Solving Systems of Equations in Three Variables

2x+3y+4z=2
2x-10y-4z=-10

should be

2x+3y+4z=2
2x-10y-4z=-4

More when I finish eating.

2x+3y+4z=2
5x-2y+3z=0
x-5y-2z=-2

the equations should be

2x+3y+4z=2
5x-2y+3z=0
x-5y-2z=-4


sorry it was a typo.

1) Solve one of the equations for any of the variables. "solve" means to multiply, divide, add, subtract or do whatever you need to do to get it so that "1 variable = the rest."

2) Substitute the solved variable in to all the other equations

3) Repeat until one equation has only 1 type of variable in it

4) simplify that equation so that you have the exact value of that variable

5) subsitute this value into all the other equations and repeat from step 1 until there are no variables left

Or the easy way, multiply one equation by a number, and add or subtract this equation from another, so that one of the variables dissappear. Repeat, solve for the final remaining variable, and substitute.

Example:

If you are working in three linear equations in three unknowns you should be able to write these equations as a linear expression. So take

2x+3y+4z=2
5x-2y+3z=0
x-5y-2z=-4

and write it as

|2 3 4||x| | 2|
|5 -2 3||y|=| 0|
|1 -5 -2||z| |-4|

Let

|2 3 4|
|5 -2 3|=A
|1 -5 -2|

|x|
|y|=B
|z|

&

| 2|
| 0|=C
|-4|

Then

AB=C

and you can solve for B by finding the inverse of A (A^-1) and multiplying it by C

or

B=A^-1*C

Since
|x|
|y|=B
|z|

ya got the solution

Rap

hmmm.. Negate an z so that u get an equation in x and y.
repeat with another two different ones..
solve the two new simultanious equations..
bingo!

here it is..

2x + 3y + 4z = 2
5x - 2y + 3z = 0
x - 5y - 2z = -4 now multiply by (2)

To get..
2x + 3y + 4z = 2
2x - 10y - 4z = -8

Therefore 4x - 7y = -6 is one equation

and again...
2x + 3y + 4z = 2
5x - 2y + 3z = 0 multiply by (2)
x - 5y -2z = -4 now multiply by (3)

To get another..
10x -4y + 6z = 0
3x - 15y - 6z = -12

and 13x - 19y = -12 is the second simultanous equation

Solve...
13x - 19y = -12 multiply by 4
4x - 7y = -6 multiply by -13

so...
52x - 76y = -48
-52x + 91y = 78 (thats got to be wrong )

haha what do you know...
15y = 30
y = 2

Hence 4x - 7y = -6
slot in the value for y... 4x - 2(7) = -6
4x - 14 = -6
4x = 8
x = 2

Lastly slot in both values in one of the original equations to ascertain (i love that word) the value of z..

2x + 3y + 4z = 2
then 2(2) + 3(2) + 4z = 2
4 + 6 + 4z = 2
4z = -8
therefore z = -2

To check...
4 + 6 - 8 = 2 True
10 - 4 -6 = 0 True
2 - 10 + 4 = -4 which is True

Therefore the solutions are (x = 2, y = 2, z = -2[/i][/u])