# How to Solve the System of Quadratic Equations

If ** x² – 5xy + 6y² = 0** and

**, then solve the system of quadratic equations**

*xy*–*y*² = 2x^2-5xy+6y^2=0

xy-y^2=2

(x, \ y)=?

## Solution to the system of quadratic equations

Let

*x*² – 5*xy* + 6*y*² = 0………………*eq*(1)

*xy* – *y*² = 2………………………….*eq*(2)

## Apply quadratic equation formula in equation 1

*x*² – 5*xy* + 6*y*² = 0

6*y*² – 5*xy* + *x*² = 0

\begin{aligned} y&=\frac {5x \pm \sqrt{(-5x)^2-4 \times 6 \times x^2}}{2 \times 6} \\ \\ &=\frac {5x \pm \sqrt{25x^2-24 x^2}}{12} \\ \\ &=\frac {5x \pm x}{12} \\ \\ \Rightarrow y&=\frac{x}{2} \ \ \& \ \ y=\frac{x}{3}\\ \\ \Rightarrow x&=2y \ \ \& \ \ x=3y\\ \end{aligned}

When *x* = 2*y*

From equation 2

*xy* – *y*² = 2

⇒ (2*y*)*y* – *y*² = 2

⇒ 2*y*² – *y*² = 2

So, *y*² = 2

⇒ *y* = ±√2

⇒ * x *= 2

*y*=

**±2√2**

When *x* = 3*y*

From equation 2

*xy* – *y*² = 2

⇒ (3*y*)*y* – *y*² = 2

⇒ 3*y*² – *y*² = 2

So, 2*y*² = 2

*y*² = 1

⇒ *y* = ±1

⇒ * x *= 3

*y*=

**±3**

So the solution to the system of quadratic equations are **( x, y) = (√2, 2√2), (-√2, -2√2), (3, 1) & (-3, -1) **