Three Variables X=, Y=, Z=?

Question

3x + 0y + 3z = -6
4x - 4y + 4z = 12
-1x + 4y - 1z = -18

Answer 1

You can use a matrix method to ease the pain. Recognize that you have 3 variables to solve on the left and the values on the right, and then you can leave out the variable names.

3 0 3 | -6
4 -4 4 | 12
-1 4 -1 | -18

Divide by 3 across the first row:

1 0 1 | -2
4 -4 4 | 12
-1 4 -1 | -18

For each other row, multiply each column in the first row by the number you find in the first column of the current row, and substract that from the current column in the current row. This goes for the values on the right as well.

1 0 1 | -2
0 -4 0 | 20
0 4 0 | -20

Notice that the first column is now "normalized". That is, there is only one column without a 0, and that is in the first row. Now, let's do exactly the same thing for the second row.

First divide by -4:

1 0 1 | -2
0 1 0 | -5
0 4 0 | -20

And now for each other row, using the second column as base:

1 0 1 | -2
0 1 0 | -5
0 0 0 | 0

Normally, you would do the same thing with the third row, but lookee what we got here...all 0's. What has happened, is that one of our functions has been shown to be redundant. We actually only have 2 interesting functions. Putting our matrix back into a form we can read:

x + z = -2
y = -5

Try it!

**Edit
I see a lot of people posting things like x *must* be -1. If you want proof that x and z are not fixed, solve the equations with x = -1 and z = -1. Then solve with x = 0 and z = -2. Then solve with x = -3 and z = 1. They all work. Now, y is indeed fixed at -5. Kinda interesting, huh?

Answer 2

Deduct the 1st equation from the 2nd:
4x - 4y + 4z = 12
-(3x +0y + 3z = -6) to get x = 4y - 1z + 18

Substitute the value of x to the 2nd equation

Thus , 4(4y-1z + 18) - 4y + 4z = 12
16y - 4z + 72 - 4y + 4z = 12
12y = 12 - 72
y = -60
y = -5

Subtitute the value of y to 3rd equation
-1x + 4 (-5) -1z = -18
- 1x -20 -1z = -18
The only value for x and z to satisfy this equation is -1
Thus x = -1 and z = - 1

Substitute the value of x = -1, y = -5 and z = -1 to the 3 equation:
Thus 3(-1) + 0(-5) + 3 (-1) = -6
-3 + 0 - 3 = - 6
- 6 = -6
2nd Equation : 4(-1) - 4(-5) + 4(-1) = 12
- 4 + 20 - 4 = 12
- 8 + 20 = 12
12 = 12
3rd Equation : -1(-1) + 4(-5) - 1(-1) = -18
1 - 20 + 1 = -18
-18 = -18

So x = -1 , y = -5 and z = -1

Answer 3

x = -1, y= -5, z = -1 OR x= 0, y = -5, z = -2 or x = -2, y = -5, z = 0
First, find an equation for z. You can change the third equation to -z = -18 + x -4y, which can be changed to z= 18 - x + 4y. Then plug the value of z into the first equation, which then looks like 3x + 3(18-x+4y)= -6. Solve the parantheses and get 3x + 54 -3x +12y = -6. Combine like terms and you get y as being -5. Now plug y into the third equation to get -x + -20 -z = -18, which simplified gives you - x - z=2. There are three different combinations of x and z that work here, along with the other two equations. Hopefully one of these is the answer you are looking for.

Answer 4

OK, what you have to do is when you ignore the first equation, you get:
4x - 4y + 4z = 12
-1x + 4y -1z = -18
Then the Y's fall out an your left with:
3x + 3z = -6
Then you add in the another equation, and by the way 0y does not exist:
3x + 3z = -6
3x + 3z = -6
And if you times the top, or the bottom, by negative 1, the equations fallout.
I don't know where to go from here!!

Answer 5

the easiest way to solve this is by using matrix equations

3 0 3 x -6
4 -4 4 * y = 12
-1 4 -1 z -18
This is a 3 x 3 matrix times a 3 x 1 matrix ( x. y. z ) equal to a 3 x 1 matrix ( -6, 12, -18 )
To solve for the x, y, z martix..... you multiply the inverse of the 3 x 3 matirx times the 3 x 1 ( -6, 12, -18 ) matrix the resulting 3 x 1 matrix will be your answer. This can easily be done in a graphing calculator.

Answer 6

I must be getting old and tired, but rather than throw my work out here it is. Then someone else can give you a better answer:

3x + 0y + 3z = -6
4x - 4y + 4z = 12
-1x + 4y - 1z = -18

(3x + 0y + 3z = -6) * 4=
12x + 12z= -24

(4x - 4y + 4z = 12) *3 =
12x -12y +12z = 36

+(12x + 12z= -24)
-(12x -12y +12z = 36)

12x + 12z= -24
-12x +12y -12z = -36
- - - - - - - - - - - - -
0x +12y +0z = -60
y = -5

4x - 4y + 4z = 12
4x – (-5) *4 + 4z = 12
4x +20 + 4z = 12
4x = -8 -4z
x = -2 –z

Answer 7

3x+3z= -6,so x+z= -2-------------------------------(1)
4x-4y+4z=12 ,so x-y+z=3--------------------------(2)

Therefore ,replace (1)into (2)
U will get -y+(x+Z)=3
-y+(-2)=3
y= -5
Then ,-x+4y-z= -18
Replace y as -5 and u will get : -x+4(-5)-z= -18
-x-z = 2----------------(3)

Then ,replace (1) to (3)

x+z= -2 and -x-z=2
x= -2-z so ....

-x-z=2
(-2-z)-z=2
-2-2z=2
z= -2

so x ===== -2-z
so x= -2-(-2)
x= 0


DONE !!!!!!! If there is an error ,should be the positive signs and negative signs but THIS is sure the way how to solve the prob

Answer 8

first of all the system clearly reduced to three unknown and two equation .so, one varible is free as follows.
on solving by gaussian ellimination method we have, -x +4y +-z= -18,
12y=60.
thus y=-5, treat z as free varible say z=0.
then x= 2.. thus the solution set is{2 ,-5, 0 }. which is a particular soultion. really the solution depends on z.

Answer 9

1) x+z = -2
x = -2 - z
2) x = y - z +3
-2 - z = y - z +3
y = -5
3) 2+z+4(-5) = -18
2+z-20 = -18
z = 0
Since z = 0, you do not have 3 unknowns
1) x = -2
2) -2 = y + 3
y = -5

Answer 10

Start with two of them, and add them but before you add them, multiply each side of the equation by the same number. So that one of the terms goes away. Keep doing this till you have a solution. I had to work many of this type of equations in circuit theory. It was a pain.