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## Algebra (all content)

### Course: Algebra (all content)>Unit 5

Lesson 9: Systems with three variables

# Solving linear systems with 3 variables

Sal solves a system of three variables using elimination. Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

Solve this system. And once again, we have three equations with three unknowns. So this is essentially trying to figure out where three different planes would intersect in three dimensions. And to do this, if we want to do it by elimination, if we want to be able to eliminate variables, it looks like, well, it looks like we have a negative z here. We have a plus 2z. We have a 5z over here. If we were to scale up this third equation by positive 2, then you would have a negative 2z here, and it would cancel out with this 2z there. And then if you were to scale it up by 5, you'd have a negative 5z here, and then that could cancel out with that 5z over there. So let's try to cancel out. Let's try to eliminate the z's first. So let me start with this equation up here. I'll just rewrite it. So we have-- I'll draw an arrow over here-- we have x plus 2y plus 5z is equal to negative 17. And then to cancel out or to eliminate the z's, I'll multiply this equation here times 5. So I'm going to multiply this equation times 5. So we have to multiply both sides by 5. So 3x times 5 is 15x, y times 5 is plus 5y, and then negative z times 5 is negative 5z-- that's the whole point and why we're multiplying it by 5-- is equal to 3 times 5, which is equal to 15. And so if we add these two equations, we get x plus 15x is 16x, 2y plus 5y is 7y, and 5z minus 5z or plus negative 5z, those are going to cancel out. And that is going to be equal to negative 17 plus 15 is negative 2. So we were able to use the constraints in that equation and that equation, and now we have an equation in just x and y. So let's try to do the same thing. Let's trying to eliminate the z's. But now I'll use this equation and this equation. So this equation-- let me just rewrite it over here. We have 2x minus 3y plus 2z is equal to negative 16. I just rewrote it. And now, so that this 2z gets eliminated, let's multiply this equation times 2. So let's multiply it times 2, so we'll have a negative 2z here to eliminate with the positive 2z. So 2 times 3x is 6x, 2 times y is plus 2y, and then 2 times negative z is negative 2z is equal to 2 times 3 is equal to 6. And now we can add these two equations. 2x plus 6x is 8x, negative 3y plus 2y is negative y, and then these two guys get canceled out. And then that is equal to negative 16 plus 6 is negative 10. So now we have two equations with two unknowns. We've eliminated the z's. And let's see, if we want to eliminate again, we have a negative y over here. We have a positive 7y. Well, we could eliminate the y's if we multiplied this times 7 and add the two equations. So let's do that. So let's multiply this times 7. 7 times 8 is 56, so it's 56x minus 7y is equal to 7 times negative 10, is equal to negative 70. And now we can add these two equations. I'm now trying to eliminate the y's. So we have 16x plus 56x. That is 72x. So we have 72x, these guys eliminate, equal to negative 72, negative 2 plus negative 70. Divide both sides by 72, and we get x is equal to negative 1. And now we just have to substitute back to figure out what y and z are equal to. So let's go back to this equation right over here. So we have 8x minus y is equal to negative 10. If x is equal to negative 1, that means 8 times negative 1, or negative 8 minus y is equal to negative 10. We can add 8 to both sides. And so we have negative y is equal to negative 2, or multiplying both sides by negative 1, y is equal to 2. let me square that off. So x is equal to negative 1, y is equal to 2. We now just have to worry about z, and we can go back to any of these up here. So I'll just use this last one. The numbers seem lower. So if we substitute back into this last equation right over here, we have 3 times x, which is 3 times negative 1 plus y, which is 2, minus z is equal to 3. So it's negative 3 plus 2 minus z is equal to 3. And this is negative 1 minus z is equal to 3. And then add 1 to both sides, we get-- these cancel out-- negative z is equal to 4. Multiplying both sides by negative 1, you get z is equal to negative 4. So we're done. Let's verify that these solutions, or the solution of x is negative 1, y is equal to 2, z is equal to negative 4, actually satisfies all three of these constraints. So let's substitute into this first one. So you have x plus 2y plus 5z. So x is negative 1 plus 2 times y, so plus 4, plus 5z. So minus 20 has to be equal to negative 17. And this is negative. This right here is positive 3 minus 20 is indeed equal to negative 17. So it satisfies the first constraint. Let's look at the second one. 2 times x, 2 times negative 1. That's negative 2 minus 3 times y. That's minus 6-- 3 times 2 is 6-- plus 2 times z. z is negative 4, so that's 2 times negative 4 is negative 8. That needs to equal negative 16. Negative 2 minus negative 8 is-- sorry, negative 2 minus negative 6 is negative 8. Subtract another 8, you get negative 16. So it meets the second constraint. And then, finally, let's look at the last constraint. We have 3 times x, so negative 3 plus y, so plus 2, minus z. So minus negative 4 is the same thing as plus 4. That needs to be equal to 3. So negative 3 plus 2 is negative 1 plus 4 is indeed equal to 3. So we found the point of intersection in three dimensions of these three planes. x is negative 1, z is negative 4, y is 2. And we were able to verify that it does indeed meet all of the constraints.