Algebra (all content)
Intro 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)
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:
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.
4z = 3x+5y- 36
z = ¼ [3x+5y- 36]
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]
−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)
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.