Solve the system 3^x+5^y=4 9^x+25^y=10

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}.