Question 359472
1.{{{x^2+y^2=24}}}
2.{{{y=2x+3}}}
Substitute eq. 2 into eq. 1 and solve for {{{x}}}.
{{{x^2+(2x+3)^2=24}}}
{{{x^2+4x^2+12x+9=24}}}
{{{5x^2+12x+9=24}}}
{{{5x^2+12x-15=0}}}
Use the quadratic formula,
{{{x = (-12 +- sqrt( 12^2-4*5*(-15) ))/(2*5) }}} 
{{{x = (-12 +- sqrt( 144+300 ))/10 }}} 
{{{x = (-12 +- sqrt( 444 ))/10 }}}
{{{x = (-12 +- 2*sqrt( 111 ))/10 }}}
{{{x = (-6 +- sqrt( 111 ))/5}}}
Then from eq. 2,
{{{y=2x+3}}}
{{{y=(-12 +- 2*sqrt( 111 ))/5+15/5}}}
{{{y=(3 +- 2*sqrt( 111 ))/5}}}
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({{{x[1]}}},{{{y[1]}}})= ({{{ (-6 + sqrt( 111 ))/5}}},{{{ (-3 + 2*sqrt( 111 ))/5}}})
({{{x[2]}}},{{{y[2]}}})=({{{ (-6 - sqrt( 111 ))/5}}},{{{ (-3 - 2*sqrt( 111 ))/5}}})
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{{{drawing(300,300,-6,6,-6,6,circle(0.907,4.81,0.3),circle(-3.31,-3.61,0.3),grid(1),graph(300,300,-6,6,-6,6,sqrt(24-x^2)),graph(300,300,-6,6,-6,6,-sqrt(24-x^2),2x+3))}}}