Question 983847
{{{x+2y+3z=1}}}..........eq.1
{{{3x-y+z=1}}}..........eq.2
{{{4x+2y+z=3}}}..........eq.3
--------------------------------------

start with

{{{x+2y+3z=1}}}..........eq.1.....both sides multiply by {{{3}}}
{{{3x-y+z=1}}}..........eq.2
-------------------------------


{{{3x+6y+9z=3}}}..........eq.1
{{{3x-y+z=1}}}..........eq.2
---------------------------------subtract eq.2 from eq.1

{{{3x+6y+9z-(3x-y+z)=3-1}}}
{{{cross(3x)+6y+9z-cross(3x)+y-z=2}}}
{{{7y+8z=2}}}.......solve for {{{y}}}

{{{7y=2-8z}}}
{{{y=(2-8z)/7}}.......eq.1a

now, go to

{{{x+2y+3z=1}}}..........eq.1...both sides multiply by {{{4}}}
{{{4x+2y+z=3}}}..........eq.3
-------------------------------------
{{{4x+8y+12z=4}}}..........eq.1 
{{{4x+2y+z=3}}}..........eq.3
-------------------------------------subtract eq.3 from eq.1

{{{4x+8y+12z-(4x+2y+z)=4-3}}}
{{{cross(4x)+8y+12z-cross(4x)-2y-z=1}}}
{{{6y+11z=1}}}......solve for {{{y}}}

{{{6y=1-11z}}}
{{{y=(1-11z)/6}}}.......eq.2a

since eq.1a and eq.2a have same left sides, then right sides are equal too

{{{(2-8z)/7=(1-11z)/6}}}...........solve for {{{z}}}

{{{(2-8z)6=7(1-11z)}}}

{{{12-48z=7-77z}}}

{{{77z-48z=7-12}}}

{{{29z=-5}}}

{{{highlight(z=-5/29)}}}

go to {{{y=(1-11z)/6}}}.......eq.2a, substitute {{{-5/29}}} for {{{z}}} and find {{{y}}}

{{{y=(1-11(-5/29))/6}}}

{{{y=(1+55/29)/6}}}

{{{y=(29/29+55/29)/6}}}

{{{y=(cross(84)14/29)/cross(6)1}}}

{{{highlight(y=14/29)}}}

go to {{{x+2y+3z=1}}}..........eq.1 substitute {{{-5/29}}} for {{{z}}},{{{14/29}}} for {{{y}}} and find {{{x}}}

{{{x+2(14/29)+3(-5/29)=1}}}..........eq.1

{{{x+28/29-15/29=1}}}

{{{x+13/29=1}}}

{{{x=1-313/29}}}

{{{x=29/29-13/29}}}

{{{highlight(x=16/29)}}}


so, your solutions are:

{{{highlight(x=16/29)}}}

{{{highlight(y=14/29)}}}

{{{highlight(z=-5/29)}}}