Question 766249
<pre>
 x²+y² = 9, 
   |x| = |y|

Square both sides of the second equation

    x² = y²

Substitute y² for x² in the first equation:

 y² + y² = 9
     2y² = 9
      y² = {{{9/2}}}
       y = {{{""+-sqrt(9/2)}}} = {{{"" +- sqrt(9)/sqrt(2)}}} = {{{"" +- 3/sqrt(2)}}} = {{{"" +- 3/sqrt(2)}}}{{{""*""}}}{{{sqrt(2)/sqrt(2)}}} = {{{"" +- 3sqrt(2)/2}}}

And since x² = y², x also = {{{"" +- 3sqrt(2)/2}}} 

So the four possible solutions are:

({{{3sqrt(2)/2}}},{{{3sqrt(2)/2}}}), ({{{3sqrt(2)/2}}},{{{-3sqrt(2)/2}}}), ({{{-3sqrt(2)/2}}},{{{3sqrt(2)/2}}}), ({{{-3sqrt(2)/2}}},{{{-3sqrt(2)/2}}})

All four check, so all four are solutions.

The graph of x²+y²=9 is this circle
{{{drawing(200,200,-4,4,-4,4,graph(200,200,-4,4,-4,4), circle(0,0,3) )}}}

and the graph of |y|=|x| is the pair of graphs y = |x| and y = -|x|

{{{drawing(200,200,-4,4,-4,4,graph(200,200,-4,4,-4,4),
line(-5,5,0,0),line(5,5,0,0) )}}} and{{{drawing(200,200,-4,4,-4,4,graph(200,200,-4,4,-4,4),
line(-5,-5,0,0),line(5,-5,0,0) )}}}

which together are

{{{drawing(200,200,-4,4,-4,4,graph(200,200,-4,4,-4,4),
line(-5,5,5,-5),line(5,5,-5,-5) )}}}

Put them together on the same graph and you get:

{{{drawing(200,200,-4,4,-4,4,graph(200,200,-4,4,-4,4),circle(0,0,3),
line(-5,5,5,-5),line(5,5,-5,-5) )}}}

The four points of intersection are 

({{{3sqrt(2)/2}}},{{{3sqrt(2)/2}}}), ({{{3sqrt(2)/2}}},{{{-3sqrt(2)/2}}}), ({{{-3sqrt(2)/2}}},{{{3sqrt(2)/2}}}), ({{{-3sqrt(2)/2}}},{{{-3sqrt(2)/2}}})

Edwin</pre>