You may also use the following substitution, such that:

`3^x = 4 - 5^y`

Substituting `4 - 5^y` for `3^x` in the bottom equation yields:

`9^x + 25^y = 10 => 3^(2x) + 5^(2y) = 10`

`(3^x)^2 + 5^(2y) = 10 => (4 - 5^y)^2 + 5^(2y) = 10`

Expanding the square yields:

`16 - 8*5^y + 5^(2y) + 5^(2y) = 10`

`2*5^(2y) - 8*5^y + 6 = 0 => 5^(2y) - 4*5^y + 3 = 0`

You need to come up with the substitution `5^y = t` , such that:

`t^2 - 4t + 3 = 0 => t^2 - 3t - t + 3 = 0`

`t(t - 3) - (t - 3) = 0 => (t - 3)(t - 1) = 0

`=gt {(t - 3 = 0),(t - 1 = 0):} =gt {(t = 3),(t = 1):}` `

`{(5^y = 3),(5^y = 1):} => {(y = ln 3/ln 5),(y = 0):}`

`3^x = 4 - 3 => 3^x = 1 => x = 0`

`3^x = 4 - 1 => 3^x = 3 => x = 1`

**Hence, evaluating the solutions to the system yields `x = 0, y = log_5 3` and **`x = 1, y = 0.`

We'll note the terms from the first equation as:

3^x = u and 5^y = v.

We notice that if we square the terms from the first equation, we'll obtain the terms from the second equation:

(3^2)^x = u^2 and (5^2)^y = v^2

We'll re-write the system using the new variables u and v:

u + v = 4 (1)

u^2 + v^2 = 10

But u^2 + v^2 = (u+v)^2 - 2uv

u^2 + v^2 = 16 - 2uv

16 - 2uv = 10

2uv = 16 - 10

uv = 3 (2)

We'll write u with respect to v, from (1):

u = 4 - v (3)

We'll substitute (3) in (2):

(4 - v)*v = 3

We'll remove the brackets:

4v - v^2 - 3 = 0

We'll multiply by -1 and we'll re-arrange the terms:

v^2 - 4v + 3 = 0

We'll apply the quadratic formula:

v1 = [4+sqrt(16 - 12)]/2

v1 = (4+2)/2

v1 = 3

v2 = 1

5^y = 3

We'll take logarithms both sides:

ln 5^y = ln 3

y*ln 5 = ln 3

y = ln 3/ln 5

5^y = 1

5^y = 5^0

y = 0

Now, we'll calculate u and then, x:

u1 = 4 - v1

u1 = 4 - 3

u1 = 1

u2 = 4 - v2

u2 = 4 - 1

u2 = 3

We'll calculate x:

3^x = 1

x = 0

3^x = 3

x = 1

The solutions of the system are: {1 ; 0} and {0 ; ln 3 /ln 5}.