Question 1111112
.
It is straightforward.


<pre>
x^2 + 3y^2 = 48
2xy        = 24    ====>  xy = {{{24/2}}} = 12.


x = {{{12/y}}}   x^2 = {{{144/y^2}}}

{{{144/y^2}}} + + {{{3y^2}}} = 48

144 + 3y^4 = 48y^2

3y^4 - 48y^2 + 144 = 0

y^4 - 16y^2 + 48 = 0

{{{y^2}}} = {{{(16 +- sqrt(16^2 - 4*48))/2}}} = {{{(16 +- 8)/2}}}.


1)  y^2 = {{{(16 + 8)/2}}} = 12  ====>  y is irrational  ===>  the solution does not work;


2)  y^2 = {{{(16 - 8)/2}}} = 4  ====>  y = +/-2  ====>  x = {{{12/y}}} = +/- 6.


<U>Answer</U>.  Two solutions are  (x,y) = (6,2);  (x,y) = (-6,-2).
</pre>

Solved.