# What is the range and domain of the relations : x^2/16 - y^2/9 = 1

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`x^2/16-y^2/9 = 1`

If we rearrange this as below, we can talk about domain more precisely.

`y^2/9 = x^2/16-1`

For y to have real values, y^2 must be equal or greater than zero.

`y^2 =gt 0`

Therefore,

`x^2/16-1 =gt 0`

x^2/16 =>1

`x^2 =gt 16`

`x^2-4^2 =gt 0`

`(x-4)(x+4) =gt 0`

The solutions for this inequality is `x lt= -4` and `x =gt 4`

Therefore the domain is, `{x inR;x in (-oo, -4) uu x in (4, oo)}`

The range is `y in R`

The function `x^2/16 - y^2/9 = 1` is the equation of a hyperbola.

**The domain of the function is `(-oo, -4)U(4, oo)` . The range of the function is the set of real numbers R.**