Question 1059326
More than one way.



R1 plus R3: 
{{{-2x+2y+3z+2x+3y+3z=0+5}}}

{{{5y+6z=5}}}



R2 plus R3:
{{{-2x-y+z+2x+3y+3z=-3+5}}}

{{{2y+4z=2}}}

{{{y+2z=1}}}



Work with the system:
{{{system(5y+6z=5,y+2z=1)}}}
-
Multiply the second equation here by 5 and subtract from first equation.
{{{system(5y+6z=5,5y+10z=5)}}}
-
{{{5y+6z-5y-10z=5-5}}}

{{{-4z=0}}}

{{{highlight(z=0)}}}
-
I would continue using this system to find y, also using elimination step.  This time, multiply second equation by 3...


{{{system(5y+6z=5,3y+6z=3)}}}


{{{5y+6z-3y-6z=5-3}}}

{{{2y=2}}}

{{{highlight(y=1)}}}


Choosing to continue in Elimination would be more effort than is justified.  Take your now found values for y and z, and take either of the original equations and solve for x.