Question 969638
Can you decide what to "substitute", and where?  You are given two equations and plainly told, "solve by the substitution method".  


What is that y= something.... ?



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You very well CAN expect two solutions or two points!  (...Until you analyze the system more carefully).  Trust your skills, and hopefully, trust your concepts, too.




{{{system(x^2+y^2=12,y=sqrt(x))}}}-----Note that {{{x>=0}}} is a requirement for the second equation.


{{{x^2+(sqrt(x))^2=12}}}


{{{x^2+x-12=0}}}, looks factorable...


{{{(x-3)(x+4)=0}}}------remember the necessary condition for x.
Solution for this quadratic now factored equation is
{{{x=3}}}  or {{{x=-4}}}.


Again, remember the necessary condition for that second equation of the system:
{{{highlight(x=3)}}}  or {{{cross(x=-4)}}}.