SOLUTION: x²-xy=3 2y²+2xy=9 solve for the value of x and y

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Question 1117424: x²-xy=3
2y²+2xy=9
solve for the value of x and y
Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.
x%5E2+-+xy   = 3     (1)

2y%5E2+%2B+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%5E2+%2B+2xy%29 = 3%2A%28x%5E2-xy%29,   or, equivalently


3x%5E2+-+5xy+-+2y%5E2 = 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)  ====>  %282y%29%5E2+-+%282y%29%2Ay = 3  ====>  2y%5E2 = 3  

              ====>  y%5E2 = 3%2F2  ====>  y = +/- sqrt%283%2F2%29 = +/- sqrt%286%29%2F2.


              If y = sqrt%286%29%2F2   <--->  x = sqrt%286%29,    and both equations (1) and (2) are satisfied.

              If y = -sqrt%286%29%2F2  <--->  x = -sqrt%286%29,   and both equations (1) and (2) are satisfied.



2)  3x+y = 0  ====>  y = -3x  ====>  substitute it into eq(1)  ====>  x%5E2+-+x%2A%28-3x%29 = 3    ====>  4x%5E2 = 3  

              ====>  x%5E2 = 3/4  ====>  x = +/- sqrt%283%29%2F2.

              
              If x = sqrt%283%29%2F2   <--->  y = -%283%2Asqrt%283%29%29%2F2,    and both equations (1) and (2) are satisfied.

              If x = -sqrt%283%29%2F2  <--->  y = %283%2Asqrt%283%29%29%2F2,     and both equations (1) and (2) are satisfied.


Answer.  The system (1), (2) has four solutions  (x,y) = (sqrt%286%29,sqrt%286%29%2F2),  (-sqrt%286%29,-sqrt%286%29%2F2),  (sqrt%283%29%2F2,%28-3%2Asqrt%283%29%29%2F2),  (-sqrt%283%29%2F2,%283%2Asqrt%283%29%29%2F2).