Find x and y: log 4 (x+y) = 2 log 3 x + log 3 y = 2 + log 3 7

We'll impose the constraints of existence of
logarithms:

x>0

y>0

Now,
we'll take anti-logarithm for the first equation:

x+y =
4^2

x+y = 16 (1)

We'll apply
the product rule of logarithms in the second equation:

log
3 x + log 3 y = log 3 (x*y)

We'll re-write the sum from the
right side of the second equation:

2 + log 3 7 =2*1 + log 3
7

2*1 + log 3 7 = 2*log 3 3 + log 3
7

We'll apply the power rule of
logarithms:

2*log 3 3 = log 3 3^2 = log 3
9

We'll re-write the second
equation:

log 3 (x*y) = log 3 9 + log 3
7

log 3 (x*y) = log 3
(9*7)

log 3 (x*y) = log 3
(63)

Since the bases are matching, we'll apply one to one
rule:

x*y = 63 (2)

We'll write
x with respect to y, from (1) and we'll substitute in
(2):

x = 16 - y

(16 - y)*y =
63

We'll remove the
brackets:

16y - y^2 - 63 =
0

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

y^2 - 16y + 63 = 0

We'll
apply the quadratic formula:

y1 = [16+sqrt(256 -
252)]/2

y1 = (16+2)/2

y1 =
9

y2 = 7

x1 = 16 -
y1

x1 = 16-9

x1 =
7

x2 = 16-7

x2 =
9

The solutions of the symmetric system
are:

{9 ; 7} and {7 ;
9}