Transcript 11.2 Sets and Compound Inequalities 11.3 Absolute Value

11.2 Sets and Compound
Inequalities
11.3 Absolute Value Equations
11.4 Absolute Value Inequalities
11.2 Sets and Compound Inequalties
• Set – a collection or group of items.
– We usually name sets with capital letters
• Elements – the objects within each set
– If we are talking about general terms within a set we
usually use lower case letters to represent them
– When we list the group of elements we enclose them
in { } and separate each element by a comma. This
method is referred as ROSTER NOTATION.
This class as a set
• A = {Amanda, Kacy, Niki, Megan, Jon, Johnny,
Shawn, James}
• To show that an element is contained in a set
or that an object belongs to a set we use the
symbol


• Amanda
A, would be read “Amanda is an
element of set A.”
• Kyle
A, would be read “Kyle is not an
element of set A.”

Sets and Subsets
• A subset is a collection of elements that
create another set but all elements come
from a larger group.
• We must have a knowledge of the overall
group in order to determine a subset.
• If we know that A= {2, 4, 6, 8, 10}
• And we know that B = {4, 6, 8} then it is
said that B is a subset of A. The
notation/symbols used to talk about
• Subsets exist only if every element in the smaller set
exists in the larger set. “every set is actually a subset of
itself…not something this text gets into”
• If we can count the number of items in a set we say
that the set is FINITE, and the number associated with
the set is called the set’s CARDINALITY. If you cannot
count the items within the set then the set is said to be
an INFINITE set.
• If there is nothing in a set it is called an EMPTY or NULL
SET. This is denoted by the symbol ∅ “no braces”. The
empty set is considered an element of every set.
• Some sets that we should be familiar with.
• Natural Numbers (counting numbers)
– N = {1,2,3,4,5,…}
• Whole Numbers (same as N but start with 0)
– N0= {0,1,2,3,4,5,…}
• Integers (Whole numbers and their opposites)
– Z= {…-4,-3,-2,-1,0,1,2,3,4,…}
• Rational Numbers (any number written as a/b, b
cannot equal 0)
• Irrational Numbers (any number that does not
terminate or repeat in regards to decimal)
• Real Numbers denoted as R. Is the overall set
that all of the above belong to.
• We can do OPERATIONS with sets.
• The two operations that we use are union “∪” and
intersection “∩”.
• Union means that we will be taking 2 or more sets
and creating a new set, UNION refers to the new set
including all elements of the other sets.
– We will say A ∪ B, which means the new set will be comprised
of all elements that are either in A or in B. For an element to be in
A ∪ B then it has to be in A or in B.
– The key word is OR
• The other operation is an intersection “∩”
• Here we will be taking 2 or more sets and
creating a new set that is comprised of only the
elements that are in (shared by) the other sets.
So for an element to be in the new set A ∩B then
it must be in A and it must be in B. The key word
is AND
Examples
• Let A = {2, 4, 6} , B={0, 1, 3, 5} and C = {1,2,3,…}
•
•
•
•
a.) A ∪ B = {
b.) A ∩ B = {
c.) A ∩ C = {
d.) B ∪ C = {
A
2
4
8 11
A={
B={
AUB={
A∩B={
5
12
4
3
33 22
B
• Graphing compound inequalities.
• When the inequalities use the word or idea of
“and” our final solution will be where the two
individual graphs overlap.
• When the inequalities use the word or idea of
“or” we will graph both inequalities and our
final solution will be EVERYTHING that is
graphed.
Graph
{x | x  3 or x  2}
Now represent the answer in interval notation using
the correct union or intersection symbol.
Graph
{x | x  1 or x  4}
Graph
{x | x  1 and x  4}
Graph
{x | x  2 and x  3}
Graph
{x | x  3 and x  6}
Solving compound inequalites
• If the word “or” is used you will have two
separate inequalities to solve, simply solve
both and graph both on the same number
line.
4t  8  12 or 4t  8  12
• Usually when the word “and” is to be used in
the inequality you will not see it. The
inequality will be written as three parts.
• With problems like this you treat them just
like normal equations however we want to
isolate the variable in the center. What you do
to one part you have to do to all 3 parts.
5  2 x  15  3
11.3 Absolute Value Equations
• In general practices the absolute value of a
number is referencing the distance a number is
from 0. Distance always has to be positive.
• Thus when asking to solve |x|=5 for x.
• We are asking the question for what number
does x have to be so that it is a distance of 5 from
zero. In this case x could be 5 or -5.
-This idea of x having two results creates the steps
for us in order to solve absolute value equations
• We simply take the equation given and get rid
of the |abs| bars and solve the equation.
• But then we have to take the given equation
get rid of the |abs| bars, multiply one side by
a negative 1 and then solve that equation.
• Each absolute value equation will result in 2
other equations that we can solve. We cannot
solve with the |abs| bars remaining in the
equation they must be eliminated.
Solve
• |3a + 2| = 6
• This is saying that 3a+2 is a distance of 6 units from 0.
Solve
• |5x-3|+12 = 19
we must isolate the |abs| bars first
Solve
• |4x + 7| = -8
– Understanding the idea of absolute value this
would say that 4x+7 is a distance of -8 from 0.
Can distance be negative? So can this be solved?
– This is somewhat of a trick question that often
gets asked on tests/quizzes.
• When you have absolute value bars on both sides
of the equation you simply solve them the way
they are without the |abs| bars.
• Then create a second equation by removing the
bars from both sides and multiplying one side by
a negative 1 (make sure you distribute it)
• |a| = |b| this means that the values of a and b
are the same distance from 0, thus they must be
equal to one another or opposites of one another
Solve
• |6a + 4| = |4a +6|
Solve
2 1
5
x  x
3 2
6
11.4 Absolute Value Inequalities
• Every absolute value inequality needs to be
changed into two separate inequalities that do
not include the absolute value bars.
• IF |a|< 4 , this means that a is somewhere
between -4 and 4 on the number line.
• If |a| > 5, this means that a is somewhere to
the right of 5 or to the left of 5 on the number
line.
Solve and Graph
| 2 x  1| 7
• Here we are saying that 2x+1 is less than 7
units away from 0 so it could be to the left of
zero or to the right of zero. So it must appear
between -7 and 7 on the number line so we
transform the inequality into…
7  2 x  1  7
Solve |4a-3|<5
Solve |-8x+4|<12
Solve 4>|y-5|
• Things have flipped but if we re-write the
inequality so that |abs| comes first we can
better decide how to address this problem
An absolute value is greater than
• When asked to solve |x -2|> 6.
• We are stating that x-2 is further than 6 units
away from 0, and that can occur to the right of
zero and to the left of zero. So we address it as
two inequalities again.
• Take the current inequality without the |abs| and
solve.
• Repeat the process for the second inequality but
multiply the constant side by a negative one and
solve
• Combine your answers with the word “or”
Solve and Graph
| 2 x  9 | 11
Solve and Graph
1
3t   5
3
Solve and Graph
| 3a  9 | 2  14
Solve and Graph
| 2a  2 | 2  11
• After we isolate the |abs| we recognize that
the quantity is > -9. Ask yourself how often
will that quantity be greater than -9 units
away from zero. And this is your answer.
Summary of absolute value
inequalities
• Blue box on page 770 is a good graphic to
have memorized for a test or quiz if you
struggle with the concepts. I think it is much
more informative and useful to develop an
understanding of the concepts and then you
won’t forget something that you were
suppose to memorize. Memorization fails us
every once and a while.