systems of equations- with 3 equations!

[KingAce]

hi, i can easily solve a system of equations with 2 equations, but those with 3 are giving me a lot of trouble.

For example, how would I go about solving the following...
x+y+z=2
-x+3y+2z=8
4x+y=4


..I know I have to use the Gaussian Elimination method (ie. find the REF), but that process is giving me some trouble. Please Help!! Thanks!

 

[o_O]

Since you only have two variables to deal with in the 3rd equation, why don't you get rid of the z variable in the first two. This will give you another equation dealing with only x and y. So then you can solve for that. That's a start.

It'd be nice if you showed some work so we could point specifically to your problems.

 

[Loren]

The object is to get a matrix looking like this...

1 0 0 ?
0 1 0 ?
0 0 1 ?

You are starting with

(A).1..1..1..2
(B)-1..3..2..8
(C).4..1..0..4 (The dots represent spaces so that every column will be in correct alignment.)
I will take one step at a time. With practice you will be able to combine steps to shorten your work.
Interchange (A) and (C)
(A).4..1..0..4
(B)-1..3..2..8
(C).1..1..1..2

2(C)-(B) becomes (B)
(A).4..1..0..4
(B)-3..1..0..4
(C).1..1..1..2

(A)-(B) becomes (A)
(A).7..0..0..0
(B)-3..1..0..4
(C).1..1..1..2

(A)÷7 becomes (A)
(A).1..0..0..0
(B)-3..1..0..4
(C).1..1..1..2

(B)+3(A) becomes (B)
(A).1..0..0..0
(B).0..1..0..4
(C).1..1..1..2

(C)-(A) becomes (C)
(A).1..0..0..0
(B).0..1..0..4
(C).0..1..1..2

(C)-(B) becomes (C)
(A).1..0..0..0
(B).0..1..0..4
(C).0..0..1.-2

x=0
y=4
z=-2

 

[Deleted member 4993]

hi, i can easily solve a system of equations with 2 equations, but those with 3 are giving me a lot of trouble.

For example, how would I go about solving the following...
x+y+z=2
-x+3y+2z=8
4x+y=4


..I know I have to use the Gaussian Elimination method (ie. find the REF), but that process is giving me some trouble. Please Help!! Thanks!

I find it useful to use a spreadsheet (e.g. Xcel) for these problems.