Question 374726
1.{{{x^2+y^2=12}}}
2.{{{x^2+y=10}}}
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From eq. 2,
{{{x^2=10-y}}}
Substitute into eq. 1,
{{{10-y+y^2=12}}}
{{{y^2-y-2=0}}}
{{{(y-2)(y+1)=0}}}
Two solutions,
{{{y-2=0}}}
{{{y=2}}}
Then 
{{{x^2=10-2}}}
{{{x^2=8}}}
{{{x=0 +- 2*sqrt(2)}}}
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{{{y+1=0}}}
{{{y=-1}}}
Then,
{{{x^2=10+1}}}
{{{x^2=11}}}
{{{x=0 +- sqrt(11)}}}
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{{{drawing(300,300,-5,5,-5,5,grid(1),
circle(-sqrt(11),-1,0.3),
circle(sqrt(11),-1,0.3),
circle(-sqrt(8),2,0.3),
circle(sqrt(8),2,0.3),
graph(300,300,-5,5,-5,5,0,sqrt(12-x^2),10-x^2),
graph(300,300,-5,5,-5,5,0,-sqrt(12-x^2)))}}}
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{{{log(x,(2y))=3}}} is equivalent to {{{x^3=2y}}}
{{{y=(1/2)x^3}}}
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{{{log(x,(4y))=2}}} is equivalent to {{{x^2=4y}}}
{{{y=(1/4)x^2}}}
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{{{(1/2)x^3=(1/4)x^2}}}
{{{2x^3-x^2=0}}}
{{{x^2(2x-1)=0}}}
Two solutions:
{{{x=0}}}
Then
{{{y=0}}}
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{{{2x-1=0}}}
{{{2x=1}}}
{{{x=1/2}}}
Then,
{{{y=(1/2)x^3}}}
{{{y=(1/2)(1/2)^3}}}
{{{y=1/16}}}
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{{{drawing(300,300,-1,1,-1,1,grid(1),
circle(1/2,1/16,0.1),
circle(0,0,0.1),
graph(300,300,-1,1,-1,1,0,(1/2)x^3,(1/4)x^2)))}}}