I edited your original question in response to Answer #2.

If you subtract the first equation from the second one, you get `2a=2,` **so** `a=1.`

Althought it wasn't asked for, you can go further and solve for `x,` and we find that both `+-sqrt(15)` work. Finally, we can check our...

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I edited your original question in response to Answer #2.

If you subtract the first equation from the second one, you get `2a=2,` **so** `a=1.`

Althought it wasn't asked for, you can go further and solve for `x,` and we find that both `+-sqrt(15)` work. Finally, we can check our solution:

`(+-sqrt(15))^2-1=14` and `(+-sqrt(15))^2+1=16,` so it works.

It is given that x^2 - a = 14.

This equation has two variables x and a. x^2 - a can be equal to 14 for infinite number of sets of x and a. It is not possible to determine a unique solution for a from the given equation.

**As the equation x^2 - a = 14 has 2 independent variables it is not possible to determine a unique solution for a.**