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Course: Algebra (all content) > Unit 5
Lesson 9: Systems with three variablesIntro to linear systems with 3 variables
Sal discusses how we approach solving systems of three variables algebraically, and how we visualize it graphically. Created by Sal Khan and Monterey Institute for Technology and Education.
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Is it possible to solve 4-variable systems?(81 votes)
Well, yes, it is possible. However, it is very tedious to solve a system of three equations and VERY time consuming for 4 equations so I recommend you go search up matrices in algebra 2. They will help you solve the equations a lot faster.(13 votes)
what if two of the equations have three variables but the last equation has two?(19 votes)
Megan,
You can think of the third equation as having the other variable multiplied by zero if that helps you solve.
For instance if the third equation is
3x+4y = 2
It can be rewritten as
3x + 4y + 0z = 2
I hope that is of help to you.(71 votes)
Is there a term/concept that is analogous to "slope" for planes is three dimensions? Maybe "tilt"?(15 votes)- The equivalent concept in analytical geometry is called "direction cosines". Otherwise, you could find the slopes of the 3 projections onto the planes defined by the x, y, and z axes ... I think.
Hope this helps!(16 votes)
- What If the three planes are slightly overlapping? Wouldn't it have multiple answers because there would be more than one point that would work?
(It's hard to explain without a diagram, so I hope you understand my question.)(11 votes)
They can't overlap, but the intersection of two planes is a line so if all the planes intersect through the same line, that line is the answer. Hence, you may have 0, 1 or infinitely many solutions.(11 votes)
Is it necessary to plug the values back in to check, or was that just to prove that no mistakes were made in the calculations?
In other words, can solutions be extraneous or such, or was it just to double-check?(8 votes)
You don't 'have' to do this step, but it make sure that those answers are right... It's very helpful for people who made lots of mistake, like me :)(1 vote)
What if you have two equations which result in the same plane if graphed or visualized (what are infinite solutions in systems of 2 variables), but a third equation provides a solution that works. Is it a legitimate 'solution'?(3 votes)- yes, if you have two equal constraints, say
y = 2x + 3
2y = 4x + 6
It should be clear here that because these two equations say the same thing, there is an infinite number of Xs and Ys that would satisfy both of them, but if you add a third that intersects at one point;
Y = -4x + 21
using the same method you can solve for the point of intersection, which will be the unique solution to the system
-4x + 21 = 2x + 3
-4x - 2x = 3 - 21
-6x = -18
x = 3
y = 2(3) + 3
y = 9
And so these three equations, even though the first two had infinite solutions, has only one unique solution, that is x = 3, y = 9
I hope this answers your question(5 votes)
- What about if you have an equation like this:
3x+5y-4z=36
9x-3y=25
12x+2y+3z=19
what would you do if there are not consistently the same amount of variables in three equations(3 votes)- Treat the missing variable as having a coefficient of 0. Otherwise you solve them the same way.
Here is how I would solve that problem. This involves a variation on substitution that will not likely be taught in your class, but I find much easier to work with.
Take the two equations that contain z and solve for z.3x+5y-4z=36
4z = 3x+5y- 36
z = ¼ [3x+5y- 36]
12x+2y+3z=19
3z = −12x − 2y + 19
z = ⅓ [−12x − 2y + 19]
Since we now have to things that are equal to z, we set them equal to each other, eliminating the z.
¼ [3x+5y- 36] = ⅓ [−12x − 2y + 19]
To get rid of the fractions, multiply both sides by the LCM of the denominators, in this case that is 12:
(12) (¼) [3x+5y- 36] = (12) ⅓ [−12x − 2y + 19]
3 [3x + 5y - 36] = 4 [−12x − 2y + 19]
9x + 15y - 108 = −48x −8y + 76
57x + 23y = 184
Now we do a similar procedure using this and the third equation (the one that never had the z in it)
57x + 23y = 184 AND 9x-3y=25
Pick a variable to solve both equations for and then set them equal, which will give you just one variable. It doesn't matter which variable you choose, so I pick y.
57x + 23y = 184
23y = −57x + 184
y = (¹⁄₂₃) [−57x + 184]
9x-3y=25
−3y = −9x + 25
3y = 9x − 25
y = ⅓ [9x − 25]
Now setting the two equations equal and solving for x:
(¹⁄₂₃) [−57x + 184] = ⅓ [9x − 25]
The LCM of 23 and 3 is 69
(69)(¹⁄₂₃) [−57x + 184] = (69) ⅓ [9x − 25]
(3) [−57x + 184] = (23) [9x − 25]
−171x + 552 = 207x - 575
378x = 1127
x = 1127/378
x = 161/54
Now we can plug x into the second equation to find that
y = 11/18
And with those two variables solved, we can find that
z = −6(6 votes)
- I followed all the steps but at5:00of the video I still had three variables to solve for, how do I solve this?(3 votes)
- It happens, just solve for another variable and/or one of the other equations.(2 votes)
Is there a way to very simply tell that the system of three variables has no answer? That the plane doesn't intersect? That way it would save you some time if you already knew that there were no answer.(3 votes)
A zero with a slash means no answer. (Ø)(1 vote)
Why is the solution of systems of three variables only a point and not a line or another plane?(2 votes)
The solution could be a line or plane but then there would not be one unique solution - you would have an infinite number of potential solutions that would satisfy the system (since your solution set is a line or plane which has an infinite number points). In other words, if your solution set turned out to be a line or plane, your system would be dependent (has infinite solutions).(2 votes)
5:00Video transcript
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.