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

Lesson 9: Systems with three variables

# 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?
• It would have infinite solutions if you get 0=0.
• At , 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?
• 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.
• at what if your final answer was -21 = 0 ?
• 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 ?
• 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.
• how do you solve a question like:
x+y+z=56
y=2x
z=y-4
im confused
• 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
z=-3x+56
z= 2x-4
Subtracting them we will get
0=-5x+60
x=12
From Eq. 2, y=2(12)=24
Then from Eq. 3, y=24=4
y=20

Checking
12+24+20=56
• 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.
• 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.
• At what would solving for `y` really get you? Does that value for `y` define the intersection of the 2 planes defined by the first 2 equations? If not, what is the intersection of those 2 planes?
• 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.