Question 388023
1.{{{x+y=13}}}
2.{{{xy=40}}}
From eq. 1,
{{{x=13-y}}}
Substitute into eq. 2,
{{{(13-y)y=40}}}
{{{13y-y^2=40}}}
{{{y^2-13y+40=0}}}
{{{(y-5)(y-8)=0}}}
Two solutions:
{{{y-5=0}}}
{{{y=5}}}
Then from eq. 2,
{{{x(5)=40}}}
{{{x=8}}}
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{{{y-8=0}}}
{{{y=8}}}
{{{x(8)=40}}}
{{{x=5}}}
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{{{drawing(300,300,-2,10,-2,10,grid(1),
circle(5,8,0.3),circle(8,5,0.3),
graph(300,300,-2,10,-2,10,0,13-x,40/x))}}}