# Is y=x^2+1, y=2^(x^2) a solvable system?

## Is $y = {x}^{2} + 1$, $y = {2}^{{x}^{2}}$ a solvable system?

##### 1 Answer
May 23, 2017

Yes. It is a solvable system.

#### Explanation:

Making ${x}^{2} = y - 1$ and substituting

$y = {2}^{y - 1}$

A natural solution is $y = 1 \Rightarrow x = 0$ and after a close view also $y = 2 \Rightarrow x = \pm 1$