One number is three more than twice another, the sum of the numbers is 12. find the two numbers

The numbers are:  "3" and "9" .

Explanation:

Let "x" represent one of the two (2) numbers.

Let "y" represent the other one of the two (2) numbers.

x = 2y + 3  ;

x + y = 12 .

Method 1)

x = 12 − y ; 

Plug this into "x" for "2y + 3 = x" ;

→ 2y + 3 = 12 − y ; 

Add "y" to each side of the equation; & subtract "3" from each side of the equation ;

→ 2y + 3 + y − 3 = 12 − y + y − 3 ;

to get:  3y = 9 ;

Divide each side of the equation by "3" ; 
to isolate "y" on one side of the equation; & to solve for "y" ;

             3y / 3 = 9 / 3 ;

               y = 3 .

Now:  x = 12 − y ;  Plug in "3" for "y" ; to solve for "x" ;

    →   x = 12 − 3 = 9

So;  x = 9,  y = 3 .

Method 2) 

When we have:

x = 2y + 3  ;

x + y = 12 .

 →  y = 12 − x ;

Substitute "(12−x)" for "y" in the equation:

" x = 2y + 3 " ;

→ x = 2(12 − x) + 3 ; 

Note the "distributive property of multiplication" :

  a(b + c) =  ab + ac ;

  a(b − c) = ab − ac ;

As such:

     →  2(12 − x) = 2(12) − 2(x) = 24 − 2x ; 

So; rewrite:

        x = 2(12 − x) + 3 ;
as:

  →  x = 24 − 2x + 3 ;

  →  x = 27 − 2x ;

Add "2x" to each side of the equation:

 →  x + 2x = 27 − 2x + 2x ;

 →  3x = 27 ;

Divide each side of the equation by "3" ;
    to isolate "x" on one side of the equation; & to solve for "x" ;

      3x / 3 = 27 / 3 ;

        x = 9 .

Note:  "y = 12 − x" ;  Substitute "9" for "x" ; to solve for "y" ;

→  y = 12 − 9 = 3 ;

→  y = 3 .

So,  x = 9 ; and y = 3.

The numbers are:  "3" and "9" .

To check our answers:
Let us plug these numbers into the original equations;
   to see if the equations hold true ; (i.e. when, "x = 9" ; and "y = 3"

  → x + y = 12 ; 

   → 9 + 3 =? 12 ?  Yes!

 
   → x = 2y + 3 ;

   → 9 =? 2(3) + 3 ?? ; 

   → 9 =?  6 + 3 ?  Yes!!