Question 868618
{{{x^2 - y = 3}}}
{{{x - y = -3}}}
From eq. 2,
{{{y=x+3}}}
Substitute into eq. 1,
{{{x^2-(x+3)=3}}}
{{{x^2-x-3=3}}}
{{{x^2-x-6=0}}}
{{{(x-3)(x+2)=0}}}
Two solutions:
{{{x-3=0}}}
{{{x=3}}}
then
{{{y=3+3=6}}}
and
{{{x+2=0}}}
{{{x=-2}}}
then
{{{y=-2+3}}}
{{{y=1}}}

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{{{drawing(300,300,-4,6,-2,8,grid(1),
circle(-2,1,0.3),
circle(3,6,0.3),
graph(300,300,-4,6,-2,8,x+3,x^2-3))}}}