SOLUTION: What does y, x, and z equal? 2x-2y-4z=-2 3x-3y-6z=-3 -2x+3y+z=7

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Question 1198522: What does y, x, and z equal?
2x-2y-4z=-2
3x-3y-6z=-3
-2x+3y+z=7
Found 4 solutions by Alan3354, MathLover1, greenestamps, math_tutor2020:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
What does y, x, and z equal?
2x-2y-4z=-2
3x-3y-6z=-3
-2x+3y+z=7
==============
There are several ways to solve these.
What method would you prefer?
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

.........1)
.........2)
..........3)
----------------------------------
start with
.........1)
..........3)
------------------------------add


.......1a)

.........2)
..........3)
---------------------------------add


.........2a)

go to
.........1), substitute and




=>The solutions to the system of equations are:




Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The response from tutor @MathLover1 shows that she knows how to find the answer; but it does little to help the student LEARN HOW.

For this particular problem, I would start by factoring out the common factor in each of the first two equations to find that they are equivalent:

x-y-2z=-1 [1]

This means there are only two equations with three unknowns; so we are going to have a solution that is a family of equations instead of a unique solution.

Use equation [1] and the third given equation to eliminate one of the variables. Arbitrarily I chose to eliminate x:
  -2x+3y+ z= 7
   2x-2y-4z=-2
  -------------
       y-3z= 5 --> y=3z+5

Substitute y=3z+5 in [1] to find x in terms of z:

x-(3z+5)-2z=-1
x-3z-5-2z=-1
x=5z+4

ANSWER:
x=5z+4
y=3z+5
z=z

Or a better format for the answer would be to use a parameter t:

z=t
x=5t+4
y=3t+5
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

There are a number of ways to solve a linear system of equations.
I'll use a matrix approach. Specifically I'll use row reduction.

Given system of equations
2x-2y-4z=-2
3x-3y-6z=-3
-2x+3y+z=7

That system translates into this augmented matrix

Technically there should be a vertical line separating the fourth column from the 3x3 block on the left, but it's not that important in my opinion.
The 3x3 block of values represent the coefficients on the left hand side of each equation. The right most column represents the right hand side values of the original equations.

The goal is to make that left-hand 3x3 block to be the 3x3 identity matrix
We use row reductions to get this done.
If we cannot get the identity matrix, then we can try to get "close enough" so to speak.

I'll express each matrix as a table to help separate out the values better.

Notation like 0.5*R1 --> R1 means we multiply everything in row 1 (aka R1) by 0.5, and then we update row 1 accordingly.

Something like R2 - 3*R1 --> R2 means we first triple everything in row 1 temporarily, then subtract that result from R2. The result will replace R2.

Let me know if you have any questions about the steps shown below.

You can use the Linear Algebra Toolkit as a handy calculator that shows you the step-by-step process to solving row reduction problems like this. I recommend using it to check your answers rather than do your entire homework for you.

2-2-4-2
3-3-6-3
-2317


1-1-2-10.5*R1 --> R1
3-3-6-3
-2317


1-1-2-1
0000R2 - 3*R1 --> R2
-2317


1-1-2-1
-2317R2 <--> R3
0000


1-1-2-1
01-35R2 + 2*R1 --> R2
0000


10-54R1+R2 --> R1
01-35
0000


Unfortunately, we cannot get the 3x3 identity matrix. This is due to the row of 0s at the bottom. A row of nothing but zero signals "infinitely many solutions". The system is dependent.

The first line 1, 0, -5, 4 translates back to 1x+0y-5z = 4, aka x-5z = 4, and then solves to x = 5z+4

The second line 0,1,-3,5 becomes 0x+1y-3z = 5, aka y-3z = 5
That solves to y = 3z+5

To summarize:
x = 5z+4
y = 3z+5
where z is any real number.
As the tutor @greenestamps pointed out, common practice is to use a parameter for the free variable.
The letter t is a common choice, but I'll go with k to avoid possible mixups with a plus sign.

------------------------------------

Answer:
There are infinitely many solutions of the form
(x,y,z) = (5k+4,3k+5,k)
where k is any real number.

Examples:
(x,y,z) = (9,8,1) when k = 1
(x,y,z) = (14,11,2) when k = 2
(x,y,z) = (19,14,3) when k = 3

All of these solutions are along the same straight line. This is because we effectively have 2 flat planes intersecting, instead of 3. Note how equation (2) is a scaled copy of equation (1).
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