[MOVED] solving systems of eqns in three variables

[tretre]

The Question is:

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

This is what I did so far:

2x + 3y + 4z = 2. . .multiplied by -1. . .->. . .-2x - 3y - 4z = -2
. .x - 5y - 2z = -4. . .multiplied by 2. .. .->. .. .2x - 10y - 4z = -8

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .->. . . . .-13y - 8z = -10

5x - 2y + 3z = 0. . .multiplied by -1. . .->. . .-5x + 2y - 3z = 0
. .x - 5y - 2z = -4. . .multiplied by 5. . .->. . .5x - 25y - 10z = -20

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .->. . . . .-23y - 13z = -20

. . . ..13y + 8z = 10. . .multiplied by 1
+. . .-23y - 13z = -20
---------------------------
. . . .-10y - 5z = -10

But now I can't figure out what y and z is. When I tried solving for y and z, the numbers don't fit the equation. Can you help me solve for y and z? Thanks.

 

[Denis]

Re: solving systems of equations in three variables

Your way is kinda messy, hard to follow; try this way:

NUMBER the equations:
2x + 3y + 4z = 2 [1]
5x - 2y + 3z = 0 [2]
x - 5y - 2z = -4 [3]

-2x - 3y - 4z = -2 [1] * -1
2x - 10y - 4z = -8 [3] *2
=================
-13y - 8z = -10 : multiply that by -1:
13y + 8z = 10 [4]

-5x + 2y - 3z = 0 [2] * -1
5x - 25y - 10z = -20 [3] * 5
==================
-23y - 13z = -20 : multiply that by -1:
23y + 13z = 20 [5]

You did ok to this point; then you seemed to get a flat tire!

184y + 104z = 160 [5] * 8
169y + 104z = 130 [4] * 13
=================== subtract:
15y = 30
y = 2

You ok now? Carry on...

 

[tretre]

oh ok . .. thanks !!