The numbers are: Â "9" and "12" .
Explanation:
Let: Â "x" be the "first number" ; AND:
Let: Â "y" be the "second number" .
From the question/problem, we are given:
   2x + 5y = 78 ;  → "the first equation" ; AND:
   5x − y = 33 ;  → "the second equation" .
From "the second equation" ; which is:
  " 5x − y = 33" ;Â
→ Add "y" to EACH side of the equation;Â
        5x − y + y = 33 + y ;
to get: Â 5x = 33 + y ;Â
Now, subtract: "33" from each side of the equation; to isolate "y" on one side of the equation ; and to solve for "y" (in term of "x");
      5x − 33 = 33 + y − 33 ;
to get:  " 5x − 33 = y " ;  ↔  " y = 5x − 33 " .
Note:  We choose "the second equation"; because "the second equation"; that is;  "5x − y = 33" ;  already has a "y" value with no "coefficient" ; & it is easier to solve for one of our numbers (variables); that is, "x" or "y"; in terms of the other one; & then substitute that value into "the first equation".
Now, let us take "the first equation" ; which is:
 "  2x + 5y = 78 " ;
We have our obtained value; " y = 5x − 33 " .
We shall take our obtained value for "y" ; which is: "(5x− 33") ; and plug this value into the "y" value in the "first equation"; and solve for "x" ;
Take the "first equation":
Â
   →  " 2x + 5y = 78 " ;  and write as:
Â
   →  " 2x + 5(5x − 33) = 78 " ;
Note the "distributive property of multiplication" :
   a(b + c) = ab + ac ; AND:
   a(b − c) = ab − ac .
So; using the "distributive property of multiplication:
→  +5(5x − 33)  = (5*5x) − (5*33) =  +25x − 165 .
So we can rewrite our equation:
     →  " 2x + 5(5x − 33) = 78 " ;
by substituting the:  "+ 5(5x − 33) " ;  with:  "+25x − 165" ; as follows:
      →  " 2x + 25x − 165 = 78 " ;
→ Now, combine the "like terms" on the "left-hand side" of the equation:
       +2x + 25x = +27x ;Â
Note: Â There are no "like terms" on the "right-hand side" of the equation.
  →  Rewrite the equation as:
     →  " 27x − 165 = 78 " ;
   Now, add "165" to EACH SIDE of the equation; as follows:
     →   27x − 165 + 165 = 78 + 165 ;
    →  to get:    27x = 243  ;
   Now, divide EACH SIDE of the equation by "27" ; to isolate "x" on one side of the equation ; and to solve for "x" ;
        27x / 27  =  243 / 27 ;Â
    →  to get:   x = 9 ; which is "the first number" .
Now; Â Â Let's go back to our "first equation" and "second equation" to solve for "y" (our "second number"):
   2x + 5y = 78 ; (first equation);
  Â
   5x − y = 33 ; (second equation);Â
Start with our "second equation"; to solve for "y"; plug in "9" for "x" ;
→ 5(9) − y = 33 ; Â
  45 − y = 33; Â
 Â
Add "y" to each side of the equation:
Â
  45 − y + y = 33 + y ;  to get:
  45 = 33 + y ; Â
↔ y + 33 = 45 ;  Subtract "33" from each side of the equation; to isolate "y" on one side of the equation ; & to solve for "y" ; Â
Â
 → y + 33 − 33  = 45 − 33 ;
to get: Â y = 12 ;
So;Â x = 9 ; and y = 12 . Â The numbers are: Â "9" and "12" .
 To check our work:
1) Â Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;Â
→ 5x − y = 33 ;  → 5(9) − 12 =? 33 ?? ;  → 45 − 12 =? 33 ?? ;  Yes!
2)  Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;
→ 2x + 5y = 78 ; → 2(9) + 5(12) =? 78?? ; → 18 + 60 =? 78?? ; Yes!
So, these answers do make sense!