You should use substitution equation x = 3 - y to solve the second equation such that:

`2^(3-y) + 2^y = 6`

You need to use the properties of exponentials such that:

`(2^3)/(2^y) + 2^y = 6`

You should bring the terms of equation to a common denominator such that:

`8 + 2^(2y) = 6*2^y`

You should come up with the substititution `2^y = a` such that:

`8 + a^2 = 6a`

You need to move the terms to the left side such that:

`a^2 - 6a + 8 = 0`

You should use quadrtic formula such that:

`a_(1,2) = (6+-sqrt(36 - 32))/2`

`a_1 = (6+2)/2 =gt a_1 = 4`

`a_2 = (6-2)/2 =gt a_2 = 2`

You need to solve for y the equations such that:

`2^y = 4 =gt 2^y = 2^2 =gt y = 2`

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

You need to use the first equation x = 3 - y to find x such that:

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

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

**Hence, the solutions to the system of simultaneous equations are x = 1 ; y = 2 and x = 2 ; y = 1.**