3^x+5^y = 4...(1)

9^x+25^5 = 10....(2).

We put 3^x = a and 5^y = b, then 9^x = (3^x)^2= 3^2x = (3^x)^2= a^2. And 25^y = 5^2y = (5^y)^2 = b^2.

Therefore a+b = 4 and a^2+b^2 = 10.

Therefore (a+b)^2 = 4^2= 16 . 2ab= (a+b)^2- (a^2+b^2) = 16-10 = 6.

Therefore (a-b)^2 = (a+b)^2 - 4ab= 16 - 2(2ab) = 16-2(6) = 4.

Therefore (a-b)^2 = 4. So a-b = sqer4 = 2.

a+b = 4...(3)

a-b = 2....(4).

(3)+(4) gives: 2a = 4+2 = 6. So a = 6/2 = 3.

(3)-(4) gives: 2b = 4-2 = 2. So b = 2/2 = 1.

a = 3^x = 3= 3^1. So x = 1.

b = 1. So 5^5 = 1= 5^0. So y = 0.

Therefore x = 1 and 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}.**