality. 2x + 3y -6

At test point (-4, -3): ( ) ( )2 -4 + 3 -3 -6

-17 -6

which is true. Therefore, shade the area in the same region as this

test point.

y

First Test

Point (0, 0)

Any point in shaded

region is a member of the solution set.

Points on line are

included in solution set in this example

(solid line).

Second

Test Point

(0, 0)



REVIEW SECTION 7 – RULES OF EXPONENTS

Vocabulary

Base of a power: The number that is used as a factor. For example, 2 is

the base of 32 .

Exponent: In a power, the number that indicates how many times the base

is used as a factor. For example, 3 is the exponent in 32 .

Exponential form: The power of a number written using a base and an

exponent. The expression 32 is the exponential form of2 2 2 .

Note: An exponent goes with the base immediately preceding it unless

parentheses indicate otherwise.

Example 1: 23 means 3y y y 2 is the exponent of the base y.

( )2

3 means 3 3y y y 2 is the exponent of the base 3y.

Product of Powers: When two powers with the same base are multiplied,

add the exponents.

m n m na a a + =

Example 2: 2 5 2 5 7x x x x+ = = Think: x is being used as a

factor a total of 7 times.

Example 3: 2 3 2 3 53 3 3 3+ = =

Example 4: + = = =5 1 5 1 5 66 3 6 3 6 3 18c c c c c c

Power of a Power: When a power is raised to a power, multiply exponents.

( )nm mna a=

Example 5: ( )67 7 6 42x x x= =

Example 6: ( )43 3 4 122 2 2= =


Power of a Product: When a product is raised to a power, raise each factor

to the power and then multiply.

( )m m mab a b=

Example 7: ( ) ( )5 5 5 52 2 32x x x− = − = −

Example 8: ( ) ( ) ( ) ( )3 3 332 5 2 5 6 156 6 216x y x y x y= =

Quotient of Powers: When two powers with the same base are divided,

subtract the exponents.

−

−

=

=

If ,

1If ,

mm n

n

m

n n m

am n a

a

an m

a a

Example 9: 8

8 5 3

5

x x x x x x x x xx x

x x x x x x−• • • • • • •

= = =• • • •

Example 10: 5

8 8 5 3

1 1x

x x x−= =

Example 11: 3 2 2

3

4

16 4

a b a

ab b= −

−

Power of a Quotient: When a quotient is raised to a power, raise the

numerator and the denominator to that power. Then divide and simplify.

m m

m

a a

b b

=

Think: Where are the most factors of x?

How many extra factors?


Remember: An exponent is applied

only to the base immediately preceding

it, unless otherwise indicated by

parentheses.

Remember: An exponent is applied

only to the base immediately preceding

it, unless otherwise indicated by

parentheses.

Example 12: ( ) ( )

5 5

5 54 204 5 4

2 2 32 32

3 2433 3x xx x

= = =

Zero Exponents: If a is a real number not equal to zero, then 0 1a = .

(The expression 00 has no meaning!)

Example 13: 025 1=

Example 14: 05 5 1 5x = =

but ( ) ( )( )0 0 05 5 1x x= =

Negative Exponents: If a is a nonzero real number and n is a positive

integer, then 1n

na

a− = . Negative exponents mean reciprocals! All rules for

positive exponents also hold for negative exponents.

Example 15: 3

3

1 12

2 8− = =

Example 16: ( ) ( ) ( )− − −− = =2 1 21 2x x x

Example 17: ( ) ( )4 42 4 2 8

8

162 2 16x x x

x− − −= = =

Example 18: 9

9 2 11

2

xx x x

x −= =

Example 19: − −= = =5

5 12 7

12 7

1xx x

x x

Example 20: ( )− −

−= =

5 3)58

3

4 42

2 2

x xx

x

Note: Simplified expressions must

Note: Negative exponents do not mean

negative numbers!


only have positive exponents.

REVIEW SECTION 8 – OPERATIONS WITH POLYNOMIALS

Vocabulary

Monomial: An expression that is either a numeral, a variable, or the product

of a numeral and one or more variables. Also called terms.

Examples: 13, m, 8c, 2xy, 25p q

Coefficient: In a monomial, the number preceding the variable. When a

monomial does not have a written coefficient, its coefficient is 1.

Polynomial: A sum of monomials. Example: 2 23 2x x y+ + −

Binomial: A polynomial that has exactly two terms. Example: 2 5x +

Trinomial: A polynomial that has three terms. Example: 2 22a ab b− +

Similar or Like Terms: Two monomials that have exactly the same variable

names. Examples: 2 23 and 7 , 5 and xy xy ab ab−

Addition of Polynomials: To add two or more polynomials, combine the

coefficients of the like-terms.

Example 1: ( ) ( )2 2 2 2 2 22 5 6 8 6 10 11 5x xy y x xy y x xy y+ − + + + = + −

Subtraction of Polynomials: To subtract polynomials, change the second

polynomial to its opposite and then add to the first polynomial.

Example 2: ( ) ( )2 2 2 23 6 2 5 5 6 3x xy y x xy y− − − − − + + − =


( ) ( )2 2 2 23 6 2 5 5 6 3x xy y x xy y− − − + − − + = 2 24 11 8 2x xy y− − −

Multiply a Polynomial by a Monomial: Multiply each term of the polynomial

by the monomial, apply the rules for exponents as appropriate.

Example 3: ( )2 3 22 3 2 1 6 4 2x x x x x x− − + = − + −

Multiply a Binomial by a Binomial: Multiply each term of the first binomial

by each term of the second binomial using the distributive property. “FOIL”

(First times First, Outer product, Inner Product, Last times Last)

( )( )a b c d+ +

Example 4:

Example 5:

Example 6:

F

O I

L

( )( )2

2

3 2 2 5 3 2 3 5 ( 2) 2 2 (5)

6 15 4 10

6 11 10

x x x x x x

x x x

x x

− + = + + − −

= + − −

= + −

( )2

2

4 ( 4)( 4)

( )( ) 4 4 (4)(4)

8 16

x x x

x x x x

x x

+ = + +

= + + +

= + +

( )2

2

5 2 (5 2)(5 2)

(5 )(5 ) 2(5 ) 2(5 ) ( 2)( 2)

25 20 4

x x x

x x x x

x x

− = − −

= − − + − −

= − +


REVIEW SECTION 9 – FACTORING POLYNOMIALS

When a polynomial contains only one variable, the degree of a polynomial is

the highest exponent on x.

Factoring polynomials – to write a polynomial as the multiplication of

polynomials of lower degree.

A polynomial that has coefficients that are integers and whose greatest

common factor is 1 is called a prime polynomial. A polynomial is factored

completely when each polynomial factor is prime.

Guidelines for Factoring

1) Factor out the greatest common factor, GCF, first! The greatest

common factor is the greatest factor common to every term in the

polynomial. For variable terms, use the greatest exponent common to

every term in the polynomial.

Example 1: The greatest monomial factor of 5 36 4 8 is 2x x x x− +

since ( )5 3 4 26 4 8 2 3 2 4x x x x x x− + = − + .

2) Look for special patterns:

Difference of Squares: ( )( )2 2a b a b a b− = − +

To factor, take the square root of each term and put one “+” and one

“-“ sign between the terms.

Example 2: ( )( )225 121 5 11 5 11x x x− = − +

3) General Method of Factoring:

For 2x bx c+ + , find two factors of c whose sum is b.

Example 3a: ( )( )2 6 8 2 4x x x x+ + = + + since 2(4) = 8 and 2 + 4 = 6.

Note the following factoring patterns:

when b is positive: ( )( )? ?x x+ +

If c is positive

when b is negative: ( )( )? ?x x− −


If c is negative: ( )( )? ?x x− +

REVIEW SECTION 10- RADICAL EXPRESSIONS

Radical Expression: An expression of the form a . The symbol, , is

called a radical sign and the expression under the radical sign, “a”, is called

the radicand.

Square root: If 2a b= , then a is a square root of b.

(Note: Negative numbers do not have square roots in the set of real

numbers.)

Symbols:

64 8= indicates the nonnegative or principal square root of 64

64 8− = − indicates the negative square root of 64

64 8 = indicates both square roots of 64

I. Simplifying Radicals: To simplify a radical expression, find the square

root of any factors of the radicand that are perfect squares. (If you cannot

see any squares, you can use prime factorization of the radicand.)

Product Property of Square Roots

For any nonnegative real numbers a and b:

ab a b= Example: 4 9 4 9 =

Quotient Property of Square Roots

For any nonnegative real number a and any positive real number b:

a a

b b= Example:

36 36

4 4=

Example 1: 256 4 64 4 64 2 8 16= = = =

Example 2: 50 25 2 25 2 5 2 5 2= = = =

Example 3: 2 80 2 16 5 2 4 5 8 5= = =


Example 4: 169 169 13

100 10100

= =

II. Multiplying Radicals:

Example 5: 4 2 3 4 2 3 4 6 = =

Example 6: 2 3 3 48 2 3 3 48 6 144 6 12 72 = = = =

III. Dividing Radicals:

Example 7: 7 54 7 54 9 3

6 28 6 28 4 2 = = =

Example 8: 24 33 24 33

36 611 2 11 2

= = =

IV. Rationalizing Denominators: A radical is not considered to be in simplest

form if there is a fraction under the radical sign or a radical in the

denominator. Rationalizing is the process of eliminating a radical from the

denominator. (Remember: ( )2

a a= ).

Example 9: 1 1 1 3 3

3 33 3 3

= = =

Example 10: 5 5 2 5 2

22 2 2

= =

V. Adding and Subtracting Radicals: Only radicals with like radicands can be

combined.

Example 11: 8 18 4 2 9 2 2 2 3 2 5 2+ = + = + =

Example 12: 6 24 6 4 6 6 2 6 6− = − = − = −

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