Algebra (all content)
Solving linear systems with 3 variables: no solution
Sal solves a system with three variables that turns out to have no solution. Created by Sal Khan.
Want to join the conversation?
- So... how do we determine if it is an infinite solution?(12 votes)
- It would have infinite solutions if you get 0=0.(8 votes)
- At1:05, Sal says you can pair these first two, last two, etc.
My question is can you pair them in any order as long as you use all 3 equations?(6 votes)
- yes, and it is important to use all three equations no matter what order you pair them in because you need to use all the constraints. Sal probably did it in that order because it was the most convenient.(9 votes)
- at4:09what if your final answer was -21 = 0 ?(2 votes)
- Same thing, it can never be true, it's nonsensical. The lines do not intersect; they're parallel.(1 vote)
- How do you do the simultaneous equations without planes and graphs ?(3 votes)
- Matrices are the easiest way. Especially if you have a graphing calculator. Plug it in, and tell it RREF. and the column on the right is the answer.(7 votes)
- how do you solve a question like:
im confused(4 votes)
- I used substitution in the for the first 2 equations.
x+2x+z=56 (from Eq. 2)
(Eq. 4) 3x+z=56
Then use substitution again for the 2nd and 3rd equation
(Eq. 5) 2x=z+4 (Eq. 3 can be rewritten as y=z+4)
Rewriting Eq. 4 & 5
Subtracting them we will get
From Eq. 2, y=2(12)=24
Then from Eq. 3, y=24=4
- Are there any videos with three variables and two equations? I have many homework problems like this, and I understand that 3 variables with 2 equations = infinitely many solutions so real number a can be substituted for one of the variables (it doesn't matter which one, does it?) and the other two variables can be expressed in terms of a, but seeing a problem like this solved out step-by-step would be so much nicer and my book does such a bad job of showing examples.(4 votes)
- Would a valid short cut be to look for two equations that cancel more than 1 variable?(1 vote)
- Well yes, that's what he stumbles onto at the beginning, and cancels out the x and the z variables by adding multiples of the first 2 equations.(3 votes)
- At2:40what would solving for
yreally get you? Does that value for
ydefine the intersection of the 2 planes defined by the first 2 equations? If not, what is the intersection of those 2 planes?(3 votes)
- While the y value in itself does not define the line of intersection between the first two planes, it partly defines the line because, as the following steps will show, that line lies on the plane y=c, where c is that value.
When you substitute y=-5/14 into 2x-4y+z=3, the further simplification produces the equation 2x+z=22/14. Solving for z gives us z=22/14-2x. So the points that represent the intersection of the first two equations are in the form (x,-5/14,22/14-2x)=(x,0,-2x)+(0,-5/14,22/14)=x(1,0-2)+(0,-5/14,22/14), which is a translation of the all vectors parallel to (1,0,-2) by the vector (0,-5/14,22/14).
You can solve for the second line by subtracting a multiple of the first equation from the third to solve for the second y value. Substitute that value into the first equation and from there follow the steps I showed you to derive the second line.(0 votes)
- can I also do column modifications , like 2 * z - column ? or is it not possible?(1 vote)
- When working with equations you must multiply the whole equation by your chosen factor. It is not a correct step to multiply one term in an equation.
One way to analyze your question: we are considering multiplying one term on the left side of an equation by 2. What will we do with the right side to maintain equality?(2 votes)
- -8 is less than 4(X -2 ) less than or equal to 9(1 vote)
Determine whether the system has no solutions or infinite solutions. So let's think about how we can go about doing this. So if at any point we might not have to solve this entirely if we somehow get something that's nonsensical which will tells us there's no solutions. Or we might have to go further and see if it's one or infinite solutions, although it looks like one solution isn't an option here, given how this question is phrased. So the way that you would proceed to solve three equations with three unknowns is you would try to eliminate variables one by one. And so first we could try to eliminate the x variables. And we could do that, we can essentially create two equations with two unknowns. The two unknowns will be y and z. If we can pair up these equations and eliminate x with each of these pairings. So for example, we can pair these first two. We can pair the last two. And that's all we would need to have to eliminate the x's and still have two equations. And have all of the information of these three equations. But then the third pairing would be the first and the third equation, But we only have to do two of these pairings. Now, just to show you what I mean by these pairings, what I want to do is take these first two. I'm going to pair this first pairing right over here, and I'm going to use them to eliminate the x terms. And over here I have 2x, over here I have 8x. If I could turn this 2x into a negative 8x I could add both sides of these equations to each other and the x terms would cancel out. And so the best way to turn this 2x into a negative 8x is to multiply this top equation times negative 4. When I say multiply, I'm saying multiply the whole equation, both sides of it, by negative 4. So 2x times negative 4 is negative 8x. Negative 4y times negative 4 is plus 16y, are positive 16y. z times negative 4 is negative 4z And that's equal to 3 times negative 4 which is negative 12. And then I can rewrite this equation right over here. It's 8x minus 2y plus 4z is equal to 7. And now I can add these both equations. On the left hand side, these guys cancel out, 16y minus 2y-- and that was the whole point behind multiplying the top equation by negative 4-- 16y minus 2y is 14y. Negative 4z plus 4z. These guys actually cancel out as well. So actually with that one pairing, by multiplying by negative 4 we were actually able to cancel out two variables. So you get 14y is equal to negative 12 plus 7 is equal to negative 5. And you can actually solve for y. And we don't know if this one will actually have solutions. But if we assume it's going to have a solution, you could actually solve for y right over here. You could divide both sides by 14. But let's worry about that a little bit later. Let's take the second pairing right over here. So once again you have an 8x. We want to eliminate the x's. So this one you have an 8x, here you have a negative 4x. If you multiply this times 2, this is going to become a negative 8x and it can cancel with this top one. So the top equation is 8x minus 2y plus 4z is equal to 7. When I say "top equation" I'm talking about this one right over here, the top one in this pairing. And this bottom equation, I'm going to multiply times negative 2. I'm going to multiply it times negative 2-- sorry times positive 2. I'm going to multiply it times positive 2. Negative 4x times positive 2 is negative 8x. So I'm going to multiply it by 2. So 2 times negative 4x is negative 8x. 2 times y is plus 2y. 2 times negative 2z is negative 4z. And then 2 times negative 14 is negative 28. And now we can add the left hand sides and add the right hand sides. These cancel out, those cancel out, those cancel out. We actually end up with nothing on the left hand side. You get 0 plus 0 plus 0. And on the right hand side you get 7 plus negative 28 is negative 21. Well, this is a nonsensical answer. 0 can never be equal to negative 21. No matter what x, y, or z you pick, 0 cannot be equal to negative 21. And that's because these second two equations right over here, if you view them as planes in three dimensions, these right over here do not intersect. If you visualize them in three dimensions, they're actually parallel planes. And since these last two definitely do not intersect, we can say that this system has no solutions. It doesn't matter if this first equation intersects one or both of these. The fact that these two don't intersect tells us that there's no unique point x, y, z coordinate, a point in three dimensions that satisfies all three of them. Because there's no unique x, y, z that can satisfy these two, because these are parallel planes. They do not intersect.