Question 829515
1.{{{x^2+y^2=13}}}
2.{{{x-2y=1}}}
From eq. 2,
{{{x=1+2y}}}
Substitute into eq. 1,
{{{(2y+1)^2+y^2=13}}}
{{{4y^2+4y+1+y^2=13}}}
{{{5y^2+4y+1=13}}}
{{{5y^2+4y-12=0}}}
{{{(5y-6)(y+2)=0}}}
Two solutions
{{{5y-6=0}}}
{{{5y=6}}}
{{{y=6/5}}}
Then from eq. 2,
{{{x=1+2(6/5)}}}
{{{x=1+12/5}}}
{{{x=17/5}}}
Also,
{{{y+2=0}}}
{{{y=-2}}}
Then from eq. 2,
{{{x=1+2(-2)}}}
{{{x=1-4}}}
{{{x=-3}}}
The intersection of a line with a circle,
({{{17/5}}},{{{6/5}}}) and ({{{-3}}},{{{-2}}})
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{{{ graph( 300, 300, -5,5,-5,5, sqrt(13-x^2), -sqrt(13-x^2), (x-1)/2) }}}