Question 964853
From eq. 2,
{{{x^2=3-y}}}
Substitute into eq. 1,
{{{3-y+y^2=5}}}
{{{y^2-y-2=0}}}
{{{(y-2)(y+1)=0}}}
Two solutions:
{{{y-2=0}}}
{{{y=2}}}
Then,
{{{x^2=3-2}}}
{{{x^2=1}}}
{{{x=0 +- 1}}}
(1,2) and (-1,2)
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{{{y+1=0}}}
{{{y=-1}}}
{{{-x^2+3=-1}}}
{{{-x^2=-4}}}
{{{x^2=4}}}
{{{x=0 +- 2}}}
(2,-1) and (2,1)
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