# What are x,y in equations x+2^y=3; x^2+4^y=5?

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You may write x in terms of y using the first equation such that:

`x = 3 - 2^y`

The reason of expressing x in terms of y is to substitute `3 - 2^y` for x in the second equation such that:

`(3 - 2^y)^2 + 4^y = 5`

You should expand the square such that:

`9 - 6*2^y + 2^(2y) + 2^(2y) - 5 = 0`

`2*2^(2y) - 6*2^y + 4 = 0`

You should come up with the substitution `2^y = t` such that:

`2*t^2 - 6*t + 4 = 0`

You need to divide by 2 such that:

`t^2 - 3t + 2 = 0`

You should use quadratic formula such that:

`t_(1,2) = (3+-sqrt(9-8))/2 `

`t_(1,2) = (3+-1)/2`

`t_1 = 2 ; t_2 = 1`

You need to solve for y the equations `2^y = 2` and `2^y = 1` such that:

`2^y = 2 =gt y = 1`

`2^y = 1 =gt 2^y = 2^0 =gt y = 0`

You need to find x such that:

`y = 1 =gt x = 3 - 2 =gt x = 1`

`y = 0 =gt x = 3 - 1 =gt x = 2`

**Hence, evaluating solutions to simultaneous equations yields `x=1;y=1` and `x=2;y=0` .**

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