Question

need help with solving linear equations with 3 variables

Answer 1

Solve x + 10y – 2z = 20
5x – 3y + 4z = 8
2x + y = 6

Answer 2

watch
http://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/matrices_elimination/v/matrices--reduced-row-echelon-form-1

Answer 3

@phi i don't think this video is talking about the same thing i'm trying to figure out.

Answer 4

no wait nevermind. i think i got it

Answer 5

It is more advanced than what you have to do... but the idea is the same
you start with
x + 10y – 2z = 20
5x – 3y + 4z = 8
2x + y = 6

and write it as
1 10 -2 20
5 -3 4 8
2 1 0 6

Answer 6

and then find the inverse and then multiply the inverse by 20, 8, 6. correct?

Answer 7

yes, that works, but you only find the inverse of
1 10 -2
5 -3 4
2 1 0

Answer 8

Finding the inverse of a 3 x 3 is a pain if you are doing it by hand

Answer 9

was going to go through using elimination but the numbers are very ugly.

though the final answers for x,y and z are simple

Answer 10

The good thing about elimination is it is a formula you follow. The bad thing is you can get ugly numbers.
But here are the gory details.

1 10 -2 20
5 -3 4 8
2 1 0 6

step 1: zero out the 5 in the 2nd row.
multiply the first row by -5: -5 -50 10 -100 (notice the big numbers, ugh!)
add this to the 2nd row to get : 0 -53 14 -92

1 10 -2 20
0 -53 14 -92
2 1 0 6

step 2: zero out the 2 in the 3rd row.
multiply the first row by -2: -2 -20 4 -40
add this to the 3rd row to get: 0 -19 4 -34


1 10 -2 20
0 -53 14 -92
0 -19 4 -34

those are ugly numbers.

Answer 11

1 10 -2 20
0 -53 14 -92
0 -19 4 -34

next, move to the 2nd column, 2nd row.

step 3:
make the -53 =1. We do this by dividing that row by -53. we get

1 10 -2 20
0 1 -14/53 92/53
0 -19 4 -34

step 4
zero out the -19 in the 3rd row. to do this, multiply the SECOND row by 19 to get
19 -14*19/53 19*92/53
(I don't want to use decimals and lose precision, but you can do so with a calculator)
add that to the last row to get

1 10 -2 20
0 1 -14/53 92/53
0 0 -54/53 -54/53

step 5: make the -54/53 in the 3rd row a 1. we do this by multiplying that row by -53/54. we get
1 10 -2 20
0 1 -14/53 92/53
0 0 1 1


the last row looks nice. It means z = 1
we could continue with elimination, but we can solve the problem using
*back substitution"

Answer 12

the 2nd row
0 1 -14/53 92/53

means y - 14/53 z = 92/53
but we know z=1, so this is
y - 14/53 = 92/53
solve for y= 92/53 + 14/53 = 106/53 =2 (a miracle, it came out so nice)

Answer 13

now we move to the first row
1 10 -2 20
which means

1x + 10y -2z = 20
but we know y=2 and z= 1 so this is
x +20 -2 = 20
x = 2

the final answer is x=2, y=2, z=1