How to solve an equation on a given relation

To solve for x+y you can simply do:

x + y /. Solve[x + y == 3 + 2 x + 2 y + 38/a, {x, y}][[1]]

-((38 + 3 a)/a)

To get an expression of the form x + y == ... you can use Reduce and post-process the output into desired form:

red = Reduce[{z == 3 + 2 x + 2 y + 38/a, z == x + y, a != 0}, z, {x, y}]

a != 0 && z == (-38 - 3 a)/a

red /. { a != 0 -> True, z -> x + y}

x + y == (-38 - 3 a)/a

This works for the example you posted. If you have a different example, where this does not work, will try to fix it. This checks that the combination of a x + b y can factor out to a common x+y term, otherwise it returns no solution.

solve[x_, y_, eq0_] := Module[{lhs, rhs, eq, cx, cy, sol = {}, z},
   rhs = eq0 /. (lhs_ == rhs_) :> rhs;
   lhs = eq0 /. (lhs_ == rhs_) :> lhs;
   eq = rhs - lhs == 0;
   cx = Last@CoefficientList[rhs - lhs, x];
   cy = Last@CoefficientList[rhs - lhs, y];
   If[cx == cy,
    sol = (x + y) /. (First@Solve[(eq /. (cx x + cy y) -> z), z] /. z -> ( x + y));
    sol = sol/cx
    ];
   x + y == sol
   ];

Now call it as

ClearAll[x, y, a];
eq = x + y == 3 + 2 x + 2 y + 38/a;
solve[x, y, eq]

eq = x + y == 3 + 4 x + 4 y + 38/a;
solve[x, y, eq]

eq = 2 x + y == 3 - 2 x - 3 y + 38/a;
solve[x, y, eq]

This returns no solution, since can't collect x+y into one term

eq = 2 x + y == 3 - 2 x - 1 y + 38/a;
solve[x, y, eq]