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)