This is video number two for sections
3.5 and 11.1 --
solving inequalities.
So take a look at this first example --
we're being asked here to
solve this for x.  It kind of looks like
an equation,
and you know when you solve an equation
for x you're just trying to find
the value of x that makes the equation
true.  It's the same thing here -- I just
want to know what x can I plug in here
so that this will come out true.  When you
solve an equation
generally you're getting one answer, but
do you see here that
you're going to get a whole ton of
answers because
whatever you plug into this left side
for x,
it doesn't have to come out to be
perfectly equal to negative 1,
it just has to come out less than
negative 1.  Well I'm thinking there's
lots of ways I could do that.
For example, what if I picked
negative 6, and I plugged that in for x:  3
times negative 6 is negative 18,
negative 18 plus 5 is negative 13, and
negative 13
is certainly less, it's lower than
negative 1.  So I guess
x equals negative 6 is a solution to
this equation.
What about x equals negative 10?
Well 3 times negative 10 is negative 30,
plus 5 is negative 25, that's what i get
on the
left.  Is that less than negative 1?  Sure,
absolutely.
So you can see there's lots and lots of
x's
that could make this thing true.  Well I
don't want to sit here guessing and
checking
all day, you know, I want to know -- how can
I find them all in a systematic way?
And so what we really need to figure out
is --
do all the same tricks that we use for
solving equations, do all of those work
for these inequalities, can I do those
normal things?
Well let's test some of them out.  So
here's a here's a straightforward enough
inequality: 3 is less than 7.  I'm sure
you would agree with that.
What if I added 2 to both sides?
Can I legally do that, does that still
result
in a true statement?  That would make this
5 is less than 9 -- well that's still true,
so it looks like that's, that's no
problem.
OK, what if I took that
and I multiplied both sides by the same
number?  So I
took say, times 2, times 2 --
that would give me 6 here and 14 here.
That's still true,
six is less than 14, so apparently that
trick works.  So, so far so good, right -- I
mean it, it kind of looks like we're
we're safe, but watch
this.  So here's that same starting
inequality --
look what happens if instead of
multiplying by
a positive number, I multiply by a
negative number.  What if I multiply both
sides by negative 3?
Then this would say: negative 9
is less than negative 21.  Hold on, that's
not true.
On a number line negative 9 is not to
the left
of negative 21.  This is,this is not
correct.
The only way that I could make it
correct
would be to flip around the direction of
the inequality symbol -- this is certainly
true:  negative 9
is bigger than negative 21.  Well we have
just discovered the one thing that is
different
in solving inequalities versus solving
equations.  The answer to the question is --
do our regular algebra tricks work?  Yes,
but if you multiply or divide (it's the
same problem), if you multiply or divide
both sides by a negative you have to
flip around
the direction of the inequality symbol
in order for the statement to remain
true.  So again the key words in this
are "by a negative".  If you multiply or
divide by a positive,
that's fine, we tried that up here -- but
when you multiply or divide by a
negative,
you've got to flip around, you've got to
reverse the direction of that inequality.
So let's take this now and apply it back
to our
starting problem here.  So we were given
this inequality.
We want to solve for x -- we found a couple
of solutions over here, we want to find
all the solutions.  So let's try to do
that.
Well you just pretend this is an
equation--  you just do your regular
solving steps
with the exception of that one thing we
found.  So
of .course what am I going to do?  I'm
going to subtract 5 from both, both sides --
subtraction is just fine, no flipping of
the symbol necessary
when you do subtraction.  The only time to
be concerned is when you start
multiplying and dividing.  Like here I am
dividing --
is this a time where I need to flip
around the inequality symbol?
No I'm fine because I just divided by
a positive number -- 3 is positive, no
problem.
So this just stays x is less than
negative 2.
This by the way is the solution
to the problem.  We were looking for -- hey
what x could I plug in here
that would make a true statement?  And
we're saying -- you know what if you pick
any number that's lower than negative 2,
remember we had negative 6 and negative
10, if you pick any number like that
you're going to make a true statement.
That's the answer to the problem.
When you're doing the homework they may
ask you to draw a number line picture
of the answer as well -- we've practiced
that already.  I would
draw some scale and make a number line
and then I would be
coloring in all numbers below
negative 2.  There's no equals so I would
use an open circle, I'm not
including the negative 2.  And then I
would shade everything
to the left, everything lower than
negative 2.  So there would be a number
line picture.  Or
they might ask you to do an interval
notation
answer.  Well that's where you look at the
number
farthest to the left -- of course this
arrow means it goes on forever.
Since it's going on forever to the left
we call that
negative infinity.  So negative infinity,
all the way to the left, comma -- what's the
farthest red number to the right?  It's
negative 2.
Infinity always gets a parentheses;
numbers
only get a bracket if we're including
the end -- we're not including negative 2
and so we just use a parentheses.  There
would be the interval notation way of
expressing the answer.
OK now that we've got the basic idea,
let's, let's try another example.  This
example looks a little bit more
complicated --
all of these are going to work in a
similar manner.  You just pretend
that this is an equals sign -- you do
exactly the same steps that you would
otherwise do
unless you are multiplying or dividing
by a negative.
Let's see, where should I start?  I think
the first thing that strikes me on this
problem
is that that parentheses is not helping
me very much so
I think I will distribute the 3
to both parts.
And then the next thing that I'm not
really excited about is this fraction
here.  I think it's always nice to clear
the fractions
when you're solving an equation, or in
this case an inequality.
The LCD is definitely a 2, and so I would
have to go through and multiply both
sides by 2.  Or another way to say that is
multiply each term by two.  So
I see two terms on the left
and I see two terms on the right -- they
all need to be multiplied by two.  Now
anytime you're multiplying you got to
think about your exception, but
2 is positive so I'm safe, no problem
there.
Here the fractions cancel, I'm left just
with x --
1 times 2 is 2, and then on this side I'm
going to get
6 x plus 6.
OK and then I'm gonna continue
solving -- it looks like I need to move
some terms around now using
subtraction or addition.
One thing I will say is this -- you know
how with equations you really have the
choice whether you could put the x's on
the left or whether you could put the
x's on the right.
When you're doing an inequality I would
strongly advise you to always put your
x's on the left.
It's not that it's impossible to put
them on the right -- it's just
more difficult to read from right to
left when you're dealing with an
inequality, so
I think you will have more success if
you always keep your
x's on the left, which of course means
I'll have to move this
2 over to the right.
And then let's see what that gives me -- so
I've got negative
5x here, and on the
right side I've got 8.  So we're almost
done --
obviously divide both sides by negative
five and
I'm all finished.  And you know it's very
easy to miss this but here is the first
time in this problem
where my exception comes into play.  I
just divided
both sides by a negative number.  Well the
only way
this statement remains true then is if I
flip around the direction of the
inequality symbol -- so less than or equal
turns into
greater than or equal.  Five does not go
in evenly and so this is as good as I've
got:
x is bigger than or equal to negative
eight fifths.  What if i want to write
this in the other two ways, like what if
I want to draw a number line picture of
my answer?
Now the eight-fifths is not very helpful --
I'd probably want to either
grab my calculator and get a decimal
from that --
negative eight-fifths on the calculator
turns out to be negative 1.6,
or I could change it to a mixed number --
so five goes into eight once with three
left over,
so eight-fifths is the same as one and
three fifths.  Probably
one of these two things is helpful
because that tells me -- okay I need to go
between negative one and negative two.
I think the decimal is probably the most
helpful -- and let's see, negative one point
five would be exactly halfway, so
negative 1.6 is a little bit
past the halfway point, so somewhere like
right here.
I'm going to go ahead and I'm going to
put the closed
circle because I can tell from my answer
that I want to include
the negative eight-fifths.  Snd then I'm
going to shade
everything above, so negative
eight-fifths and
higher is what we got.  The interval
notation way of writing this would be
negative eight fifths,
is the number farthest to the left, comma
infinity,
to the right, bracket around the negative
eight fifths because it's included,
and infinity is not a number -- it always
gets a parentheses.
OK let's try one more example.
This final example is a little bit
different looking.
This one is called a compound inequality
because if you look at it you can see
there are actually two less thans.
Now wait a minute -- something in the
middle of two less thans, we've seen that
before.
You know what this is?  It's a between
inequality.
In fact you could even read this -- would
you try to figure out
when 3x plus 1 comes out
between negative 5 and 13?
I just want to know what x's will make
that happen.
Well the great thing here is you can
solve one of these
in the same way we were solving the
other ones.  You just have to do the same
step to all three sides,
so to speak.  So think of it as an
equation with three sides, even though
that sounds kind of funny.
As long as you do the same step, the same
algebra step, to all
three parts, all three sides, you're
legal.  So remember the goal is just to
get the x by itself.
So I'm concentrated here on the middle.
What would i do?  In some ways it's
even helpful, if you're really kind of
stumped, if you,
if you cross out this part with your
finger,
then it should be pretty obvious to you
what the solving step would be -- clearly
it would be
subtract 1 from both sides.  We're just
saying you're going to do that
to all three parts, so subtract 1 here,
subtract 1 here,
subtract 1 here,  What's that going to
give me?  Negative 6 --
the middle part is really what I'm
focused on -- that's now
3x, and of course this part over here is
12.
What would be my next step to get the x
by itself?  Well I'm going to have to
divide everybody by
three.  Again as long as you're going to
all three
sides, you're going to get the right
answer.  And you know, I think I'm done.
Here then is the final statement -- x is
now by itself,
it's an x surrounded by two less thans,
that is literally between.  This says x
is between negative two and four.
Apparently if you pick any x between
negative two
and positive four and you stick it up in
this original
statement, you will get a true statement.
Let's wrap up by writing this in each of
the other two ways.
The number line picture is pretty
straightforward -- I'm not going to have
any arrows on this one
because I'm only coloring in numbers
that are
in between negative two and four.  I do
have the equals sign,
so I'm including the negative two and
the four.
Remember this equals corresponds to
these closed circles.  And then I'm
coloring everything in between.
So here would be the number line picture.
The interval notation
would simply be number farthest to the
left,
comma number farthest to the right, and
then what do I want to around them?  Well
they're both included, so that's the
bracket -- the bracket
also goes with the equals sign
in the inequality symbol.  This would mean
all numbers
from negative 2 to 4 including the ends.
