Question 1049309
please solve this one 
<pre><b>
{{{system(2y^2+5xy=1,3x-2y=-6)}}}

Solve the second equation for x

{{{3x-2y}}}{{{""=""}}}{{{-6}}}
{{{3x}}}{{{""=""}}}{{{-6+2y}}}
{{{x}}}{{{""=""}}}{{{(-6+2y)/3}}}

Substitute {{{((-6+2y)/3)}}} for x in the
first equation:

{{{2y^2+5xy}}}{{{""=""}}}{{{1}}}
{{{2y^2+5((-6+2y)/3)y}}}{{{""=""}}}{{{1}}}
{{{6y^2+5(-6+2y)y}}}{{{""=""}}}{{{3}}}
{{{6y^2+5y(-6+2y)}}}{{{""=""}}}{{{3}}}
{{{6y^2-30y+10y^2}}}{{{""=""}}}{{{3}}}
{{{16y^2-30y}}}{{{""=""}}}{{{3}}}
{{{16y^2-30y-3}}}{{{""=""}}}{{{0}}}
{{{y}}}{{{""=""}}}{{{(-b +- sqrt(b^2-4ac ))/(2a) }}} 
{{{y}}}{{{""=""}}}{{{(-(-30) +- sqrt((-30)^2-4(16)(-3) ))/(2(16)) }}} 
{{{y}}}{{{""=""}}}{{{(30 +- sqrt(900+192))/32 }}}
{{{y}}}{{{""=""}}}{{{(30 +- sqrt(1092))/32 }}}
{{{y}}}{{{""=""}}}{{{(30 +- sqrt(4*273))/32 }}}
{{{y}}}{{{""=""}}}{{{(30 +- 2sqrt(273))/32 }}}
{{{y}}}{{{""=""}}}{{{(2(15 +- sqrt(273)))/32 }}}
{{{y}}}{{{""=""}}}{{{(15 +- sqrt(273))/16 }}}


Two values of y, {{{y}}}{{{""=""}}}{{{(15 + sqrt(273))/16 }}} and {{{y}}}{{{""=""}}}{{{(15 - sqrt(273))/16 }}}

Substituting the first in

{{{3x-2y}}}{{{""=""}}}{{{-6}}}

{{{3x-2((15 + sqrt(273))/16)}}}{{{""=""}}}{{{-6}}}

{{{3x-((15 + sqrt(273))/8)}}}{{{""=""}}}{{{-6}}}

{{{24x-(15 + sqrt(273))}}}{{{""=""}}}{{{-48}}}

{{{24x-15 - sqrt(273)}}}{{{""=""}}}{{{-48}}}

{{{24x - sqrt(273)}}}{{{""=""}}}{{{-33}}}

{{{24x}}}{{{""=""}}}{{{-33 + sqrt(273)}}}

{{{x}}}{{{""=""}}}{{{(-33 + sqrt(273))/24}}}

So one solution is

{{{(matrix(1,3,x,",",y))}}}{{{""=""}}}{{{(matrix(1,3,(-33 + sqrt(273))/24,",",(15 + sqrt(273))/16))}}}

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Substituting the second in

{{{3x-2y}}}{{{""=""}}}{{{-6}}}

{{{3x-2((15 - sqrt(273))/16)}}}{{{""=""}}}{{{-6}}}

{{{3x-((15 - sqrt(273))/8)}}}{{{""=""}}}{{{-6}}}

{{{24x-(15 - sqrt(273))}}}{{{""=""}}}{{{-48}}}

{{{24x-15 + sqrt(273)}}}{{{""=""}}}{{{-48}}}

{{{24x + sqrt(273)}}}{{{""=""}}}{{{-33}}}

{{{24x}}}{{{""=""}}}{{{-33 - sqrt(273)}}}

{{{x}}}{{{""=""}}}{{{(-33 - sqrt(273))/24}}}

So the other solution is

{{{(matrix(1,3,x,",",y))}}}{{{""=""}}}{{{(matrix(1,3,(-33 - sqrt(273))/24,",",(15 - sqrt(273))/16))}}}

Edwin</pre></b>