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## Algebra (all content)

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

Lesson 9: Systems with three variables

# 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.

## Want to join the conversation?

• Is it possible to solve 4-variable systems? • what if two of the equations have three variables but the last equation has two? •   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.
• Is there a term/concept that is analogous to "slope" for planes is three dimensions? Maybe "tilt"? • 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.) • 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? • 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'? • 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

• 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 • 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- 36z = ¼ [3x+5y- 36]12x+2y+3z=193z = −12x − 2y + 19z = 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 + 7657x + 23y = 184Now 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=25Pick 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 + 253y = 9x − 25y = 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 - 575378x = 1127x = 1127/378x = 161/54Now we can plug x into the second equation to find that y = 11/18And with those two variables solved, we can find that z = −6``
• I followed all the steps but at of the video I still had three variables to solve for, how do I solve this?   