SOLUTION: Solve each system of equations: {3x^2+4xy+5y^2=8 and x^2+3xy+2y^2=0 When I try to solve it, it turns out to be really messy. PLease use elimination or substituion in your answe

Algebra ->  Systems-of-equations -> SOLUTION: Solve each system of equations: {3x^2+4xy+5y^2=8 and x^2+3xy+2y^2=0 When I try to solve it, it turns out to be really messy. PLease use elimination or substituion in your answe      Log On


   

Question 1021116: Solve each system of equations:
{3x^2+4xy+5y^2=8 and x^2+3xy+2y^2=0
When I try to solve it, it turns out to be really messy. PLease use elimination or substituion in your answer. Thank you
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2%2B4xy%2B5y%5E2=8+ (Equation A)
+x%5E2%2B3xy%2B2y%5E2=0 (Equation B)
The key lies in factoring Eqn. B: +x%5E2%2B3xy%2B2y%5E2=0<==> (x+y)(x+2y) = 0.
We discuss 2 cases:
(i) When x+y = 0, or y = -x. Substitute this into Eqn.A
==> 3x%5E2-4x%5E2%2B5x%5E2+=+8==> 4x%5E2+=+8 ==> x%5E2+=+2
==> x+=sqrt%282%29 or x+=+-+sqrt%282%29
==> the solutions (sqrt%282%29, -sqrt%282%29) and (-sqrt%282%29, sqrt%282%29).
(ii) The case of x+2y = 0, or x = -2y, is handled in a similar manner.
After doing so we get the pair of solutions (-4sqrt%282%29%2F3, 2sqrt%282%29%2F3) and (4sqrt%282%29%2F3, -2sqrt%282%29%2F3).
Therefore the solutions are (sqrt%282%29, -sqrt%282%29), (-sqrt%282%29, sqrt%282%29), (-4sqrt%282%29%2F3, 2sqrt%282%29%2F3), and (4sqrt%282%29%2F3, -2sqrt%282%29%2F3).