Question 328152
1.{{{2xy=3}}}
2.{{{4x^2-8y^2=1}}}
From eq. 1,
{{{y=3/(2x)}}}
{{{y^2=9/(4x^2)}}}
Substitute into eq. 2,
{{{4x^2-8(9/(4x^2))=1|||
{{{4x^2-18/x^2=1}}}
{{{4x^4-18=x^2}}}
{{{4x^4-x^2-18=0}}}
{{{(4x^2-9)(x^2+2)=0}}}
One solution:
{{{4x^2-9=0}}}
{{{4x^2=9}}}
{{{x^2=9/4}}}
{{{x=0 +- 3/2}}}
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Now find y.
{{{y=3/(2x)}}}
{{{y=3/(2*(3/2))
{{{y=1}}}
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(-3/2,-1)
(3/2,1)
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{{{drawing(300,300,-10,10,-10,10,circle(-3/2,-1,.3),circle(3/2,1,.3),graph(300,300,-10,10,-10,10,3/(2x),-sqrt((4x^2-1)/8),sqrt((4x^2-1)/8)))}}}