# Find All Possible Natural Roots of a System of Equations

# Algebra math problem: How to solve system of two equations?

## If *x *+* y *=* **x*² and **x − y = ****1/***x*² then find all possible natural roots of *x* and *y*

*x*+

*y*=

*²*

*x***x − y =**

x+y=x^2

x-y=\dfrac{1}{x^2}

(x, y)=??

## Solution

Let

*x *+* y *=* x*

^{2}………………………

*eq*(1)

*x − y = *1/*x*^{2}……………………*eq*(2)

Add equation 1 and equation 2, then

*x *+* y * +* x* *− y *= *x*^{2} + 1/*x*^{2}

⇒ 2*x* = *x*^{2} + 1/*x*^{2}

⇒ 2*x*^{3}* *=

*x*

^{4}+ 1

⇒ *x*^{4} − 2*x*^{3}* *+ 1 = 0

⇒ *x*^{4} − *x*^{3}* *−

*x*^{3}

*+ 1 = 0*

⇒ *x*^{4} − *x*^{3} − *x*^{3} + 1 = 0

⇒ * x*

^{3}

*(*

*x*− 1) − (

*x*^{3}

*– 1) = 0*

⇒ *x*^{3}(*x* − 1) − (*x* − 1) (*x*^{2} + *x* + 1^{2}) = 0

⇒ (*x* − 1) (*x*^{3}* − x*^{2} − *x* − 1) = 0

Here, *x* − 1 = 0* *or *x*^{3}* − x*^{2} − *x* − 1 = 0

When *x *− 1 = 0, then

*x* = 1 → Natural Solution

*x*^{3}* − x*^{2} − *x* − 1 → Has no natural Solution

So ** x = 1** is the only natural solution

From equation 1

When *x* = 1

*x *+* y *=* x*

^{2}

⇒ 1* *+* y *=* *1

⇒ *y *=* *0

Then, natural roots of **( x, y)** is

**(1, 0)**