Transcript Addition Property (of Equality)

Algebra 1
A Review and Summary
Gabriel Grahek
In the next slides you will review:
Solving 1st power equations in one variable
A. Special cases where variables cancel to get {all
reals} or
B.
Equations containing fractional coefficients
C.
Equations with variables in the denominator –
(throw out answers that cause division by zero)
1st Power Equations
• Any type of equation that has only
one variable.
x+3=8
x+3-3=8
x=5
Note that the variable
can be on both sides.
4(x2-2)+8=36
(4x2-8)+8=36
4x2-8=28
4x2=36
x2=9
x=3
1st Power Equations
3x-4=12+x
3x-4+4=12+4+x
3x=16+x
3x-x=16+x-x
2x=16
x=8
4y+16=24
4y+16-16=24-16
4y=8
y=2
1st Power Equations
• Special cases- Ø and {all reals}
x-(4-3)=x
4x+4=2(2x+2)
x-1=x
4x=4x
-1=0
x=x
Ø
1st Power Equations
•Equations containing fractional
coefficients and with variables in the
denominator.
5(7)  5
x2 4
 16
2
x4
 16
2
x  4  32
x 8
2
x 10
30
2
2
x  10
30  2( x 2  10)
2
30  2 x 2  20
50  2 x 2
25  x 2
5 x
In the next slides you will review:
Review all the Properties and
then take a Quiz on identifying
the Property Names
Addition Property (of Equality)
Example:
If a, b, and c, are any real numbers, and a=b, then a+c=b+c
and c+a=c+b
If the same number is added to equal numbers, the sums are
equal.
Multiplication Property (of Equality)
Example:
If a, b, and c are real numbers, and a=b, then ca=cb and
ac=bc.
If equal numbers are multiplied by the same number, the
products are equal.
Reflexive Property (of Equality)
Example:
For all real numbers a, b, and c:
a=a
Symmetric Property (of Equality)
Example:
For all real numbers a, b, and c:
If a=b, then b=a.
Transitive Property (of Equality)
Example:
For all real numbers a, b, and c:
If a=b and b=c, then a=c.
Associative Property of Addition
Example:
For all real numbers a, b, and c:
(a+b)+c=a+(b+c)
Example: (5+6)+7=5+(6+7)
Associative Property of Multiplication
Example:
For all real numbers a, b, and c:
(ab)c=a(bc)
Example:
 2  3 4  2 3  4
Commutative Property of Addition
Example:
For all real numbers a and b:
a+b=b+a
Example: 2+3=3+2
Commutative Property of Multiplication
Example:
For all real numbers a, b, and c:
ab=ba
Example:
45  5 4
Distributive Property (of
Multiplication over Addition)
Example:
For all real numbers a, b, and c:
a(b+c)=ab+ac
and
(b+c)a=ba+ca
Prop of Opposites or Inverse
Property of Addition
Example:
For every real number a, there is a real number -a such that
a+(-a)=0 and (-a)+a=0
Prop of Reciprocals or Inverse Prop. of
Multiplication
Example:
If we multiply a number times its
reciprocal, it will equal one. For example:
1
x 1
x
Identity Property of Addition
Example:
There is a unique real number 0 such that for every real number a,
a + 0 = a and 0 + a = a
Zero is called the identity element of addition.
Identity Property of Multiplication
Example:
There is a unique real number 1 such that for every real number a,
a · 1 = a and 1 · a = a
One is called the identity element of multiplication.
Multiplicative Property of Zero
Example:
For every real number a,
a · 0 = 0 and 0 · a = 0
Closure Property of Addition
Example:
Closure property of real number addition states that the sum of any two real
numbers equals another real number.
Closure Property of Multiplication
Example:
Closure property of real number multiplication states that the product of any two
real numbers equals another real number.
Product of Powers Property
Example:
This property states that to multiply powers having the same base, add the
exponents.
That is, for a real number non-zero a and two integers m and n, am × an = am+n.
Power of a Product Property
Example:
This property states that the power of a product can be obtained by finding the
powers of each factor and multiplying them.
That is, for any two non-zero real numbers a and b and any integer m, (ab)m = am
× bm.
Power of a Power Property
Example:
This property states that the power of a power can be found by multiplying the
exponents.
That is, for a non-zero real number a and two integers m and n, (am)n = amn.
Quotient of Powers Property
Example:
This property states that to divide powers having the same base, subtract the
exponents.
am
mn

a
That is, for a non-zero real number a and two integers m and n,
.
n
a
Power of a Quotient Property
Example:
This property states that the power of a quotient can be obtained by finding the
m
powers of numerator and denominator and dividing them.
a
That is, for any two non-zero real numbers a and b and any integer m,   
 
b
am
bm
Zero Power Property
Example:
a 1
0
Any number raised to the zero power is equal to “1”.
Negative Power Property
Example:
Change the number to its reciprocal.
x
2
1
 2
x
Zero Product Property
Example:
Zero - Product Property states that if the product of two factors is zero, then at
least one of the factors must be zero.
If xy = 0, then x = 0 or y = 0.
Product of Roots Property
For all positive real numbers a and b,
a  b  a b
That is, the square root of the product is the same as the product of the
square roots.
Quotient of Roots Property
For all positive real numbers a and b, b ≠ 0:
a
a

b
b
The square root of the quotient is the same as the quotient of the square roots.
Root of a Power Property
x4  x2
Example:
( x 2 )2
7 7
2
Power of a Root Property
2
Example: ( 9)  9
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
1. a + b = b + a
Answer:
Commutative Property (of Addition)
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
2. am × an = am+n
Answer:
Product of Powers
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
3. For every real number a,
a · 0 = 0 and 0 · a = 0
Answer:
Multiplicative Property of Zero
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
4. the sum of any two real numbers equals another real number.
Answer:
Closure Property of Addition
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
5. There is a unique real number 1 such that for every real number a,
a · 1 = a and 1 · a = a
Answer:
Identity property of Multiplication
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
6.
a 1
0
Answer:
Zero Power Property
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
m
7.
a
mn
a
n
a
Answer:
Quotient of Powers Property
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
8. (ab)m = am
Answer:
Power of a Product
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
9.
x
2
1
 2
x
Answer:
Negative Power Property
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
10.
1
x 1
x
Answer:
Prop of Reciprocals or Inverse Prop. of Multiplication
Now you will take a quiz!
Look at the sample problem and give
the name of the property illustrated.
Click when you’re ready to see the answer.
11. (ab)c=a(bc)
Answer:
Associative Property of Multiplication
Solving Inequalities
Solving Inequalities

Remember the Multiplication Property
of Inequality! If you multiply or divide
by a negative, you must reverse the
inequality sign.
-2x < 8
x > -4
Solution Set: {x: x > -4}
Graph of the Solution:
-4
Solving Inequalities




Open endpoint for these symbols: > <
Closed endpoint for these symbols: ≥ or ≤
Conjunction must satisfy both conditions
Conjunction = “AND”
{x: -5 < x ≤ 8}
Click to see solution graph
-5
8
Solving Inequalities




Open endpoint for these symbols: > <
Closed endpoint for these symbols: ≥ or ≤
Disjunction must satisfy either one or both of the
conditions
Disjunction = “OR”
{x: x < -6 or x ≥ 8}
Click to see solution graph
-6
8
Solving Inequalities – Special Cases




Watch for special cases
No solutions that work: Answer is Ø
Every number works: Answer is
{reals}
When the disjunction goes the same
way you use one arrow.
{x: x > -6 or x ≥ 8}
Click to see solution graph
-6
8
Solving Inequalities – Special Cases
Watch for special cases:



No solutions that work: Answer is Ø
Every number works: Answer is {reals}
{x: -2x < -4 and -9x ≥ 18}
Click to see solution
Ø
Solving Inequalities

Now you try this problem
2x > 6 or -16x ≤ 32
Click to see solution and graph
-2 < x or x ≤ 3
-2
3
Solving Inequalities

Now you try this problem.
4x-8 < 12 and -x < 10-4
Click to see solution and graph
-6 < x < 5
-6
5
Type a sample problem here. Blah blah
blah. You can duplicate this slide.
Click when ready to see the answer.
Type the answer here. Set to fade-in on
click
Type any needed explanation or tips here.
Set to fade-in 3 seconds after the answer
appears above.
In the next slides you will review:
Linear equations in two variables
Lots to cover here: slopes of all
types of lines; equations of all types
of lines, standard/general form,
point-slope form, how to graph, how
to find intercepts, how and when to
use the point-slope formula, etc.
Remember you can make lovely
graphs in Geometer's Sketchpad and
copy and paste them into PPT.
Linear Equations






y 2  y1
x 2  x1
Slope=
Point-Slope Formula= y  y  m( x  x )
Slope-Intercept
y  mx  b
Formula=
Midpoint Formula=  x 2 x , y 2 y 


Standard/ General
Form= Ax+Bx=C
Distance Between
Two Points Formula= d  ( x  x )  ( y  y )
1
1
2
1
1
2
2
1 2
2
1 2
Slope

(9,12) and (13, 20)
20  12
13  9
8
4

Pt-Slope Formula
(9,12) and (13, 20)
8
y  12  ( x  9)
4
y  12  2 x  18
y  2x  6
Would be negative if it
had a negative sign in
front of it. It would then
be a falling line and not a Use when you only
rising line.
have solution points.
Midpoint
Distance
(9,12) and (13, 20)

(9,12) and (13, 20)
 9  13 12  20 
,


2 
 2
 22 32 
 , 
 2 2 
11,16 

Use to find the middle
point on a line.
d  (13  9)  (20  12)
2
2
d  (4) 2  (8) 2
d  16  64
d  80
d  4 5
Use to find the Distance
between to points.
Equations in Two Variables


The pairs of numbers that come out for
each variable can be written as an (x,y)
value. (ordered pair)
You give the solutions in alphabetical
order of the variables. So, it would be
(a,b) and not (b,a).
Standard Form
ax+by=c
 All linear equations can be written in this
form.
 A, b, and c are real numbers and a and b
are non-zero. A, b, and c are integers.
 To change to slope intercept:
Ax+bx=c
bx=ax+c

How to graph


To graph the slope-intercept form: you can
take the y intercept and use the slope to
determine the points on the line.
To graph the standard form you have to
change it to slope-intercept, explained in
the last slide, and then graph it.
To find the yintercept
F(x)=mx+b
Set x to 0
F=b
Example:
F(x)=4x+6
F(0)=4(0)+6
F=6

To find the xintercept
F(x)=mx+b
Set f(x)=0
0=mx+b
Divide out
Example:
F(x)=4x-8
0=4x-8
8=4x
2=x
In the next slides you will review:
Linear Systems
A. Substitution Method
B. Addition/Subtraction
Method (Elimination )
C. Check for understanding of
the terms dependent,
inconsistent and consistent
When two line share solution points…

Null set (if they are parallel)
This

One point (if they cross)
This

will be called an INCONSISTENT SYSTEM
will be called a CONSISTENT SYSTEM
Infinite Set or All Pts on the Line (if same line is
used twice)
This
will be called a DEPENDENT SYSTEM (It is also
consistent. Dependent is the better name for it than
consistent.)

Solution: Null set (if they are parallel)
INCONSISTENT
SYSTEM
4
2
-2
-4

One point (if they cross)
CONSISTENT
SYSTEM
4
2
-2

Infinite Set or All Pts on the Line (if
same line is used twice)
DEPENDENT
SYSTEM
-4
4
2
-2
-4
The SOLUTION of a
SYSTEM is the
INTERSECTION SET
Where do the two lines
Meet
intersect.
Two Different Equation on the
same graph are called a SYSTEM
OF EQUATIONS.
Think about:
 y  2x  5
2x  y  8



3
y

x

2
 x  y 1

2
Method 1: To estimate the solution of a
system, you have to find out where they
intersect.
5
fx =
gx =4 -2x+5

3
x-2

2
3
 y  2x  5


3
 y  2 x  2
2
Solution to this
system is:
1
{(2, 1)}
2
4
Example: You can use Trace on a graphing calculator
to help you estimate the solution of a system. It can
find where they intersect.
3
fx =
2
gx = 1
-2

1
x+1

3
2
-1
1

y   3 x  1

 y  9  x2

9-x2
4
Example: You can use Trace on a graphing calculator
to help you estimate the solution of a system. It can
find where they intersect
3
1

y   3 x  1

 y  9  x2

fx = 9-x2
A: (-2.38, 1.79)
2
gx = 1
-2

1
x+1

3
2
4
B: (3.00, 0.00)
-1
Solution to this system appears to include TWO pts:
{ (-2.38, 1.79), (3, 0) }
Summary of Method 1: Estimate the
SOLUTION of a SYSTEM on a graph.
(Goal: Find intersection pts.)
Disadvantages:
Might only give an estimate.
It might not be possible to graph some equations yet.
Advantages:
If the graph is easy, this is a good way to check.
It is good for a quick answer.
Method 2: Substitution Method
(Goal: Replace one variable in one equation with
the set from another.)
 x  y  15

4x  3y  38

Step 1: Look for a variable with a coefficient of one.
Step 2: Move everything else to the other side.
Equation A now becomes: y = 15-x
Step 3: SUBSTITUTE this expression into that variable in Equation B
Equation B now becomes 4x – 3( 15-x ) = 38
Step 4: Solve the equation.
Step 5: Back-substitute this coordinate into Step 2 to find the other
coordinate. (Or plug into any equation but step 2 is easiest!)
Method 2: Substitution
(Goal: replace one variable
with an equal expression.)
Step 1: Look for a variable with a
coefficient of one.
Step 2: Isolate that variable
Equation A now becomes:
y = 3x + 1
Step 3: SUBSTITUTE this
expression into that
variable in Equation B
Equation B becomes
7x – 2( 3x + 1 ) = - 4
Step 4:
Solve for the remaining
variable
Step 5:
Back-substitute this
coordinate into Step 2
to find the other
coordinate. (Or plug
into any equation but
step 2 is easiest!)
 3x  y  1

7x  2y  4
y  3x  1
7x  2(3x  1)  4
7x  6x  2  4
x  2  4
x  2  y  3(2)  1
y  5
Solution : {( 2, 5)}
Example: Substitution
(Goal: replace one variable with the
set of another equation.)
Step 1: Look for a variable with a
coefficient of one.
Step 2: Move everything else to the
other side.
Step 3: SUBSTITUTE this expression
into that variable in
Equation B
Step 4:
Solve for the remaining
variable
Step 5:
Back-substitute this
coordinate into Step 2 to
find the other coordinate.
(Or plug into any equation
but step 2 is easiest!)
 x  5y  8

2x  3y  3
A : x  5y  8  x  5y  8
B : 2(5y  8)  3y  3
10y  16  3y  3
13y  16  3
13y  13
y  1  x  5 y  8
Example: Substitution
(Goal: replace one variable
with an equal expression.)
Step 1: Look for a variable with a
coefficient of one.
Step 2: Isolate that variable
Step 3: SUBSTITUTE this
expression into that
variable in Equation B
Step 4:
Solve for the remaining
variable
Step 5:
Back-substitute this
coordinate into Step 2
to find the other
coordinate. (Or plug
into any equation but
step 2 is easiest!)
 x  5y  8

2x  3y  3
A : x  5y  8  x  5y  8
B : 2(5y  8)  3y  3
10y  16  3y  3
13y  16  3
13y  13
y  1  x  5y  8
x  5(1)  8  3
Solution :{(3,1)}
Method 2 Summary:
Substitution Method
(Goal: replace one variable with an equal
expression.)
Disadvantages:
Avoid this method when it requires messy fractions  Avoid IF no
coefficient equals one.
Advantages:
This is the algebra method to use when degrees of the equations
are not equal.
Method 3: Elimination Method
or Addition/Subtraction Method
(Goal: Combine equations
to cancel out one variable.)
 5x  3y  5

3x  2y  16
Step 1: Look for the LCM of the coefficients on either x or y. (Opposite
signs are recommended to avoid errors.)
Here: -3y and +2y could be turned into -6y and +6y
Step 2: Multiply each equation by the necessary factor.
Equation A now becomes: 10x – 6y = 10
Equation B now becomes: 9x + 6y = -48
Step 3: ADD the two equations if using opposite signs (if not, subtract)
Step 4: Solve the equation.
Step 5: Back-substitute this coordinate into any equation to find the other
coordinate. (Look for easiest coefficients to work with.)
Method 3: Elimination or Addition/Subtraction Method
(Goal: Combine equations
to cancel out one variable.)
Step 1: Look for the LCM of the coefficients
on either x or y. (Opposite signs
are recommended to avoid
errors.)
Here: -3y and + 2y could be
turned into -6y and + 6y
Step 2: Multiply each equation by the
necessary factor.
A becomes: 10x – 6y = 10
 5x  3y  5

3x  2y  16
 10x  6y  10
+
9x  6y  48
B becomes: 9x + 6y = -32
Step 3: ADD the two equations if using
opposite signs (if not, subtract)
Step 4:
Solve for the remaining variable
Step 5:
Back-substitute this coordinate
into any equation to find the other
coordinate. (Look for easiest
coefficients to work with.)
19x  38
x  2  Are we done ?
Method 3: Elimination or Addition/Subtraction Method
 5x  3y  5

3x  2y  16
(Goal: Combine equations
to cancel out one variable.)
Step 1: Look for the LCM of the coefficients
on either x or y. (Opposite signs
are recommended to avoid
errors.)
+
Step 2: Multiply each equation by the
necessary factor.
Step 3: ADD the two equations if using
opposite signs (if not, subtract)
Step 4:
Solve for the remaining variable
Step 5:
Back-substitute this coordinate
into any equation to find the other
coordinate. (Look for easiest
coefficients to work with.)
 10x  6y  10

9x  6y  48
19x  38
x  2  5x  3y  5
5( 2)  3y  5
 10  3y  5
 3y  15
y  5
Solution :{( 2, 5)}
Example: Elimination or Addition/Subtraction Method
(Goal: Combine equations
to cancel out one variable.)
Step 1: Look for the LCM of the coefficients
on either x or y. (Opposite signs
are recommended to avoid
errors.)
Step 2: Multiply each equation by the
necessary factor.
+
Step 3: ADD the two equations if using
opposite signs (if not, subtract)
Step 4:
Solve for the remaining variable
Step 5:
Back-substitute this coordinate
into any equation to find the other
coordinate. (Look for easiest
coefficients to work with.)
 2x  3y  2

5x  7y  34
Example: Elimination or Addition/Subtraction Method
(Goal: Combine equations
to cancel out one variable.)
Step 1: Look for the LCM of the coefficients
on either x or y. (Opposite signs
are recommended to avoid
errors.)
+
Step 2: Multiply each equation by the
necessary factor.
Step 3: ADD the two equations if using
opposite signs (if not, subtract)
Step 4:
Step 5:
Solve for the remaining variable
Back-substitute this coordinate
into any equation to find the other
coordinate. (Look for easiest
coefficients to work with.)
 2x  3y  2

5x  7y  34
10x  15y  10

10x  14y  68
29y  58
y  2  2x  3(2)  2
2x  6  2
2x  8
x  4
Solution :{( 4, 2) }
Method 3 Summary:
Elimination Method
or Addition/Subtraction Method
(Goal: Combine equations to
cancel out one variable.)
Disadvantages:
Avoid this method if degrees and/or formats of the equations do not match.
Advantages:
Similar to getting an LCD, so this is intuitive, and uses only integers until
the end of the problem.
Three Methods
Method 1: Graphing Method
Method 2: Substitution Method
Method 3: Elimination Method
or Addition/Subtraction Method
 x  3y  11 8x  3y  11
 x  y 1
A. 
B. 
C. 
2
7x  4y  6  7x  4y  6
7x  4y  6
Factoring
1.
2.
3.
4.
5.
6.
Factor GCF  for any # terms
Difference of Squares  binomials
Sum or Difference of Cubes 
binomials
PST (Perfect Square Trinomial) 
trinomials
Reverse of FOIL  trinomials
Factor by Grouping  usually for 4
GCF
3x2-9x2
3x2(1-3)
Take out what the two side share
in common to simplify.
GCF
3(1-6)2+6(1-6)2
3(1-6)2(1+3)
You can take out a glob and then combine with the other globs.
Hint: Remember that a “glob”
can be part of your GCF.
Difference of
Squares
25x2-4x2
(5x-2)(5x+2)
Recall these binomials are called conjugates.
IMPORTANT!
Remember that the difference
of squares factors into
conjugates . . .
The SUM of squares is PRIME –
cannot be factored.
2
2
a + b  PRIME
2
2
a –b  (a + b)(a – b)
Sum/Difference of
Cubes
x3-y3
(x-y)(x2+xy+y2)
Sum/Difference of Cubes
3
x
-
(x - y) (
Cube roots w/ original
sign in the middle
3
y
)
Sum/Difference of
Cubes
3
x
(x - y)
-
2
(x
3
y
2
y )
+
Sum/Difference of
Cubes
3
x
(x - y)
2
(x
-
3
y
+ xy +
The opposite of the product
of the cube roots
2
y )
Sum/Difference of
Cubes
(x - 2)
3
x
-8
2
(x
+ 2x + 4)
Cube roots of each Squares of those cube
roots &
Special Casest
1 step:
*
Diff of Squares
nd
* 2 step: Sum/Diff of
Cubes
6
6
x – 64y
(
(
)(
)(
)
)
)(
)(
Special Case
* 1ststep: Diff of Squares
* 2nd step: Sum/Diff of
Cubes
6
6
x – 64y
3
(x
(
)
)(
3
3
3
– 8y ) (x + 8y )
)(
)(
Special Case
* 1ststep: Diff of Squares
* 2nd step: Sum/Diff of
Cubes
6
6
x – 64y
3
(x
3
3
3
– 8y ) (x + 8y )
(x–2y)(x2+2xy+4y2) (x+2y)(x22xy+4y2)
PST
x2-10x+25
(x-5)(x-5)
(x-5)2
Recall PST test:
If 1st & 3rd terms are squares and the
middle term is twice the product of
their square roots.
PST
2
x -24+16
(3x-4)(3x-4)
Conjugates
Reverse FOIL (Trial
& Error)
12x2-45x+42
(3x-6)(4x-7)
Conjugates
Reverse FOIL
(Trial & Error)
Hint: don’t forget to read the “signs”
ax2 + bx + c  ( + )( + )
ax2 – bx + c  ( – )( – )
ax2 + bx – c  ( + )( – )
positive product has
larger value
ax2 – bx – c  ( + )( – )
negative product has
larger value
Example 11: Factor by
Grouping
(4 or more terms)
a(x-y)+x-y(x-y)
3
(x-y) (a)
Example 11: Factor by
Grouping
(4 or more terms)
8x-2y+16x-4y
2(4x-y)+4(4x-y)
(4x-y)2(2+4)
2X2
Factor by Grouping 3
X1
2
2
x +xy+y -4x2
(x-y)(x-y)-4x2
[(x-y)-2x][(x-y)-2x]
• 3X1
(PST)
Factor by Grouping 3
X1
x2-10+25-4x2
(x-5)(x-5)-4x2
[(x-5)-2x][(x-5)-2x]
Rational Expressions
Rational Numbers



Thinking back to when you were dealing with whole-number fractions, one of the first
things you did was simplify them: You "cancelled off" factors which were in common
between the numerator and denominator. You could do this because dividing any
number by itself gives you just "1", and you can ignore factors of "1".
Using the same reasoning and methods, let's simplify some rational expressions.
Simplify the following expression:



To simplify a numerical fraction, I would cancel off any common numerical factors.
For this rational expression (this polynomial fraction), I can similarly cancel off any
common numerical or variable factors.
The numerator factors as (2)(x); the denominator factors as (x)(x). Anything divided
by itself is just "1", so I can cross out any factors common to both the numerator and
the denominator. Considering the factors in this particular fraction, I get:



Then the simplified form of the expression is:
Additoin and Subtraction of
Rational Numbers

3
20

+
5
20
=
8
20
=
(2)(4)
(4)(5)
=
2
5
Notice the steps we have done to solve this
problem. We first combined the numerators since the
denominators are the same. Then we factored both the
numerator and denominator and finally we cross
cancelled.
Multiplication And Division of
Rational Numbers

x2 - y2
(x - y)2
is a rational expression.
To simplify, we just factor and cancel:

(x - y)(x + y)
(x - y)2
=
x +y
x-y
Quadratic Equations
Quadratic Equations



A quadratic equation is an equation that can be
written in this form.
ax2+bx+c=0
The a,b, and c here represent real number
coefficients. So this means we are talking about
an equation that is a constant times the variable
squared plus a constant times the variable plus a
constant equals zero, where the coefficient a on
the variable squared can't be zero, because if it
were then it would be a linear equation.
Completing the Square













a² + 2ab + b²=(a + b)².The technique is valid only when 1 is the
coefficient of x².
1) Transpose the constant term to the right:x² + 6x = −2
2) Add a square number to both sides. Add the square of half the
coefficient of x. In this case, add the square of 3:x² + 6x + 9 = −2 + 9.
The left-hand side is now the perfect square of (x + 3).
(x + 3)² = 7.
3 is half of the coefficient 6.
This equation has the form
a² = b which implies a = ± .
x = −3 ± .
That is, the solutions to
x² + 6x + 2 = 0
are the conjugate pair,
−3 + , −3 − .
Therefore,x + 3 = ±
Quadratic Formula
Quadratic Formula

on multiplying both c and a by 4a, thus making the denominators
the same (Lesson 23),




This is the quadratic formula.
Discriminant




The radicand b² − 4ac is called the
discriminant. If the discriminant is
a) Positive:The roots are real and
conjugate.
b) Negative: The roots are complex and
conjugate.
c) Zero:The roots are rational and equal
-- i.e. a double root.
Factoring




Problem 2. Find the
quadratic by factoring.
a) x² − 3x + 2
(x − 1)(x − 2)
x = 1 or 2.
roots of each
b) x² + 7x + 12
(x + 3)(x + 4)
x = −3 or −4.
In the next slides you will review:
Functions
D. Quadratic functions –
explain everything we know about how to
graph a parabola
Functions

A function is an operation on numbers of some
set (domain) that gives (calculates) one
number for every number from the domain. For
example, function 3x is defined for all numbers
and its result is a number multiplied by 3. The
notation is y=f(x). x is called the argument of
the function, and y is the value of the function.
The inverse function of f(x), is a different
function of y that finds x that gives f(x) = y.
Functions

















We have seen that a function is a special relation. In the same sense, real function is a special function. The special about
real function is that its domain and range are subsets of real numbers “R”. In mathematics, we deal with functions all the
time – but with a difference. We drop the formal notation, which involves its name, specifications of domain and codomain, direction of relation etc. Rather, we work with the rule alone. For example,
f(x) =x2+2x+3
This simplification is based on the fact that domain, co-domain and range are subsets of real numbers. In case, these
sets have some specific intervals other than “R” itself, then we mention the same with a semicolon (;) or a comma(,) or
with a combination of them :
f(x) =\
(x+1) 2−1;x<−2,x≥0
Note that the interval “ x<−2,x≥0 ” specifies a subset of real number and defines the domain of function. In general,
co-domain of real function is “R”. In some cases, we specify domain, which involves exclusion of certain value(s), like :
f(x) =
11−x,x≠1
This means that domain of the function is R−{1{ . Further, we use a variety of ways to denote a subset of real numbers
for domain and range. Some of the examples are :
x>1: denotes subset of real number greater than “1”.
R−{0,1{ denotes subset of real number that excludes integers “0” and “1”.
1<x<2: denotes subset of real number between “1” and “2” excluding end points.
(1,2]: denotes subset of real number between “1” and “2” excluding end point “1”, but including end point
“2”.
Further, we may emphasize the meaning of following inequalities of real numbers as the same will be used frequently for
denoting important segment of real number line :
Positive number means x > 0 (excludes “0”).
Negative number means x < 0 (excludes “0”).
Non - negative number means x ≥ 0 (includes “0”).
Non – positive number means x ≤ 0 (includes “0”).
Quadratic Functions




A quadratic function is one of the form f(x) = ax2 + bx + c, where a,
b, and c are numbers with a not equal to zero.
The graph of a quadratic function is a curve called a parabola. Parabolas
may open upward or downward and vary in "width" or "steepness", but
they all have the same basic "U" shape. All parabolas are symmetric with
respect to a line called the axis of symmetry. A parabola intersects its
axis of symmetry at a point called the vertex of the parabola.
You know that two points determine a line. This means that if you are given
any two points in the plane, then there is one and only one line that
contains both points. A similar statement can be made about points and
quadratic functions.
Given three points in the plane that have different first coordinates and do
not lie on a line, there is exactly one quadratic function f whose graph
contains all three points. To graph simplify (see quadratic equations) and
plot the points.
Simplifying expressions with
exponents
Simplifying Exponents

Use the Power of a Power Property, the
Product of a Power Property, the Quotient
of a Power Property, the Power of a
Quotient Property, the
Simplifying exponents
(4
Use These
Simplifying Exponents

32
(x )  x
2 4
315=317
a 1
0
(4  5)  4  5
2
6
8
2
4 8
8
1
1
6  2 
6
36
2
2
8
2
70  1
2
2
4
4
 
   2
5
5
Simplifying expressions with
radicals
Simplifying radicals

When presented with a problem like 4 , we don’t have too much difficulty saying
that the answer 2 (since 2x2=4 ). Our trouble usually occurs when we either can’t
easily see the answer or if the number under our radical sign is not a perfect square or
a perfect cube.
A problem like 24 may look difficult because there are no two numbers that multiply
together to give 24. However, the problem can be simplified. So even though 24 is not
a perfect square, it can be broken down into smaller pieces where one of those pieces
might be perfect square. So now we have 24  4  6  2 6

Simplifying a radical expression can also involve variables as well as numbers. Just as
you were able to break down a number into its smaller pieces, you can do the same
with variables. When the radical is a square root, you should try to have terms raised
to an even power (2, 4, 6, 8, etc). When the radical is a cube root, you should try to
have terms raised to a power of three (3, 6, 9, 12, etc.). For example,
http://www.algebralab.org/l
essons/lesson.aspx?file=Alg
ebra_radical_simplify.xml
x3  x 2  x  x x
Simplifying Radicals

Use the root of a power, power of a root,
product of a root, and quotient of a root
properties to solve.
In the next slides you will review:
Minimum of four word problems of
various types. You can mix these in
among the topics above or put
them all together in one section.
(Think what types you expect to
see on your final exam.)
Suppose that it takes Janet 6 hours to paint her
room if she works alone and it takes Carol 4 hours
to paint the same room if she works alone. How
long will it take them to paint the room if they
Click to see answer.
work together?

First, we will let x be the amount of time it takes to paint the room (in
hours) if the two work together.






Janet would need 6 hours if she did the entire job by herself, so her
working rate is 16 of the job in an hour. Likewise, Carol’s rate is 1 of
the job in an hour.
4
x
1 x

6 6
x
1 x

4 4
In x hours, Janet paints
of the room and Carol paints
of
the room. Since the two females will be working together, we will add
the two parts together. The sum equals one complete job and gives us
the following equation:
x x
 1
6 4
Continued answer
x x
 1
6 4
x
 x
12     12     12 1
6
4
2 x  3 x  12
5 x  12
12
2
x
 2 hours
5
5

Multiply each term of
the equation by the
common denominator
12
Simplify
Collect like terms
Solve for x
Suppose Kirk has taken three tests and made 88, 90, and 84. Kirk’s
teacher tells the class that each test counts the same amount. Kirk
wants to know what he needs to make on the fourth test to have an
overall average of 90 so he can make an A in the class.
88  90  84  x
 90
4
88  90  84  x  360
262  x  360
Steps
x  98
88  90  84  98 360
Check :

 90
4
4
Suppose a bank is offering its customers 3%
interest on savings accounts. If a customer
deposits $1500 in the account, how much interest
does the customer earn in 5 years?




I is the amount of interest the account
earns.
P is the principle or the amount of
money that is originally put into an
account.
r is the interest rate and must ALWAYS
be in a decimal form rather than a
percent.
t is the amount of time the money is in
the account earning interest.
I  P r t
I  (1500)(.03)(5)
I  225

If we want to find out the total amount in the
account, we would need to add the interest to
the original amount. In this case, there would be
$1725 in the account
The Smith’s have a rectangular pool that measure 12 feet
by 20 feet. They are building a walkway around it of
uniform width.

The length of the larger rectangle is
, which simplifies to


Length larger rectangle =


The width of the larger rectangle is
, which simplifies to


Width larger rectangle =


The area for the larger rectangle then becomes


Area larger rectangle =


The pool itself has an area of
square feet
Answer continued
(20  2 x)(12  2 x)  68  240
(2 x  20)(2 x  12)  308
4 x 2  24 x  40 x  240  308
4 x 2  64 x  68  0
4( x  17)( x  1)  0
x  17; x  1
x 1






Rearrange the terms for easier
multiplication and find the sum
of 68 and 240.
Multiply the binomials.
Combine like terms and
subtract 308 from each side.
Factor.
Solve each factor.
Since dimensions of a pool and
a walkway around a pool
cannot be negative
our answer is that the width of
the walkway is 1 foot.
Line of Best Fit or Regression Line
Line of best fit

A line of best fit is a straight line that
best represents the data on a scatter plot.
This line may pass through some of the
points, none of the points, or all of the
points.
Line of Best Fit


You can find the line of best fit by
estimation or by using graphing
calculators.
The line of best fit is good for estimating
the average of the points on the graph.
Line of best fit (how to solve)














1. Separate the data into three groups of equal size according to the values of the
horizontal coordinate.
2. Find the summary point for each group based on the median x-value and the
median y-value.
3. Find the equation of the line (Line L) through the summary points of the outer
groups.
4. Slide L one-third of the way to the middle summary point.
a. Find the y-coordinate of the point on L with the same x-coordinate as the
middle summary point.
b. Find the vertical distance between the middle summary point and the line by
subtracting y-values.
c. Find the coordinates of the point P one-third of the way from the line L to the
middle summary point.
5. Find the equation of the line through the point P that is parallel to line L.
Line of Best Fit

Try to find the line of best fit for the points
(3,9)(4,8)(6,6)(7,5)(9,7)(11,9)(13,12)(14,
17)(13,19).