Question 372393
1.{{{xy = 3 }}}
2.{{{x + y = 4}}} 
From eq. 2,
{{{x=4-y}}}
Substitute into eq. 1,
{{{(4-y)y=3}}}
{{{4y-y^2=3}}}
{{{y^2-4y+3=0}}}
{{{(y-1)(y-3)=0}}}
Two solutions,
{{{y-1=0}}}
{{{y=1}}}
Then 
{{{x=3/1=3}}}
(3,1)
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.
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{{{y-3=0}}}
{{{y=3}}}
Then 
{{{x=3/3=1}}}
(1,3)
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{{{drawing(300,300,-5,5,-5,5,grid(1),circle(1,3,0.2),circle(3,1,0.2),graph(300,300,-5,5,-5,5,0,3/x,4-x))}}}