# solve in x y x^2+2^y=8 x+2^y+1=10

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You should write the first equation such that:

`2^y = 8 - x^2`

You may substitute `8 - x^2` for `2^y` in the second equation such that:

`x + 8 - x^2 + 1 = 10`

`- x^2 + x + 9 - 10 = 0`

`x^2 - x + 1 = 0`

If you multiply by x+1 both sides yields:

`(x+1)(x^2 - x + 1) = 0`

You may write the product as difference of cubes such that:

`x^3 + 1 = 0`

You need to consider the real root of this equation, only:

`x = -1`

You need to substitute -1 for x in the firs equation such that:

`2^y = 8 - 1 =gt 2^y = 7`

You need to take logarithms of both sides such that:

`ln (2^y) = ln 7 =gt y*ln 2 = ln 7`

`y = ln 7/ln 2`

**Hence, evaluating the solution to the system of equations yields `x = -1 , y = ln 7/ln 2.` **