Question 114564
3x+4y=13
2x-3y=-14
In matrix form, it would look like,
{{{(
 matrix( 2, 2, 
   3, 4,
   2, -3
 )
)( matrix( 2, 1, 
   x, y ))=( matrix( 2, 1, 
   13, -14 )
)}}}
[{{{A}}}][{{{x}}}]=[{{{b}}}] 
The solution is 
{{{(
 matrix( 2, 1, 
   x, y
 ))=( 
 matrix( 2, 2, 
   k, l,
   m, n 
))( 
 matrix( 2, 1, 
   13, -14 
))}}}
Where k,l,m,n are coefficients of the inverse matrix of 
{{{(
 matrix( 2, 2, 
   3, 4,
   2, -3
 )
)}}}
[{{{x}}}]=[{{{A^-1}}}][{{{b}}}] 
The inverse of a 2x2 matrix, A, where   
[{{{A}}}]={{{(
 matrix( 2, 2, 
   a, b,
   c, d
 )
)}}}
is given by
[{{{A^-1}}}]= {{{(1/(ad-bc))*(
 matrix( 2, 2, 
   d, -b,
   -c, a
 )
)}}}
Using your coefficients in the inverse matrix formula,
[{{{A^-1}}}]=  {{{(1/(3(-3)-4(2)))*(
 matrix( 2, 2, 
   3, 4,
   2, -3
 )
)}}}
[{{{A^-1}}}]=   {{{(-1/17)*(
 matrix( 2, 2, 
   -3, -4,
   -2, 3
 )
)}}}
[{{{A^-1}}}]=   {{{(1/17)*(
 matrix( 2, 2, 
   3, 4,
   2, -3
 )
)}}}
The solution is therefore
{{{(
 matrix( 2, 1, 
   x, y
 ))=(1/17)*( 
 matrix( 2, 2, 
   3, 4,
   2, -3 
))( 
 matrix( 2, 1, 
   13, -14 
))}}}
Let's carry out each calculation separately, 
{{{x=(1/17)*(3(13)+(4)(-14))}}}
{{{x=(1/17)*(39-56)}}}
{{{x=(1/17)*(-17)}}}
{{{highlight(x=-1)}}}
{{{y=(1/17)*(2(13)+(-3)(-14))}}}
{{{y=(1/17)*(26+42)}}}
{{{y=(1/17)*(68)}}}
{{{highlight(y=4)}}}
{{{(
 matrix( 2, 1, 
   x, y
 ))=(
 matrix( 2, 1, 
   -1, 4
 ))}}}