Question 1131435
using inverse matrix:

{{{3x - 2y = 5}}}
{{{x - y = 0}}}


{{{A=matrix(2,2,3,-2,1,-1)}}}

and 

{{{B=matrix(2,1,5,0)}}}



{{{A }}}is invertible and 

{{{A^-1=matrix(2,2,1,-2,1,-3)}}}


Then you get 

{{{X=A^1*B }}}={{{(matrix(2,2,1,-2,1,-3))}}}*{{{(matrix(2,1,5,0))}}}


={{{matrix(2,1,1*5+(-2)*0,1*5+(-3)*0)}}}


= {{{(matrix(2,1,5 + 0,5 + 0))}}}

= {{{(matrix(2,1, 5,5))}}}:

{{{x= 5}}}
{{{y=5}}}


or using Cramer's rule:

∆ = {{{ matrix(2,2,3,	-2,
1,	-1)= -1}}}


∆ [1]= {{{ matrix(2,2,5,	-2,
0,	-1)= -5}}}


∆ [2]={{{  matrix(2,2,3,	5,
1,	0)= -5}}}


{{{x}}}=∆ [1]/∆ ={{{-5/-1=5}}}
{{{y}}}=∆ [2]/∆ ={{{-5/-1=5}}}