How to Solve the System of Quadratic Equations

Solve the system of quadratic equations “2x² – 3xy + 4y² = 3 and x² + xy – 8y² = -6”
solve the system of quadratic equations
2x² – 3xy + 4y² = 3
x² + xy – 8y² = – 6
then (x, y) =?
Solutions
Let
2x² – 3xy + 4y² = 3……………eq(1)
x² + xy – 8y² = – 6……………..eq(2)
Multiply equation 1 with equation 2, then
2(2x² – 3xy + 4y²)= 2 × 3
⇒ 4x² – 6xy + 8y² = 6…………..eq(3)
Add equation 2 and equation 3
x² + xy – 8y² + 4x² – 6xy + 8y² = – 6 + 6
⇒ 5x² – 5xy = 0
⇒ 5x(x – y) = 0
Thus, x = 0 or x = y
When x = 0
From equation 2
x² + xy – 8y² = – 6
⇒ 0 + 0 – 8y² = – 6
⇒ -8y² = – 6
⇒ y² = 6/8
thus, y = ±√3/2
When x = y
From equation 2
x² + xy – 8y² = – 6
⇒ y² + y² – 8y² = – 6
⇒ – 6y² = – 6
⇒ y² = 1
so, y = ±1
thus, x = y = ±1
That is the solutions are (x, y) = (0, √3/2 ), (0, – √3/2 ), (1, 1) & ( – 1, – 1)