Transcript Alegebra II - University High School

Algebra II
Chapter 1
By: Matt Raimondi
1-1 Expressions and Formulas
• Order of operations
– First, always try to simplify expressions in brackets, parentheses,
and fraction bars.
– Next, evaluate all numbers raised to a power or under a radical.
– Do all multiplication/ division from left to right.
– Then do all adding and subtracting from left to right.
– An easy way to remember this is by remembering:
– Please Excuse My Dear Aunt Sally
– Parentheses, Exponents, Multiplication, Division, Addition,
Subtraction
1-1 Example
•
2+5{2-3*2+(9-6)} – 32
2+5{2-3*2+(3)} – 32
2+5{2-6+(3)} – 32
2+5{-1} – 32
2+(-5) – 32
2-5-9
= -12
•
•
•
•
•
•
First we have to simplify the
brackets.
We subtract 9-6
Multiply 3*2
Evaluate what is left in the
brackets.
Multiply –1 by 5 to get rid of the
brackets.
Evaluate 32
Subtract 5 and 9 from 2.
1-1 Problems
• 1. 2*5-4+6/3
• 2. (12/4/3-1)100
• 3. 5-2(2+6/3)
• 4. 1+4*6+5-10/2
• 5. (5*2-5^2)+10
• 6. 2^2 +5*4/2-8
Answers:
1. 6
4. 25
2. 0
5. -5
3. -3
6. 6
1-2 Properties of Real Numbers
• Types of numbers
– Real Numbers – All numbers that you can encounter
in real life.
• Rational – Rational numbers can be expressed as
a fraction, repeating decimal, or a terminating
decimal. ie .5, ½, 1.33333333…..
• Irrational – Numbers that cannot be expressed like
rational numbers. ie π, √5, √2
– Integers – { ….-3,-2,-1,0,1,2,3…..}
– Wholes – {0,1,2,3,4……}
– Naturals – {1,2,3,4,5…..}
1-2 Properties cont.
Addition
Multiplication
Commutative
A+B=B+A
A*B=B*A
Associative
(A+B)+C=A+(B+C) (A*B)*C=A*(B*C)
Identity
A+0=A
A*1=A
Inverse
A+(-A)=0
If A ≠0
A* 1/A=1=1/A*A
Distributive
A(B+C)=AB+AC and (B+C)A=BA+CA
1-2 Examples
• Identify all of the groups the following numbers belong to.
• 1) √13
(reals and irrationals)
• 2) 7
(reals, rationals, integers, wholes, and naturals)
• 3) ½
(reals and rationals)
• 4) 0
(reals, rationals, integers, and wholes)
• 5) -1.978
(reals and rationals)
1 reals, rationals, integers, wholes, naturals
2)reals, rationals, integers 3)reals irrationals
4) reals, rationals
5) reals, rationals
6)reals, irrationals
7) reals,
rationals, integers, wholes, naturals
• 7) √9
• 1) 28
2) -1
3) π
• 4) 2.79
5) 1/3
6) √12
• Identify all of the groups the following numbers belong to.
1-2 Problems
1-2 Problems cont.
• Identify the following properties
5+2=2+5
(2*12)*3=2*(12*3)
4*1=4
5(a+b)=5a+5b
6+0=6
½ * 2=1
10*5=5*10
15+ (-15)=0
2) Associative
6) Multiplicative inverse
3) Multiplicative identity
7) Commutative
4) Distributive
8) Addition inverse
1)
2)
3)
4)
5)
6)
7)
8)
1) Commutative
Addition identity
•
•
•
•
•
•
•
•
5)
1-3 Measures of Central Tendency
• The median of a set of numbers is the
middle term. If there is an even number of
terms, then add the two middle terms and
divide by two.
• The mode of a set of numbers is the term
that appears most frequently.
• The mean of a set of numbers is the
average of the set of numbers.
1-3 Example
• Use the following set of test scores to find the mean, median and
mode.
• 58,74,82,95,63,84,75,82,64,99,84,84,48
• To find the median we should first rearrange the scores in order to
see which term is the middle term. We then get
48,58,63,64,74,75,82,82,84,84,84,95,99
• The middle term is 82, so the median is 82.
• To find the mode, look and see which term appears the most. 84 is
the only score that appears 3 times. So, the mode is 84
• To find the mean, we need the algebraic average of the terms. So
we add up all of the terms and divide by how many there are. There
are a total of 13 terms. The total divided by 13 gives up
approximately 76%. So the mean of the test scores is 76%.
1-3 Problems
Find the mean
1)
3)
10,8,2,17,3
2)12,14,5,2,6,40,5
12,8,5,5,32,7,10,3,7,11
Find the mode
4) 12,2,3,10,17,19,2,14 5) 3,10,9,3,3,10,4,18,10,9,3,16
Find the median
50,78,58,90,90,63,84,71,70,100
99,75,50,86,88,63,64,70,76,99,73
•
1) 8
2) 12
3) 10
4) 2
5)3
6) 74.5
7) 75
6)
7)
1-4 Practice Equations
Evaluate the following:
Solve for the unknown variable:
1) X=5
9x+7-3x
6) 30-10x=0
2) x=2
2-x+10x
7) 2x+5=-15
4) x=9
(x+3)/(x-5)
8) 2x-8=5x+19
9) 6w-2(2-w)=5
5) x=2
y=4
xy+3y-2x
10) 2x+.5=20-9x
1) 37
2)20
8)x=-9
3)51
9)w=9/8
4)3
3) y=3
y(2+5y)
5)16
6)x=3
10)x=20.5/11=41/22
7)x=-10
1-5 Solving Absolute Value
Equations
• The absolute value of a number is the number
of units it is from zero on the number line.
– For example, the absolute value of -3 is 3 because it
is 3 units away from zero on the number line.
– The absolute value of positive numbers and zero is
always the number.
– The absolute value of negative numbers is the same
as the number without the negative sign.
– The absolute value is never a negative number. This
is because a distance cannot be negative.
1-5 Solving Absolute Value
Equations
Look at the absolute values of the following •
numbers:
Abs(2)=2
Abs(17.94)=17.94
Abs(-1/2)=1/2
Abs(-90)=90
Abs(0)=0
Abs(22.2)=22.2
Abs(-x)=x
Lets take a look at an example
equation:
Abs(x+3)=5
X+3=5 and
x+3=-5
-3 -3
-3 -3
X=2 and
x=-8
Now to check it we test each
solution
Abs(2+3) and abs(-8+3)=5
Abs(5)=5 and abs(-5)=5
Since both of the equations are
true, 2 and –8 are both
answers.
1-5 Problems
•
Solve the equations
1)
Abs(x)=4
5)
Abs(-10)=x
2)
Abs(x-1)=12
6)
Abs(3x)=6
3)
Abs(x+4)=9
7)
4)
Abs(x)+3=7
Explain why this equation
cannot be solved.
Abs(x+2)=-7
1)-4,4
2)13,-11
3)5,-13
4)-4,4
5)10
6)2,-2
7)the absolute value can never be a negative number
1-6 Solving Inequalities
• Solving inequalities is not much different than solving
regular equations.
• Numbers can be added and subtracted the same as in
equalities.
• Multiplication and division is the same with only one
exception. Whenever a term is multiplied or divided by a
negative term, it reverses the sign. So, a greater than
sign turns into a less then sign.
• After solving an inequality, make sure to test your
answer and see if it is true. Errors often occur when
changing signs.
1-6 Examples
1)
5x-5 < 10
5x < 15
x<3
5(0) -5 <10
-5 <10
2)
14 < 1-3x
13 < -3x
-13/3 > x
14 < 1-3(-10)
14 < 1+30
14 < 31
• After getting x is less than
three, we should check it with
a value less than three. Zero
is an easy number to use.
• As you can see, zero is a
possible solution to the
inequality.
• This equation has a step with
division by a negative number.
Notice the sign change. Lets
try –10 to check it.
1-6 Problems
1)
5+x < 2
2) 7x+2>16
3)
5-x < -12
4) 14+3x < -8
5)
3x+9-5x > -41
6) 2x+17 < 53
2) x >2
Solve the following inequalities
1) x <-3
•
3) x > 17
4) x < -22/3 5) x < 25
6) x < 18
1-6 Graphing Inequality Examples
•
•
•
•
•
After finding the solution to an
inequality, graphing generally
follows. Graphs of inequalities only
lie on the real number line.
Greater than and less than
statements have open circles
because they do not include the
solution.
Greater than or equal to and less
than or equal to statements have
closed circles because they include
the solution.
A little trick to remember which way
the arrow goes is to make sure
your variable is on the left side.
Then just look which way the point
of the inequality sign points and the
graph goes that way.
For example lets look at
x > -2 and x ≤ 4
-10 -8
-10
-9
-8
-7
-10 -8
-6 -4
-6
-5
-4
-3
-6 -4
-2
-2
-1
-2
2
1
2
3
2
4
4
5
6
6
4
7
8
8
6
9
10
10
8
10
1) Solid @ 4, line to the right
2)solid @ -1, line to the left
3) open @ 5,
line to the left
4) open @ -3, line to the left
5) open @ 6, line to the
left
6) solid @ -4, line to the right
•
1)
2)
3)
4)
5)
6)
Solve and graph the following:
x≥4
x ≤ -1
x<5
-3 > x
2x-5 < 7
12 +x ≥ -3x - 4
1-6 Graphing Inequality Problems
1-7 Solving Absolute Value
Inequalities
• When you solve an absolute value
inequality, you will end up with two parts to
your answer.
– If the inequality began as a “greater than”
statement, you will have an “or” statement in
your final answer.
– If it began as a “less than” statement, you will
have an “and” statement in your final answer.
1-7 Examples
Abs(x+2) < 8
X+2 < 8 and x+2 > -8
x <6
and x > -10
So x is greater than –10 and
less than 6
When an absolute value
inequality has an “and” statement
the solution is a contained set. In
the first example, x is between
–10 and 6.
Abs(x-5) > 12
X-5 > 12 or x-5 < -12
X > 17 or
x <-7
So x is less than –7 or greater
than 17.
On the other hand, when an
absolute value inequality is an
“or” statement, the set of numbers
is not contained. In the second
example, x could be less than –7
or greater than 17.
1-7 Graphing Absolute Value
Inequalities
• Graphing absolute value inequalities is the
same as graphing regular inequalities but
with two parts.
– If the solution is an “or” statement the graph
will have an open area in the middle that the
graph doesn’t touch. It will continue to the left
and right on the number line.
– When the solution is an “and” statement it will
have a closed set. The solution will be
between two numbers on the line.
1-7 Graphing Examples
•
•
•
•
Lets look at the first example from
earlier. We found the answer to
be:
x < 6 and x > -10
So this will be a closed set
between -10 and 6.
Now lets look at the second
example from before. We found
the solution to be:
X > 17 or
x <-7
So this will be an open set
containing numbers greater than
17 or less than -7.
-12
-8
-4
-8
-4
4
8
4
8
12
16
20
1-7 Problems
State if the following are “and” or “or” statements:
1) Abs(x+4) > 5
2) 2) Abs(2+x) ≤ 19
3) 3) Abs(x) >2
Solve the following inequalities
6) Abs(2x+2) ≤ 18
2) and
5) Abs(x-4) < 12
1) or
and x ≥ -10
3) or
4) x ≤-5/2 or x ≥5/2
4) Abs(4x) ≥ 10
5) x >-8 and x < 16
6) x≤8
Answers:
1) Open dots @ 1 and -9, open set
2) closed dots @ 17 and -21, closed set
3) open dots @ 2 and 2, open set
4) closed dots @ -5/2 and 5/2, open set
5) open dots @ -8 and 16, closed set
6) closed dots @ 8 and -10, closed set
• Graph the problems from the last page
1-7 Graphing Problems