Question 1117424
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{{{x^2 - xy}}}   = 3     (1)

{{{2y^2 + 2xy}}} = 9     (2)



Right side of (2) is 9 = 3*3.  Replace one factor of 3 by the left side of (1).  You will get


{{{2y^2 + 2xy)}}} = {{{3*(x^2-xy)}}},   or, equivalently


{{{3x^2 - 5xy - 2y^2}}} = 0.    (3)  


Now factor left side of (3) to get

(x-2y)*(3x+y) = 0.


Thus EITHER  x-2y = 0  OR  3x+y = 0.


Lets consider each case separately.



1)  x-2y = 0  ====>  x = 2y   ====>  substitute it into eq(1)  ====>  {{{(2y)^2 - (2y)*y}}} = 3  ====>  {{{2y^2}}} = 3  

              ====>  {{{y^2}}} = {{{3/2}}}  ====>  y = +/- {{{sqrt(3/2)}}} = +/- {{{sqrt(6)/2}}}.


              If y = {{{sqrt(6)/2}}}   <--->  x = {{{sqrt(6)}}},    and both equations (1) and (2) are satisfied.

              If y = {{{-sqrt(6)/2}}}  <--->  x = {{{-sqrt(6)}}},   and both equations (1) and (2) are satisfied.



2)  3x+y = 0  ====>  y = -3x  ====>  substitute it into eq(1)  ====>  {{{x^2 - x*(-3x)}}} = 3    ====>  {{{4x^2}}} = 3  

              ====>  {{{x^2}}} = 3/4  ====>  x = +/- {{{sqrt(3)/2}}}.

              
              If x = {{{sqrt(3)/2}}}   <--->  y = {{{-(3*sqrt(3))/2}}},    and both equations (1) and (2) are satisfied.

              If x = {{{-sqrt(3)/2}}}  <--->  y = {{{(3*sqrt(3))/2}}},     and both equations (1) and (2) are satisfied.


<U>Answer</U>.  The system (1), (2) has four solutions  (x,y) = ({{{sqrt(6)}}},{{{sqrt(6)/2}}}),  ({{{-sqrt(6)}}},{{{-sqrt(6)/2}}}),  ({{{sqrt(3)/2}}},{{{(-3*sqrt(3))/2}}}),  ({{{-sqrt(3)/2}}},{{{(3*sqrt(3))/2}}}). 
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