Question 878182
<pre>
I just guessed and put in an x and an equal sign in the second one.

{{{system(x^2-expr(48/9)x+expr(1/3)y+expr(1/3)=0,
expr(-5/4)x^2-expr(3/2)x+expr(1/4)y-expr(1/2)=0)}}}

Reduce the fraction {{{48/9}}} to {{{16/3}}}

{{{system(x^2-expr(16/3)x+expr(1/3)y+expr(1/3)=0,
expr(-5/4)x^2-expr(3/2)x+expr(1/4)y-expr(1/2)=0)}}}

Multiply the first one through by 3 and the
second one through by 4

{{{system(3x^2-16x+y+1=0,
-5x^2-6x+y-2=0)}}}

Solve each for y:

{{{system(y = -3x^2+16x-1,
y = 5x^2+6x+2)}}}

St the right sides equal

{{{-3x^2+16x-1}}}{{{""=""}}}{{{5x^2+6x+2)}}} 

Get 0 on the left side:

{{{0}}}{{{""=""}}}{{{8x^2-10x+3)}}}

Factor the right side:

{{{0}}}{{{""=""}}}{{{(2x-1)(4x-3))}}}

2x-1 = 0;    4x-3 = 0
  2x = 1       4x = 3
   x = {{{1/2}}};       x = {{{3/4}}}
 
Substituting     x = {{{1/2}}} in

{{{y}}}{{{""=""}}}{{{-3x^2+16x-1}}}

{{{y}}}{{{""=""}}}{{{-3(1/2)^2+16(1/2)-1}}}

{{{y}}}{{{""=""}}}{{{-3(1/4)+8-1}}}

{{{y}}}{{{""=""}}}{{{-3/4+7}}}

{{{y}}}{{{""=""}}}{{{-3/4+28/4}}}

{{{y}}}{{{""=""}}}{{{25/4}}}

So one solution is (x,y) = {{{(matrix(1,3,1/2, ",", 25/4))}}}      
  
-----

Substituting     x = {{{3/4}}} in

{{{y}}}{{{""=""}}}{{{-3x^2+16x-1}}}

{{{y}}}{{{""=""}}}{{{-3(3/4)^2+16(3/4)-1}}}

{{{y}}}{{{""=""}}}{{{-3(9/16)+12-1}}}

{{{y}}}{{{""=""}}}{{{-27/16+11}}}

{{{y}}}{{{""=""}}}{{{-27/16+176/16}}}

{{{y}}}{{{""=""}}}{{{149/16}}}

So the other solution is (x,y) = {{{(matrix(1,3,3/4, ",", 149/16))}}}     

Edwin</pre>