Question 1129923

{{{-x^2 + y= 8}}}........eq.1
{{{8y= -x }}}........eq.2
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{{{8y= -x }}}........eq.2=> {{{-8y= x }}}.......substitute in .eq.1

{{{-(-8y)^2 + y= 8}}}........eq.1

{{{-64y^2 + y= 8}}}

{{{0=64y^2-y+ 8}}}


{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{y= (-(-1) +- sqrt( (-1)^2-4*64*8))/(2*64) }}} 

{{{y= (1 +- sqrt( 1-2048))/128 }}} 

{{{y= (1 +- sqrt( -2047))/128 }}} 

{{{y= (1 +- sqrt( 2047)*i)/128 }}} 


{{{y= (1/128)(1 +- sqrt( 2047)*i) }}} 


solutions:

{{{highlight(y= (1/128)(1 + sqrt( 2047)*i)) }}} 
or
{{{highlight(y= (1/128)(1 - sqrt( 2047)*i)) }}} 


and {{{x}}} will be

eq.2=> {{{x=-8y }}}
 {{{x=-8(1/128)(1 + sqrt( 2047)*i) }}}

{{{x=-cross(8)(1/cross(128)16)(1 + sqrt( 2047)*i) }}}

{{{highlight(x=-(1/16)(1 + sqrt( 2047)*i)) }}}

or
{{{highlight(x=-(1/16)(1 - sqrt( 2047)*i)) }}}