x*-2xy+3y*=9 4x*-5xy+6y*=30*指平方由这两个式子组成的方程组,求x和y的解

Question

x*-2xy+3y*=9 4x*-5xy+6y*=30
*指平方由这两个式子组成的方程组,求x和y的解

Answer 1

Solution:x^2 - 2xy + 3y^2 = 9 (1)
4x^2 - 5xy + 6y^2 = 30 (2)
Eq(1)*5 - Eq(2)*2,we have x^2 - y^2 = 5 (3)
Eq(2) - Eq(1),we get x^2 + y^2 - xy = 7 (4)
Eq(3) + Eq(4),we obtain x^2 - xy = 12
Make x subject,x = 2(y - 1/y) (5)
Sub Eq(5) into Eq(3),4(y^2 - 2 + 1/y^2) - y^2 =5
Let u = y^2,4(u - 2 + 1/u) - u = 5
Transposing and Simplifying,3u^2 - 13u + 4 = 0
Factorizing,then u1 = 4,u2 = 1/3
Sub u = y^2 into u1 = 4,we find y1 = 2,
y2 = -2,
Sub u = y^2 into u1 = 1/3,we find y3 = 1/(square root of 3)
y4 = -1/(square root of 3)
Correspondingly,x1 = 3
x2 = -3
x3 = -4/(square root of 3)
x4 = 4/(square root of 3)
By inspection,we confirm all solutions above are correct.