Question 958358

{{{xy=4}}}...........(1)
{{{x^2+y^2=8}}}.......(2)
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{{{xy=4}}}...........(1)...solve for {{{x}}}

{{{x=4/y}}}..........substitute in (2)

{{{(4/y)^2+y^2=8}}}.......(2) solve for {{{y}}}

{{{16/y^2+y^2=8}}}

{{{(16+y^4)/y^2=8}}}

{{{16+y^4=8y^2}}}

{{{y^4-8y^2+16=0}}}

{{{(y^2)^2-8y^2+4^2=0}}}

{{{(y^2-4)^2=0}}}

{{{(y^2-4)(y^2-4)=0}}}

{{{(y-2)(y+2)(y-2)(y+2)=0}}}

solutions: two double solutions

if {{{(y-2)=0}}}=>{{{y=2}}}
if {{{(y+2)=0}}}=>{{{y=-2}}}

now find {{{x}}}

{{{x=4/y}}} substitute =>{{{y=2}}}

{{{x=4/2}}}

{{{x=2}}}

and

{{{x=4/y}}} substitute =>{{{y=-2}}}

{{{x=4/-2}}}

{{{x=-2}}}

so, 
{{{x=2}}},{{{y=2}}} 
or 
{{{x=-2}}},{{{y=-2}}}