Question 968002
<pre>
{{{system(x^2-y = 0,x + y = 2)}}}

Solve the second equation for y:

x + y = 2
    y = 2 - x

Substitute (2 - x) for y in the first equation:

{{{x^2 - y = 0}}}
{{{x^2 - (2 - x) = 0}}}
{{{x^2 - 2 + x = 0}}}
{{{x^2 + x - 2 = 0}}}
{{{(x+2)(x-1) = 0}}}
{{{x+2=0}}},  {{{x-1=0}}}
{{{x=-2}}},   {{{x=1}}}

Substitute x = -2 in

  y = 2 - x
  y = 2 - (-2)
  y = 2 + 2
  y = 4

So one point where they cross is (-2,4).

Substitute x = 1 in

  y = 2 - x
  y = 2 - (1)
  y = 2 - 1
  y = 1

So another point where they cross is (1,1).

Get some points on     x + y = 2, (-1,3), (-2,4), (0,2), (1,1), (2,0)
Get some points on     x^2 - y = 0, (-2,4), (-1,1), (0,0), (1,1), (2,4) 

{{{drawing(2400/7,400,-3,3,-2,5,

graph(2400/7,400,-3,3,-2,5,x^2),graph(2400/7,400,-3,3,-2,5,2-x),
locate(-2.8,4,"(-2,4)"),
locate(1.1,1.2,"(1,1)")


)}}}

Edwin</pre>