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Systems of three variables - Math, Class 10 Video Lecture
FAQs on Systems of three variables - Math, Class 10 Video Lecture
| 1. What is a system of three variables in mathematics? |  |
Ans. A system of three variables in mathematics refers to a set of three equations that contain three unknown variables. The goal is to find the values of these variables that satisfy all three equations simultaneously.
| 2. How can I solve a system of three variables? | |
Ans. To solve a system of three variables, you can use various methods such as substitution, elimination, or matrix methods. One common approach is to use the method of substitution, where you solve one equation for one variable and substitute it into the other two equations. This reduces the system to two variables, which can then be solved using any suitable method.
| 3. Are there any special cases in solving systems of three variables? | |
Ans. Yes, there are special cases in solving systems of three variables. One special case is when the system has no solution, which means the three equations are inconsistent and cannot be satisfied simultaneously. Another special case is when the system has infinitely many solutions, indicating that the three equations are dependent and represent the same line or plane in space.
| 4. Can a system of three variables have a unique solution? | |
Ans. Yes, a system of three variables can have a unique solution. If the three equations are independent and represent three distinct lines or planes in space that intersect at a single point, then the system has a unique solution. In this case, each variable will have a specific value that satisfies all three equations.
| 5. When is it necessary to use matrices to solve a system of three variables? | |
Ans. Matrices are often used to solve systems of three variables when dealing with large systems or when using advanced methods such as Gaussian elimination or Cramer's rule. Matrices provide a concise and organized way of representing the coefficients and constants of the system, making it easier to perform row operations and solve for the variables. However, for smaller systems, other methods like substitution or elimination may be more convenient.
Text Transcript from Video
Solve this system.
And here we have three
equations with three unknowns.
And just so you have a
way to visualize this,
each of these equations
would actually
be a plane in three dimensions.
And so you're actually
trying to figure out
where three planes in
three dimensions intersect.
I won't go into
the details here.
I'll focus more
on the mechanics.
But you can imagine if I were
to draw three-dimensional space
over here.
Now, all of a sudden, it
will an x, y, and z-axes.
So you can imagine that maybe
this first plane-- and I'm
not drawing it the way
it might actually look--
might look something like that.
I'm just drawing
part of the plane.
And maybe this plane over here,
it intersects right over there,
and it comes popping
out like this.
And then it goes
behind it like that,
keeps going in every direction.
I'm just drawing
part of the plane.
And maybe this plane
over here, maybe it
does something like this.
Maybe it intersects
over here and over here,
and so it pops out like that.
And then it goes below it like
that, and it goes like that.
I'm just doing this for
visualization purposes.
And so the intersection
of this plane, the x, y,
and z-coordinates that
would satisfy all three
of these constraints
the way I drew them,
would be right over here.
So that's what
we're looking for.
And a lot of times, three
equations with three
unknown systems, they
will be inconsistent.
You won't have a solution here
because it's very possible
to have three planes that all
don't intersect in one place.
A very simple example
of that is, well, one,
they could all be
parallel to each other.
Or they could
intersect each other,
but maybe they intersect each
other in kind of a triangle.
So maybe one plane
looks like that.
Then another plane maybe pops
out like that, goes underneath.
And then maybe the
third plane cuts in.
It does something like this,
where it goes into that plane
and keeps going out
like that, but it
intersects this plane over here.
So you see, this kind
of forms a triangle,
and they don't all
intersect in one point.
So in this situation, you would
have an inconsistent system.
So with that out of
the way, let's try
to actually solve this system.
And the trick here is
to try to eliminate
one variable at a time
from all of the equations,
making sure that you have the
information from all three
equations here.
So what we're going to
do is we could maybe,
it looks like the
easiest to eliminate--
since we have a positive
y and a negative y,
and then another positive y--
it seems like we can eliminate
the y's.
We can add these two
equations and come up
with another equation that will
only be in terms of x and z.
And then we could use these
two equations to come up
with another equation that'll
only be in terms of x and z.
But it'll have all of the x
and z constraint information
embedded in it because we're
using all three equations.
So let's do that.
So first, let's add these two
equations right over here.
So we have x plus y minus
3z is equal to negative 10,
and x minus y plus
2z is equal to 3.
So over here, if we
want to eliminate y
we can literally just
add these two equations.
So on the left-hand side, x
plus x is 2x, y plus negative y
cancels out.
And then negative
3z plus 2z, that
gives us just a negative z.
And then we have negative 10
plus 3, which is negative 7.
So using these two
equations, we got
2x minus z is equal
to negative 7,
just adding these two equations.
Now let's do these
two equations.
And we can we
reuse this equation
as long as we're using
new information here.
Now we're using the
extra constraint
of this bottom equation.
So we have x minus y
plus 2z is equal to 3,
and we have 2x plus y minus
z is equal to negative 6.
And if we want to
eliminate the y's, we
can just add these
two equations.
So x plus 2x is 3x.
Negative y plus y cancels out.
2z minus z, well,
that is just z.
And that is going to be
equal to 3 plus negative 6
is negative 3.
So if I add these
two equations, I
get 3x plus z is
equal to negative 3.
Now I have a system of two
equations with two unknowns.
This is a little bit more
traditional of a problem.
So let me write them over here.
So we have 2x minus z
is equal to negative 7,
and then we have 3x plus
z is equal to negative 3.
And the way this
problem is set up,
it gets pretty
simple pretty fast,
because if we just add these two
equations, the z's cancel out.
Otherwise, if it didn't
happen so naturally,
we would have to multiply
one of these equations
or maybe both of them
by some scaling factor.
But we can just add
these two equations up.
On the left-hand
side, 2x plus 3x
is 5x, negative z plus z
cancels out, negative 7
plus negative 3, that
is equal to negative 10.
Divide both sides of
this equation by 5,
and we get x is
equal to negative 2.
Now, we can substitute back
to find the other variables.
We could substitute
back into this equation
to figure out what
z must be equal to.
So we have 2 times x, 2
times negative 2 minus z
is equal to negative
7, or negative 4 minus
z is equal to negative 7.
We can add 4 to both
sides of this equation,
and then we get negative
z is equal to negative 7
plus 4 is negative 3.
Multiply or divide both
sides by negative 1,
and you get z is equal to 3.
And now we can go
and substitute back
into one of these
original equations.
So we have x-- we know x
is negative 2-- so we have
negative 2 plus y
minus 3 times z.
Well, we know z is
3-- minus 3 times
3 should be equal
to negative 10.
And now we just solve for y.
So we get negative
2 plus y minus 9
is equal to negative 10.
And so negative 2 minus
9, that's negative 11.
So we have y minus 11
is equal to negative 10.
So we can add 11 to both
sides of this equation,
and we get y is equal to
negative 10 plus 11 is 1.
So we're done.
We got x is equal to
negative 2, z is equal to 3,
and y is equal to 1.
Now we can actually
go back and check it,
verify that this x, y,
and z works for all three
constraints, that this
three-dimensional coordinate
lies on all three planes.
So let's try it out.
We got x is negative
2, z is 3, y is 1.
So if we substitute it-- let
me do it into each of them.
So in this first equation,
that means we have negative 2
plus 1.
Remember, y was equal to 1.
Let me write it over
here. y is equal to 1,
x is equal to negative
2, z is equal to 3.
That was the result we got.
Yep, that's the result we got.
So when we test it in this
first one, you have negative 2
plus 1 minus 3
times 3, so minus 9.
This should be equal to
negative 10, and it is.
Negative 2 plus 1 is negative
1 minus 9 is negative 10.
So it works for the first one.
Let's try it for the second
equation right over here.
So we have negative 2 minus
y, so minus 1 plus 2 times z.
So z is 3, so 2 times 3.
So plus 6 needs
to be equal to 3.
So this is negative 3 plus 6,
which is indeed equal to 3.
So it satisfies the
second equation.
And then we have the
last one right over here.
We have 2 times x, so
2 times negative 2,
which is negative 4.
Negative 4 plus y, so
plus 1 minus z, so minus 3
needs to be equal to negative 6.
Negative 4 plus 1 is negative 3,
and then you subtract 3 again.
It equals negative 6.
So it satisfies all
three equations,
so we can feel pretty
good about our answer.
And here we have three
equations with three unknowns.
And just so you have a
way to visualize this,
each of these equations
would actually
be a plane in three dimensions.
And so you're actually
trying to figure out
where three planes in
three dimensions intersect.
I won't go into
the details here.
I'll focus more
on the mechanics.
But you can imagine if I were
to draw three-dimensional space
over here.
Now, all of a sudden, it
will an x, y, and z-axes.
So you can imagine that maybe
this first plane-- and I'm
not drawing it the way
it might actually look--
might look something like that.
I'm just drawing
part of the plane.
And maybe this plane over here,
it intersects right over there,
and it comes popping
out like this.
And then it goes
behind it like that,
keeps going in every direction.
I'm just drawing
part of the plane.
And maybe this plane
over here, maybe it
does something like this.
Maybe it intersects
over here and over here,
and so it pops out like that.
And then it goes below it like
that, and it goes like that.
I'm just doing this for
visualization purposes.
And so the intersection
of this plane, the x, y,
and z-coordinates that
would satisfy all three
of these constraints
the way I drew them,
would be right over here.
So that's what
we're looking for.
And a lot of times, three
equations with three
unknown systems, they
will be inconsistent.
You won't have a solution here
because it's very possible
to have three planes that all
don't intersect in one place.
A very simple example
of that is, well, one,
they could all be
parallel to each other.
Or they could
intersect each other,
but maybe they intersect each
other in kind of a triangle.
So maybe one plane
looks like that.
Then another plane maybe pops
out like that, goes underneath.
And then maybe the
third plane cuts in.
It does something like this,
where it goes into that plane
and keeps going out
like that, but it
intersects this plane over here.
So you see, this kind
of forms a triangle,
and they don't all
intersect in one point.
So in this situation, you would
have an inconsistent system.
So with that out of
the way, let's try
to actually solve this system.
And the trick here is
to try to eliminate
one variable at a time
from all of the equations,
making sure that you have the
information from all three
equations here.
So what we're going to
do is we could maybe,
it looks like the
easiest to eliminate--
since we have a positive
y and a negative y,
and then another positive y--
it seems like we can eliminate
the y's.
We can add these two
equations and come up
with another equation that will
only be in terms of x and z.
And then we could use these
two equations to come up
with another equation that'll
only be in terms of x and z.
But it'll have all of the x
and z constraint information
embedded in it because we're
using all three equations.
So let's do that.
So first, let's add these two
equations right over here.
So we have x plus y minus
3z is equal to negative 10,
and x minus y plus
2z is equal to 3.
So over here, if we
want to eliminate y
we can literally just
add these two equations.
So on the left-hand side, x
plus x is 2x, y plus negative y
cancels out.
And then negative
3z plus 2z, that
gives us just a negative z.
And then we have negative 10
plus 3, which is negative 7.
So using these two
equations, we got
2x minus z is equal
to negative 7,
just adding these two equations.
Now let's do these
two equations.
And we can we
reuse this equation
as long as we're using
new information here.
Now we're using the
extra constraint
of this bottom equation.
So we have x minus y
plus 2z is equal to 3,
and we have 2x plus y minus
z is equal to negative 6.
And if we want to
eliminate the y's, we
can just add these
two equations.
So x plus 2x is 3x.
Negative y plus y cancels out.
2z minus z, well,
that is just z.
And that is going to be
equal to 3 plus negative 6
is negative 3.
So if I add these
two equations, I
get 3x plus z is
equal to negative 3.
Now I have a system of two
equations with two unknowns.
This is a little bit more
traditional of a problem.
So let me write them over here.
So we have 2x minus z
is equal to negative 7,
and then we have 3x plus
z is equal to negative 3.
And the way this
problem is set up,
it gets pretty
simple pretty fast,
because if we just add these two
equations, the z's cancel out.
Otherwise, if it didn't
happen so naturally,
we would have to multiply
one of these equations
or maybe both of them
by some scaling factor.
But we can just add
these two equations up.
On the left-hand
side, 2x plus 3x
is 5x, negative z plus z
cancels out, negative 7
plus negative 3, that
is equal to negative 10.
Divide both sides of
this equation by 5,
and we get x is
equal to negative 2.
Now, we can substitute back
to find the other variables.
We could substitute
back into this equation
to figure out what
z must be equal to.
So we have 2 times x, 2
times negative 2 minus z
is equal to negative
7, or negative 4 minus
z is equal to negative 7.
We can add 4 to both
sides of this equation,
and then we get negative
z is equal to negative 7
plus 4 is negative 3.
Multiply or divide both
sides by negative 1,
and you get z is equal to 3.
And now we can go
and substitute back
into one of these
original equations.
So we have x-- we know x
is negative 2-- so we have
negative 2 plus y
minus 3 times z.
Well, we know z is
3-- minus 3 times
3 should be equal
to negative 10.
And now we just solve for y.
So we get negative
2 plus y minus 9
is equal to negative 10.
And so negative 2 minus
9, that's negative 11.
So we have y minus 11
is equal to negative 10.
So we can add 11 to both
sides of this equation,
and we get y is equal to
negative 10 plus 11 is 1.
So we're done.
We got x is equal to
negative 2, z is equal to 3,
and y is equal to 1.
Now we can actually
go back and check it,
verify that this x, y,
and z works for all three
constraints, that this
three-dimensional coordinate
lies on all three planes.
So let's try it out.
We got x is negative
2, z is 3, y is 1.
So if we substitute it-- let
me do it into each of them.
So in this first equation,
that means we have negative 2
plus 1.
Remember, y was equal to 1.
Let me write it over
here. y is equal to 1,
x is equal to negative
2, z is equal to 3.
That was the result we got.
Yep, that's the result we got.
So when we test it in this
first one, you have negative 2
plus 1 minus 3
times 3, so minus 9.
This should be equal to
negative 10, and it is.
Negative 2 plus 1 is negative
1 minus 9 is negative 10.
So it works for the first one.
Let's try it for the second
equation right over here.
So we have negative 2 minus
y, so minus 1 plus 2 times z.
So z is 3, so 2 times 3.
So plus 6 needs
to be equal to 3.
So this is negative 3 plus 6,
which is indeed equal to 3.
So it satisfies the
second equation.
And then we have the
last one right over here.
We have 2 times x, so
2 times negative 2,
which is negative 4.
Negative 4 plus y, so
plus 1 minus z, so minus 3
needs to be equal to negative 6.
Negative 4 plus 1 is negative 3,
and then you subtract 3 again.
It equals negative 6.
So it satisfies all
three equations,
so we can feel pretty
good about our answer.
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Nov 28, 2024 Last updated |
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