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>> Hello there.
This is Professor Shannon
Gracie and we are now
on section 2.3,
Solving Linear Equations.
These videos are a guided,
you know, notebook
for the blitzer introductory
and intermediate algebra combo
text and here we go.
Today we'll be going over how
to solve linear equations,
solve linear equations using
fractions, identify equations
with no solution
or infinitely many solutions
and then use these scales
to solve applied problems.
So here we go.
Why don't you guys go ahead
and attack that warm up
and see how you do.
Pause the movie
and do the warm up.
On your mark, get set, go.
All right.
Let's see how you did.
We've got a coefficient
of negative 12 in front
of that z, so we are going
to need to multiply both sides
by negative 112 in order
to get a coefficient of 1
in front of the z.
So we'll end
up with 1z is equal
to negative 12
and that gives us a result
of z is equal to negative 12,
and we'll put
that in our roster notation
and call it good.
How'd you do?
Great. I knew you could do it.
Okay, so this next one I'm
thinking that I would
like the x's on the left side.
So if I were to add 7x
to both sides,
we will get a result
of 6 x's is equal
to zero x's plus 24.
That is equivalent
to 6x is equal
to zero plus 24.
6x is 24. Now we need
to multiple both sides by 16
in order to get a coefficient
of 1 in front of the x.
So we get 1x is 4 and x is 4.
All right.
Feel al warmed up?
Got the math juices flowing?
Good. So here we go.
We have solved many linear
equations at this point
in time and now what we're
going to do is just put some
steps down for, you know,
a procedure
for solving linear equations.
So first off,
we want to simplify the
algebraic expression
on each side.
So if you've got like terms
or something
that you can combine,
or if you need
to use the distributive
property, you would go ahead
and you would do that first.
Then you collect all the
variable terms on one side
and all the constant terms
on the other side.
Then we needed
to isolate the variables,
remember getting a 1 in front
of the variable was isolating
the variable, and solve.
And then it's always, always,
always a good idea
to check the proposed solution
and the original equation.
Now when I go
through these movies,
I'm going to do steps,
you know, 1 through 3 for you
and I'm going
to leave the checking
of the solutions to you.
So that will be left
to students.
All right.
So now it just is a matter
of practice.
You know what?
You have all
of the math skills needed
to solve these problems.
We covered
that in the first two sections
of the chapter.
So what we're going
to do now is you're going
to pause this movie
and you're going
to work the next six problems
and then check
and see how you did,
and see how I solved them.
All right.
So good luck.
[ Pause ]
>> Okay, let's see how this
went for you guys.
Now I am seeing here
for number one
that we have different ways a
person could go about this.
I see that we have a negative
1z and a positive 10z there.
Over here we have constants
so they're going to be
like terms.
So I'll get 9z minus 34.
So the blue is what happened
with the z's on the left side.
And that's going to equal
to 10z's minus 52.
So 2 minus 54 gave us the
negative 52.
And I'm going to condense it.
I'm going to subtract 10z
from both sides and add 34.
I decided I wanted my z's
on the left side.
So we get a result
of negative z plus zero is
equal to zero z minus,
is that going to be 18
or did I make up my own math?
I think we're good.
And it's negative
because the negative 52 has
the larger absolute value.
This will give us a result
of negative z is equal
to zero minus 18 negative z is
equal to negative 18.
And if I multiple both sides
by negative 1,
I'll get a positive 1 in front
of the z. I don't want the
opposite of z. I want z.
So z will equal to positive 18
and we're done.
How'd you do?
Awesome. I knew you could
do it.
Okay, so next step.
Why don't we go ahead
and do this one here
since it's on the same side
of the page, or the same page,
and then we'll go
down to the next items.
So what I'm seeing here is
something that we need
to use the distributive
law on.
So this gives us 3 times x
plus 3 times 2 is equal
to x plus 30.
So then we have 3x plus 6 is
equal to x plus 30.
Now again, make a decision,
what side do you want
everything to be on.
So I decided I wanted the x's
on the left.
So if I subtract x
from both sides
and I subtract 6
because I need to get the 6
out of here,
I'll be good to go.
Okay. This will give us a
result of 2x's plus zero is
equal to zero x's plus 24.
This gives us a result
of 2x is equal
to zero plus 24.
2x is equal to 24.
Now I need
to multiply both sides
by one half
to get a 1 coefficient
in front of the x. This gives
me 1 x is equal to 12,
and that's times that.
And x is 12.
How are we doing do far?
Good. All righty.
Scrolling down.
Number two, again we need
to apply the distributive law
and you got to be careful
with this minus here.
I think I'm going
to be very careful
and I'm going
to just distribute the 8
and then I'll put a grouping
symbol around it so I remember
to flip the signs.
So we're going
to get 20 is equal
to 44 minus,
going to put a grouping symbol
here and apply the
distributive laws.
I'm just doing it
where you see the yellow,
the positive 8.
So that'd be 8 times 2 minus 8
times x, all right.
So this is the same
as the yellow.
Now you could have distributed
a negative 8 through
and that would have worked
as well.
You just have to be careful
about that minus sign.
Now I'm going to go ahead
and work this inside.
We'll get 16 minus 8x's
and when I distribute the
minus, I'll get minus 16 minus
negative 8x, okay.
So this resulted in this.
>> So now we get 40--
well actually, we can go ahead
and do the 44 minus the 16
to get a result of--
is that 28 or did I make
up my own math?
I believe we got it.
28 plus 8x.
Now I've already got my x's
on that side so I'm going
to go ahead
and just take the constant
to the other side.
And we will get negative 8 is
equal to zero plus 8x's.
And I'll go up here.
This gives us a result
of negative 8 is equal to 8x.
In order to clear that 8 out,
I need to multiply
by one-eighth on both sides,
remember you got
to balance them out.
And then we get negative 1 is
equal to 1 x. Negative 1
equals x and that's what we
put in our roster notation.
How we doing so far?
You guys hanging in there?
Good. Okay, so number three,
again I see an area
where we need
to use the distributive law
and I'm going
to show you another way
to do it okay.
This is going to be
where we're applying the
negative through.
So we're going to go ahead
and do this whole negative 4
and apply it to both of those.
So we will get 5x minus 4
times x and then plus negative
4 times 9.
And that's equal
to 2x's plus 3.
So this here is equivalent
to that yellow
and you just need
to be really careful
about your negatives
if you do it that way.
So then we'll get 5x minus 4x
minus 36 equals to x plus 3.
5x minus 4x is 1x minus 36
equals 2x plus 3.
And then, I don't know,
just for the heck of it,
why don't we get the x's
on the right side?
So we're going
to be subtracting 1x
from both sides
and if I'm getting the x's
on the right,
I need to get the constants
on the left.
Whoopsie. So now we will have
zero x's minus 39 is equal
to 1x plus zero.
Zero minus 39 equals 1x.
Negative 39 equals x.
And that's what we put
in our roster notation,
negative 39 is the solution
to that equation.
Very good.
Okay, coming back
up to number 5.
Got a lot going
on in this guy.
So here's a case
where we're going to need
to use the distributive law
and over here we're going
to need to distribute the
minus through
those parenthesis.
So here we go.
2 times x minus 2 times 15
plus 3x is equal to--
now these parenthesis aren't
doing anything.
They're kind
of like a shower curtain
that's over a shower door.
It's just there for looks,
okay, to group those two
numbers, but there's nothing
that needs to be multiplied
through them.
So I'm going to go ahead
and remove those parenthesis
and then I'll apply the
distributive law
to this other area.
Do you see we'll have negative
9x plus 2 because the negative
distributes to both terms.
Okay. And here was the
equivalence for the yellow.
So this will give us 2x minus
30 plus 3x is equal to,
the 6 and the 2 can combine
to be 8, 4x minus 9x is
negative 5x.
I'll go ahead and get the--
well, let's see.
So we'll have 5x's combining
the 2x and the 3x minus 30 is
equal to 8 minus 5x.
I'll go ahead
and get these x's
on the left side this time
and the constants
on the right side.
Okay, so I'll need to add 30
to get rid of that one.
So I'll add 30 here.
And then I need to add 5x's
to zero out that.
So this will give us a result
of 10x's plus zero is equal
to 38 plus zero x's,
10x is equal to 38 plus zero,
10x is equal to 38, okay.
Now we will need
to multiply both sides
by one-tenth in order
to get a coefficient of 1
in front of our x.
So we will have 1x is equal
to 38 tenths
and that can be reduced.
2 goes into 38 19 times
and 2 goes into 10 5 times
and we're done.
19 fifths is our result.
I know that in your last math
class there was a lot
of emphasis placed
on changing improper fractions
to mixed numerals
but we're not going to worry
about that unless it makes
sense for an
application problem.
All right.
Okay. So here is number 6
on that example.
We've got 2 cases
on the right side
where we're going to need
to use the distributive law.
So we'll get 100 equals,
we're going
to get negative x minus a
negative 1 is plus 1.
And then plus 4 times x minus
4 times 6.
Here is the yellow,
the end result of the yellow,
and here is the blue.
And we'll get 100 equals
negative x plus 1 plus 4x
minus 24.
100 equals 3x's combining the
negative x and the 4x,
minus 23, combining the
positive 1
and the negative 24.
Now I'm going to go ahead
and get the constant
to the other side.
And we will have 123 is equal
to 3x plus zero.
123 is equal to 3x
and if we multiply both sides
by one-third, we'll get a 1
in front of that x. So 3 goes
into 123 41 times
and that's equal to 1x.
So 41 equals x,
and in our roster notation
we'll put 41.
That is the solution
to the equation.
Did you guys do all right
on these?
I know it's a lot
of information, but remember,
just take it step by step
and you can do it.
And with enough practice
and enough, you know,
time put in,
you can really do this.
I promise.
All right.
Okay. So here we go,
linear equations
with fractions.
Well, we've already dealt
with some of them, all right.
A lot of people believe
that equations are easier
to solve when they do not
contain fractions.
So this is going
to be something that,
you know, you'll have
to decide the best way
for you.
Personally,
I believe fractions are
your friends.
So if you want
to use fractions that's fine.
But you can remove them,
all right.
So to remove fractions,
we can multiply both sides
of the equation
by the least common
denominator of any fractions,
whoops, sorry,
in the equation.
And remember,
the LCD is the smallest number
that all denominators will
divide into.
This is often called clearing
an equation of fractions.
>> Okay. So here we go.
This is sort
of like solving linear
equations on steroids,
all right.
We've again, this is going
to be up to you how you do it.
Personally we've got,
in this case,
I'm going to tell you straight
up, I don't always think I'm
going to show you this skill
and you're going to have
to decide the best way for you
to solve these problems,
all right.
But here it goes, all right.
So using this skill,
we would need
to multiply both sides
of the equation here by 2
because 2 is the only
denominator other than,
you know, these have kind
of denominators of 1.
So this would give us 2 times
x over 2 plus 2 times 13.
It's important
to write the line
that I'm writing,
and then 2 times negative 22
is negative 44.
It's important
to show the distributive law
because the main mistake
that students make on problems
like this is they forget
to multiply the second term
by 2, all right.
So then 2 times x over 2,
the-- do you guys see the 2's
will divide out
and we'll get x plus--
well, I guess we could write
1x plus 26 is equal
to negative 44.
Subtracting 26
from both sides,
would yield a result 1x is
just x and then we'll have
plus zero.
In this one,
negative 44 decreased
by 26 is going
to give us negative 70.
So x is equal to negative 70
and we're done.
Whoops. I'll put it
in our roster notation.
Okay. All righty.
So many of you may have
noticed that problem,
it might have been easier
if we had first subtracted 13
from both sides
and then solved it using,
you know, our other methods,
but I just wanted
to show you how to do it
so you'd have an example,
okay.
Now I'm going
to show you this one.
This one's a little
more complicated.
So here in order
to clear the equation
of fractions,
we need to figure
out the smallest number
that 5, 2 and 6 all
divide into.
And what is that going to be?
Good. It's going to be 30.
So we're going
to multiply both sides
of the equation by 30.
Again, we're going
to need the distributive law
for these.
So this is 30 times z
over 5 minus 30 times one-half
is equal to,
now this one is just--
do you see that the 6 goes
into itself once
and into 30 5 times,
so we'll get 5z's.
And the result
of the distributive law is
shown here.
Okay, so here 5 goes
into itself once
and into 30 6 times.
And over here,
2 goes into itself once
and into 30 15 times.
So our next line will be 6z's
minus 15 is equal to 5z.
and then if we subtract 5z
from both sides and then,
you know, we would want
to add the 15 to both sides.
What we will get,
1z plus zero is equal
to zero z plus 15.
That will give us z is equal
to zero plus 15.
Z is equal to 15.
All right.
Okay. So now it is your turn.
You need to pause the movie
and you need to work numbers 3
and 4 and see how you do.
Okay. You can either clear the
fractions or not and--
but just make sure you get the
correct answer.
On your mark, get set, go.
You can do it!
All right.
Let's see how you did
on numbers 3 and 4.
Whoops. Okay.
So for number 3, I'm looking
at this 4, 3 and 12.
The smallest number
that they all divide into is--
you got it, 12.
So we need to use,
what was that law called?
Good, the distributive law
or property.
On that side we will get 12
times 3y's
over 4 minus 12 times
two-thirds is equal to--
now on this side 12 completely
divides out
and we'll just get 7.
And so here is the result
of using the distributive law.
All right.
And now 4 goes
into itself once
and into 12 3 times.
3 goes into itself once
and into 12 4 times.
So here we'll get 9y,
because we're doing 3 times 3y
is 9y, minus 8,
4 times 2 is 8, equals 7.
And then isolating the
variable term, we will add 8
to both sides.
9y plus zero is 15.
9y is 15. So now we need
to multiply both sides
by one-ninth in order
to get a coefficient of 1
in front of, whoops,
coefficient of 1 in front
of the y, the term with the y.
So we'll get 1y is equal to,
now let's see.
3 goes into 9 3 times.
Actually, you know what?
I'll do that in the next step.
I don't want you
to be confused.
So we'll get 15 ninths,
and we can reduce
that 15 ninths
to be five-thirds.
So y is equal to five-thirds.
Are you hanging in there?
See, fractions are
your friend.
They're not bad.
All right, number four.
Here we go.
In order to clear the
fractions here,
the least common denominator
would be 12, okay.
You'll be applying the
distributive law here.
We'll get 12 times x minus 2
all over 3 minus 12 times 4 is
equal to.
And over here,
4 goes into itself once and it
into 12 3 times
but you're going to have
to distribute the x plus 1
times the 3, okay.
So that's going to end
up being x times 3 plus 1
times 3.
That's the result of the blue.
>> And the yellow
of course was this
result here.
All right.
We can simplify that a bit.
3 goes into itself once,
into 12 4 times.
And so now we're going
to have 4 times x minus 2
minus 48 is equal
to 3x plus 3.
And now we got
to use the distributive
property yet again.
So we'll get 4 times x minus 4
times 2 minus 48 is equal
to 3x plus 3.
This is a result
of that distribution.
So then 4x minus 8 minus 48 is
equal to 3x plus 3.
Now we can combine
that negative 8 minus 48
to be 4x minus 56 is equal
to 3x plus 3.
I decided I want the x's
on the left side
so we'll subtract the 3x
from both sides and I got
to get rid of that 56.
So I'll add 56 to both sides.
And at the end of the day,
well, almost the end
of the day,
we get 1x plus zero is equal
to zero x's plus 59.
And this just gives us a
result of x is equal
to zero plus 59.
X is equal to 59.
And you box your final result.
Okay. Recognizing inconsistent
equations and identities.
This is very important.
What you got--
you have to be comfortable
enough with this topic
so that you know
when something looks strange
or doesn't look okay,
all right.
So here we go.
Let's check it out.
If you attempt
to solve an equation
with no solution,
or one that is true
for every real number,
you will eliminate
the variable.
If you keep working on it,
you will eliminate the
variable entirely.
All right.
An inconsistent equation
with no solution results
in a false statement
such as zero equals 5,
you know.
That's not true unless you
already went
and had a few margaritas
before you watched your
math movie.
All right.
Okay. An identity that is true
for all real numbers results
in a true statement
such as zero equals zero.
That is definitely true, okay.
So let's check it out.
Example 3,
why don't you guys--
well, we'll just do them
together, but if you'd
like to pause the movie
and work on them on your own,
that's totally fine.
So here do you see we need
to use our distributive law.
So we'll get 2 times x minus 2
times 5 equals 2x plus 10.
This is the result
of that distributive property.
Then this gives us 2 x's minus
10 is equal to 2 x's plus 10.
Now if I want to get the x's
on the left, I would need
to subtract 2x
from both sides.
And if I want
to get the constant
on the right, I would need
to add 10 to both sides.
This will give us a result
of zero x plus zero is equal
to zero x plus 20.
So then zero x,
remember is zero.
So we'll get zero is equal
to 20.
What do you think about that?
Is that true?
It's not true to me,
all right.
So here we go.
This is a false,
this is false.
All right.
So what that means is
that there is no number
that you could plug in for x
that would ever make this
statement true.
So the way
to say there's no number
that'll work is
to put empty roster brackets.
Now another alternative
notation that you'll see
is this.
This means the null set, okay,
or the empty set, okay.
Actually, they both mean that.
Okay. And so I'll take,
you know, either one of those.
All right.
So let's try the next one.
Notice that we're going
to need to apply the
distributive property there.
So we'll get 5x minus 5 equals
3x minus 7 plus 2 times x plus
2 times 1.
Here's the result
of the distributed property.
Cleaning it up,
we will get 2x plus 2
at the end.
And then we can combine some
like terms of the right side.
3x plus 2x is 5x.
Negative 7 plus 2 is
negative 5.
So you might be seeing
something interesting
but you can keep going
if you want.
If we want our variables
on the left, we would need
to subtract 5x
from both sides.
And if we wanted our constants
on the right,
we'd need to add 5
to both sides.
So this will give us the
results of zero x plus zero
equals zero x plus zero.
Zero equals zero.
Is that true?
You are right.
It's definitely true.
So what that means
in this case is
that it doesn't matter what
number you plug in for x,
it'll always work out.
It'll always be a solution.
So now the way
that we say this
in math ease is we write
a set.
We built a set.
It's called set
builder notation.
And you write the name
of the variable,
which is in this case,
is just x. A line like this,
which means such that.
And what is x?
X is a real number.
And yes, you have
to write that.
That is the answer.
Okay. And let me just make
it clear.
The such that is this line
right there stands
for such that.
All righty.
So let's look at number 3.
Number 3, I would go ahead
and subtract 3 from both sides
to make this a little bit
easier and we'll get x
over 2 plus 2x
over 3 plus zero is equal
to x plus zero
which will give us x over 2.
Actually, I'm going to--
oh, no.
That's fine.
X over 2 plus 2x
over 3 equals x. Now in order
to clear these fractions,
we need to multiply both sides
of the equation by 6.
>> And then again we'll be
distributing on the left side.
So we will get, let's see.
2 goes into itself, oh, no.
See that's not something I
want you guys to do.
It just ends up causing a lot
of heartache.
So we'll do it
at the next step.
So 6 times x
over 2 plus 6 times 2x
over 3 is going
to equal to 6x.
Sorry, I'm at the edge
of the page here.
And this, the left side,
that's the result
of the distribution.
So now we can say,
2 goes into itself once
and into 6 3 times.
3 goes into itself once
and into 6 twice
and we'll get 3 times x plus 2
times 2x is 4x,
and that's equal to 6x.
7x is equal to 6x.
Subtract 6x from both sides.
We will get 1x is zero x's.
So that gives us x is zero.
Now do you see how this
is different?
We got x is zero whereas
on the other ones we ended
up with a constant
on either side
of the equation.
It's okay if x is zero.
That means
that zero is the only number
that will make this a
true statement.
Whoops. All right.
Let's divide this up.
And I went a little
into this guy's space.
Okay, so let's check this out.
Now here, you know,
you can go through,
you can clear the fractions
but do you see
that they already have a
common denominator?
So if we subtract, you know,
x over 4 from both sides,
so I'm getting the x's
on the right side
of the equation, what happens?
Do you see that we're going
to get zero x
over 4 plus 3 is equal
to zero x over 4?
So that gives us zero plus 3
is zero.
3 is zero.
What do you think about that?
That's right.
It's not true unless you had
your margaritas, okay.
3 is certainly not equal
to zero.
So therefore what is
our answer?
Good, it's the--
I like using the empty set.
So it's the empty set.
All right.
Very good.
Okay, now we're
onto our last page
in this section.
Okay, so let's take a look
at an application.
So with this particular
application, the formula, p,
equals 15 plus 5 times d
divided by 11 describes the
pressure of sea water, p,
in pounds per square foot
at a depth
of d feet below the surface.
So in the first part,
let's get some room here,
the record held
for breath held diving
by Francisco Ferreras
in Cuba off Grand Bahama
Island on November 14,
1993 involved pressure
of 201 pounds per square foot.
So they're asking us what
depth, to what depth did
Francisco decent
on this venture.
He was underwater
for 2 minutes and 9 seconds.
So that's pretty crazy.
So remember,
your general formula here,
sometimes it's good
to go ahead and rewrite
that formula.
So p is 15 plus 5d over 11.
Whoop, sorry.
All right.
P is 15 plus 5 times d
over 11.
So let's see what information
we've got.
We've been given the
information
that we've got pressure, p,
of 201 pounds per square foot.
So here this p we'll plug
in 201 in for p
and then we'll, what we need
to do is isolate our d. Okay.
So to that end what we'll do
is we'll go ahead
and we'll subtract 15
from both sides.
And we get, let's see,
11 minus 5 is 6,
9 minus 1 is 8,
so 186 is equal to,
15 minus 15 is zero,
so then we'll get 5d over 11.
Now what we need
to do is we need
to multiply both sides
by 11 fifths because we need
to have a coefficient of 1
in front of that d. So here
when we do 11 times 186,
we're going to get, let's see.
Well, if you do it
on the calculator,
you're going to get 2,046,
all right, and that's going
to be over 5 equals 1 times d,
because these will all divide
out to be 1.
And then when we divide 2,046
by 5 we end up with 409
and 2 tenths is equal to d.
So let's put our answer
in words.
So Francisco dove to 409
and 2 tenths,
and the unit will be in feet.
All right.
We're all set.
So let's take a look
at our next part.
Okay. So this next part is
having us find out,
find the depth
where the pressure is 20
pounds per square foot.
So again, we'll put our
general formula for this set
that describes this equation.
And then the pressure given
this time is 20 is equal
to 15 plus 5 times d over 11.
So this is what we plugged
in here.
And again, you know,
if we subtract 15
from both sides,
we will get 5 is equal
to 5 times d over 11.
And then multiplying both
sides by 11 fifths,
nicely these divide
out this time.
So we will get 11 is equal
to d and let's just double
check our work.
I think everything is looking
good here.
Perfect. And put our
conclusion at a depth
of 11 feet.
The pressure is 20 pounds per
square foot.
And we're all set.
So don't forget
to do your homework
and have a fabulous day.