>> Hi! This is Julie Harland
and I'm Your Math Gal.
Please visit my website
at yourmathgal.com
where you could search for any
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by topic.
We're gonna solve this system
of 3 linear equations
which has 3 variables, X,
Y and Z. What we're looking
for is an ordered pair X, Y,
Z, I don't know what the
numbers are yet,
that work in all 3 equations
to make it true.
So, this is a linear system
because there are no square
terms and we've got 3
equations and 3 variables.
Now to keep track of things,
I'm gonna call this equation
A, this equation B,
and this equation C. Now you
need to already be familiar
with how to solve the system
of linear equations
with 2 variables
where you can use the
substitution method
or the addition
or elimination method
to solve.
When you have 3 equations
and 3 unknowns often it's
easier to use the addition
method so that's what I'm
gonna use on this problem.
So here's the idea.
We want to just begin
by working with 2 equations
and try to eliminate 1
of the variables,
and then we're gonna work
with different 2 equations
and eliminate the
same variable.
Alright, so we kind of glance
at it and we decide
which variable do we want
to get rid of first?
Is there any one that,
you know, looks easier working
with any 2 equations
and there's a lot
of different ways to do this.
I'm gonna go ahead and work
with equations A and B
because I know
if I just add those together
that I'm gonna eliminate the Y
right off the bat.
But it wouldn't be that hard
to eliminate the X or the Z
in fact, so--
working with the equation A.
So it's-- it's really
up to you.
But I'm gonna go ahead
and say let's start off
with just equation A and B
and write the system,
the 2 equations down.
So, we have 3X plus 2Y plus C
equals 1.
And we've got X minus 2Y plus
2C equals 4.
And I don't have
to alter either these
equations by multiplying
either by anything to get any
of the variables
to be eliminated
because I have a 2Y
and a negatives 2Y,
so I'm gonna go ahead
and just add the 2 equations
and the Y value,
I mean the Y terms will
be eliminated.
So I've got 4X plus 3C
equals 5.
So what I've done is I've come
up with 1 equation
that has only 2 variables
in it.
Right, I've eliminated the Y.
Now to keep track
of this equation,
I'm gonna call this
one equation.
I have A, B and C,
I'm gonna call this equation D
so we could refer back
to it later.
So, now what I wanna do is
work with 2 different
equations, not A and B,
so either A and C or B and C
and do the same thing.
I want to work with M
but I wanna eliminate the
Y variable.
So I could either choose
to work with A and C
or I could work with B and C.
And it's really up to you
which one.
I'm gonna go ahead and work
with B and C, so let's see.
If I put B and C down,
so I'm gonna write
down what B is
and I'm gonna be right
down what C is.
So B again is the X minus 2Y
plus 2Z equals 4
and equation C was 2X plus 4Y
plus 3C equals 9.
It's not gonna work
to just add these 2 equations
together because the Y is not
gonna be eliminated,
you will get a 2Y.
So I need to alter it,
and let's see.
I'm trying to eliminate the Y
so if multiply the top
equation by 2,
I'll have a negative 4Y
and that will work out.
So, let's do that.
I'm gonna multiply 2 times,
all the terms and both sides
of the equal side
so that's gonna give you 2X
minus 4Y plus 4Z equals 8.
And I don't need
to alter the bottom equation
at all so I'm just gonna write
it again.
2X plus 4Y pus 3Z equals 9,
and go ahead
and add those 2
equations together.
So I have 4X plus 7Y--
I'm sorry, plus 7Z equals 17.
So, I had a new equation,
right?
Where this no Y term,
and I'm gonna call this
equation E. Alright,
now remember I wanna go back
up to my 2 new equations
that don't have a Y terminate
and that's equations D and E.
So now I'm gonna put those 2
together, D
and E. I think I'll use
another color
so it doesn't get so confusing
so D was 4X plus 3Z equals 5
and E was 4X minus 7Z,
I'm sorry, plus 7Z equals 17.
And now this is--
this is a system of equations
with only 2 variables
so we could use either the
substitution method
or the addition method.
And it looks
like the addition method is
gonna be the easy one
if I just multiply one
of these equations
by negative 1.
I'm going to get the Xs
to cancel out, alright?
So I could either multiply the
top or the bottom equation
by negative 1.
I like the variables,
the coefficient of variables
to be positive
so I'm gonna go ahead
and multiply the top
by negative 1.
That seems easier to me
and let's see what we've got.
We have negative 4X,
remember you have
to multiply negative 1
and by all terms
on both sides.
So the bottom equation I don't
need to change at all,
you know, that's only
because there was a 4X in it,
a 4X but, you know,
sometimes you have
to multiply both equations
by something.
In fact look what I did,
I forgot to put the minus sign
in front of 5.
As I was saying, don't forget,
there it is, okay.
So now, oh yeah,
those 2 equations,
and I have 4Z equals 12.
Okay, that's gonna be easy
to solve.
We're just gonna divide both
sides by 3, I'm sorry,
divide both sides by 4
and Z equals 3.
I've got at least hopefully
if I didn't make any mistakes,
one of the variables solved
for in my original equations,
so Z is going to be 3.
That means Z is the solution
in every equation that you see
up here so far, you know, A,
B, C, D and E. Now once you
know what Z is,
what we do is we go back
to one of the previous
equations we just worked
with to solve for the variable
that was there.
So I go back up to either D
or E, right?
One of these 2 equations
and variable.
And I could now plug in Z
to figure out what X is, okay?
So let's do that.
How about if I just take the
top equation D
which was 4X plus 3Z equals 5
and since I know now
that Z is 3,
I'm gonna play again 3 for Z.
So I'm gonna plug on 3
and now I may be able to solve
for X. So we have 4X plus 9
equals 5.
So, if I subtract 9
from both sides,
we get negative 4
and that's going
to give you X equals negative
1, and that's based here.
We divided by 4.
So now I know the value of X
and I know the value of Z.
So by the way,
I'm trying
to get my order triple X, Y,
Z. I know that Z is 3
and I know X is negative 1.
Unless, of course,
unless I made some mistake,
that's why we're gonna check
our answers in the back.
So all we have left
to do is solve for Y, right?
So now that we know X and Z,
we have to go back
up to be top,
in one of our originally
equations up here,
and pick one of this equations
to figure out what Y is.
And it doesn't matter
which equation we choose.
So let's go ahead
and choose equation A.
That looks like it's got,
you know, easy numbers in this
so I'm gonna pick A. Alright
so I write
down the equation A. And keep
in mind, I already know what X
and Z is, right?
That was my X,
that was my value,
oh sorry, for Z3.
And now I'm gonna plug those
values into these equations
and I'll be able to solve
for Y. So we've got 3 times
negative 1 plus 2Y
>> I don't know what Y is plus
Z. Okay Z was the 3 equals 1.
Alright, so I'm plugging
in my values for X and Z back
in the original equations
so I can figure out what Y is.
And now we go ahead
and just solve this equation.
So we've got negative 3 plus
2Y plus 3 equals 1.
So, that just gives me 2Y
is 1.
And then we divide both sides
by 2.
So Y is 1/2.
Okay, now some people think,
"Oh, I must have the
wrong answer.
I got a fraction."
No, don't be prejudice,
fractions are numbers too.
So, it looks
like that's my solution
for the Y. I say it looks
like because it's really,
really easy to make a mistake.
In fact I worked
on this problem
by myself before I decided
to make the video and I kept
on making stupid
arithmetic mistakes.
Very easy to just copy a
number down wrong
or make an arithmetic mistake.
So anyway that's my value.
So it looks to me
like the answer is negative 1,
1/3, 3.
Alright, and of course this
is, the solution
to the problem.
I'm gonna use solutions sets
or you maybe you could box
around your answer.
Now, we've got
to make sure that's correct.
So I need to check
that that is going
to make all 3
of my original equations true
when I plug in those values
for X, Y and Z.
And that's the last thing you
need to do is definitely check
your answers.
Alright, so here are our 3
equations, we wanna check
by plugging in these values
for X, Y and Z to both sides
of the equation.
So let's see, we're going
to have 3 times negative 1
plus 2 times 1/2 plus 3.
So on this side,
that's negative 3 plus 1 plus
3 which is 1, and I've got 1
on the other side so yup,
you're check in that equation.
Alright, let's check the
next one.
Hey, by the way,
feel free to put this on pause
and try it on your own.
So, plugging in negative 1
for X, plugging in 1/2 for Y,
plugging in 3 for Z.
Into the next equation,
I get negative 1 minus 1 plus
6 and that's negative 2 plus 6
is 4, okay.
Checked on that one, right?
So we go onto the next
equation, going to plug in.
Negative 1 for X,
one half for Y, 3 for Z
and negative 2 plus 2 plus 9,
get 9 on both sides.
That checks.
Awesome! So the solution,
checked on all 3,
so the answer is negative 1,
1/2, 3.
And this is not the only way
you could have solve this,
there's like about 10 ways
that you could solve
this problem.
[Laughter] Depending on,
use the elimination,
which equations you put
together first,
whether you think--
you could also use
substitution,
all sorts of different
ways this.
But this is just one way
to solve this problem.
Make sure you check
because it's so easy
to make a mistake,
and just be very careful
of your positive
and negative signs,
copy in the problem
down correctly, distributing
on both sides of equation.
All of those things,
it's really easy
with 3 variables
and 3 linear equations
to make a mistake.
You need to lots of paper
and lots of space,
lots of room,
lots of organization.
[ Pause ]
>> Please visit my web site
at yourmathgal.com
where you can view all
of my videos
which are organized by topic.