The sum of twice a first number and five times a second number is 78. if the second number is subtracted from five times the first number the result is 33. find the numbers.

The numbers are:  "9" and "12" .

Explanation:

Let:  "x" be the "first number" ; AND:

Let:  "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" ; 

→ Add "y" to EACH side of the equation; 

              5x − y + y = 33 + y ;

to get:  5x = 33 + y ; 

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":
 
      →   " 2x + 5y = 78 " ;  and write as:
 
      →   " 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 ; 

Note:  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 ; 

       →   to get:    x = 9 ; which is "the first number" .

Now;    Let's go back to our "first equation" and "second equation" to solve for "y" (our "second number"):

     2x + 5y = 78 ; (first equation);
     
      5x − y = 33 ; (second equation); 

Start with our "second equation"; to solve for "y"; plug in "9" for "x" ;

→ 5(9) − y = 33 ;  

    45 − y = 33;  
   
Add "y" to each side of the equation:
 
   45 − y + y = 33 + y ;  to get:

   45 = 33 + y ;  

↔ y + 33 = 45 ;  Subtract "33" from each side of the equation; to isolate "y" on one side of the equation ; & to solve for "y" ;  
 
 → y + 33 − 33  = 45 − 33 ;

to get:  y = 12 ;

So;  x = 9 ; and y = 12 .  The numbers are:  "9" and "12" .

 To check our work:

1)  Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ; 

→ 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!

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