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jmsoftball02 · Aug 19, 2009, 03:25 PM

I have to solve this linear equation with 3 variabes:

x-2y=1
x+3y+z=0
2x-2z=18

answer has to be in (x,y,z) formation... please help

galactus · Aug 19, 2009, 04:26 PM

Use elimination method or substitution method. Here is elimination:

$x-2y=1$
$x+3y+z=0$
$2x-2z=18$

Multiply the second row by 2:

$x-2y=1$
$2x+6y+2z=0$
$2x-2z=18$

Now, add it to the third row:

$x-2y=1$
$4x+6y=18$

That eliminated z.

Now, Multiply the first row by -4:

$-4x+8y=-4$
$4x+6y=18$

Add to the second row:

$14y=14$

$y=1$

Now, back substitute to find x and z.