Question 1196171

using elimination method

{{{3x+2y+z=1}}}.....eq1
{{{x+y+z=0}}}......eq.2
{{{5x+3y-2z=-4}}}......,eq.3

start with

{{{3x+2y+z=1}}}.....eq1
{{{x+y+z=0}}}......eq.2
__________________________subtract  eq.2 from eq1

{{{3x+2y+z-x-y-z=1-0}}}...........eliminate {{{z}}}


{{{2x+y=1}}}......eq.a

now

{{{x+y+z=0}}}......eq.2
{{{5x+3y-2z=-4}}}......,eq.3
__________________________multiply   eq1 by {{{2}}}

{{{2x+2y+2z=0}}}......eq.2
{{{5x+3y-2z=-4}}}......,eq.3
__________________________add both

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

{{{7x+5y=-4}}}..................eq.b

now

{{{2x+y=1}}}......eq.a
{{{7x+5y=-4}}}..................eq.b
__________________________multiply   eq.a by {{{5}}}

{{{10x+5y=5}}}......eq.a
{{{7x+5y=-4}}}..................eq.b
__________________________subtract  eq.b from eq.a

{{{10x+5y-7x-5y=5-(-4)}}}

{{{3x=5+4}}}

{{{3x=9}}}

{{{x=3}}}

go to

{{{2x+y=1}}}......eq.a, plug in {{{x=3}}}

{{{2*3+y=1}}}

{{{6+y=1}}}

{{{y=1-6}}}

{{{y=-5}}}

now

{{{x+y+z=0}}}......eq.2, plug in {{{x=3}}} and {{{y=-5}}}

{{{3-5+z=0}}}

{{{-2+z=0}}}

{{{z=2}}}

so, solution to this system is: {{{x=3}}},{{{y=-5}}}, {{{z=2}}}