Replace symbols with values without evaluation

Using OwnValues and HoldForm:

{a = 7, b = 5, c = 6};

HoldForm[(a + b)/c] /. OwnValues@a /. OwnValues@b /. OwnValues@c

(7 + 5)/6

With[{a = a, b = b, c = c}, HoldForm[(a + b)/c]]

(7 + 5)/6

Assuming that variables are already defined and you don't want to bother with listing the symbols, and you want to have a function that does it in one go:

Attributes[hold] = {HoldAll};
hold[x_] := HoldForm@x /. Cases[Hold@x, s_Symbol :> (HoldPattern@s -> s), Infinity];

hold[(a + b)/c]

(7 + 5)/6

Hold[(a + b)/c] /. {a -> 1, b -> 3, c -> 4} /. Hold -> Defer

(*   (1 + 3)/4   *)

Hold[(a + b)/c] /. {a -> 1, b -> -3, c -> 4} /. Hold -> Defer

(*   (1 - 3)/4   *)

    Hold[(a + b)/c] /. {a -> 1, b -> -3, c -> -4} /. Hold -> Defer

(*   -(1/4) (1 - 3)   *)

Edit: answer rewrite

This question in its base form is a duplicate, but since I cannot find the original, and since you extended the question to something more unique than what I recall, I shall provide a short answer.

You asked:

I know this a probably even harder, but would it be possible to output (7+5)/(-6) instead of -(1/6)*(7+5) when c=-6?

For that kind of control see for example: Returning an unevaluated expression with values substituted in

Combining that with RuleCondition, described in Replacement inside held expression, we can use:

SetAttributes[{defer, fill}, HoldAll]

MakeBoxes[defer[args__], fmt_] := Block[{Times}, MakeBoxes[Defer[args], fmt]]

fill[expr_] := defer[expr] /. x_Symbol :> RuleCondition[x]

Now:

{a, b, c} = {7, 5, -6};

fill[(a + b)/c]

(7+5)*-(1/6)

Because this uses Defer you can use the output of fill as input and it will fully evaluate.