Question 549768
<pre>

ax + by = c
dx + ey = f

By Cramer's rule:

Delta = {{{abs(matrix(2,2,a,b,d,e))}}} = ae - bd

D<sub>x</sub> = {{{abs(matrix(2,2,c,b,f,e))}}} = ce - bf 

D<sub>y</sub> = {{{abs(matrix(2,2,a,c,d,f))}}} = af - cd

x = {{{D[x]/(Delta)}}} = {{{abs(matrix(2,2,c,b,f,e))/(ae-bd)}}} = {{{(ce-bf)/(ae-bd)}}}

y = {{{D[y]/(Delta)}}} = {{{abs(matrix(2,2,a,c,d,f))/(ae-bd)}}} = {{{(af-cd)/(ae-bd)}}}

Denominators cannot be 0, so Delta = {{{abs(matrix(2,2,a,b,d,e))}}} = ae - bd &#8800; 0.
and both {{{(ce-fe)/(ae-bd)}}} and {{{(af-cd)/(ae-bd)}}} have unique
values when that denominator is not zero.

------------------------------------------------------

In case you need to explain why Cramer's rule works:
 
To eliminate y, multiply the first equation by e and the
second equation by -b, and add the equations vertically
term by term:

     aex + bey =  ce
    -bdx - bey = -bf
-----------------------
 aex-bdx       =  ce-bf  
(ae-bd)x       =  ce-bf
       x = {{{(ce-bf)/(ae-bd)}}}

To eliminate x, multiply the first equation by -d and the
second equation by a, and add the equations vertically
term by term:

  -adx   - bdy = -cd
   adx   + aey =  af
-----------------------
       aey-bdy =  af-cd  
      (ae-bd)y =  af-cd
             y = {{{(af-cd)/(ae-bd)}}}

a and y are the same using the elimination method as they are 
using Cramer's rule.
  
Edwin</pre>