# How to solve this systems of equations? x^2-yz=1` ` y^2-xz=2 z^2-xy=4

### 2 Answers | Add Yours

Assuming that `x,y,z!=0` , you may multiply the first equation by x, the second equation by y, the third equation by z, such that:

`{(x^3 - xyz = x),(y^3 - xyz = 2y),(z^3 - xyz = 4z):}`

`{(x^3 - x = xyz),(y^3 - 2y = xyz),(z^3 - 4z = xyz):} => x^3 - x = y^3 - 2y = z^3 - 4z`

You need to factor out x,y and z, such that:

`x(x^2 - 1) = y(y^2 - 2) = z(z^2 - 4)`

You need to notice that the equations are cancelled if the factors `x^2 - 1 = y^2 - 2 = z^2 - 4 = 0` , such that:

`x^2 - 1 = 0 => x_(1,2) = +-1`

`y^2 - 2 = 0 => y_(1,2) = +-sqrt2`

`z^2 - 4 = 0= > z_(1,2) = +-2`

**Hence, since `x_(1,2) = +-1 != 0, y_(1,2) = +-sqrt2 != 0, z_(1,2) = +-2 != 0` yields that the solutions to the system of equations are **`(1,sqrt2,2),(-1sqrt2,2),(1,-sqrt2,2),(-1,-sqrt2,2),(1,sqrt2,-2),(-1,sqrt2,-2),(-1,-sqrt2,-2),(1,-sqrt2,-2).`

1)

x^2 - yz = 1

x*x - y*z = 1

x*x = 1+yz

therefore value of x = squre root of (1+yz)

2)

y^2-xz=2

y^2=2+xz

the value of y = square root of (2+xz)

3) z^2-xy=4

z^2 = 4 + xy

the value of z is = square root of (4+xy)