In order to solve the system of equations (x-3y=2, 2x+y=11 using the elimination method, which of the following steps could be used? A. Multiply the first equation by 2 and then add that result to the second equationB. Multiply the first equation by 3 and then add that result to the second equationC. Multiply the second equation by 2 and then add that result to the first equationD. Multiply the second equation by 3 and then add that result to the first equation

The correct answers are:
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  [C]:  "Multiply the second equation by 2 and then add that result to the first equation" ;  AND:
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  [D]:  "Multiply the second equation by 3 and then add that result to the first equation" .
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Explanation:
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Given:
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The first equation:  " x − y = 2 "  ;

The second equation:  " 2x + y = 11 " ;
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Consider choice [A]:  "Multiply the first equation by 2 and then add that result to the second equation" .

→ Multiply the first equation by "2" :

   2 * {x − 3y = 2} ; 

    →  2x − 6y = 4 ; 

Then, add this to the second equation:
 
      2x − 6y = 4 
 +   2x +  y = 11
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     4x  − 5y = 15 ;    → RULE OUT "Choice [A]" .
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Consider choice [B]:  "Multiply the first equation by 3 and then add that result to the second equation" ;

→ Multiply the first equation by "3" :

  3 * {x − 3y = 2} ; 

    →  3x − 9y = 6 ; 

Then, add this to the second equation:
 
      3x − 9y =   6 
 +   2x +  y   =  11
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     5x − 8y = 17  ;    → RULE OUT "Choice [B]" .
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Consider choice [C]:  "Multiply the second equation by 2 and then add that result to the first equation" ;

→ Multiply the SECOND equation by "2" :

  2 * (2x + y = 11} ; 

    →  4x + 2y = 22 ; 

Then, add this to the first equation:
 
     4x + 2y  =  22
 +   x  − 3y  =  11
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     5x − y = 33  ;    → This is a correct answer choice—since we can easily isolate "y" on one side of the equation:

   →  5x − y = 33 ; 

   ↔   -y + 5x = 33 ;

Subtract "5x" from each side of the equation;

   →  -y + 5x − 5x = 33 − 5x ; 

   →  -y  = 33 − 5x ; 

Multiply each side of the equation by "-1" ; 
   → (-1) * (-y) = (-1) * (33 − 5x) ; 

   →  y = 5x − 33 ;
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Consider choice [D]:  "Multiply the second equation by 3 and then add that result to the first equation" ;

→ Multiply the SECOND equation by "3" :
 
       3 (2x + y  = 11)  

    →  6x + 3y = 33 ; 

Then, add this to the first equation:
 
      6x + 3y  =  33 
 +    x  − 3y  =   2
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   7x            =  35 ; 

Now, we can solve for "x" ; 

  7x = 35 ;

Divide EACH SIDE of the equation by "7" ; 
  to isolate "x" on one side of the equation; & to solve for "x" ; 

7x / 7  = 35/7 ; 

x = 5 ;   →  Yes; this answer choice, [D]; is a correct step.

Furthermore, we can take:  "y = 5x − 33 " ;  from "choice [D]: 
 and plug in "5" into "x" ; to solve for "y" ;

→ y = 5(5) − 33 = 25 − 33  =  -8.
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So, the answer to this system of equations is:
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  "x = 5, y = 8 " ;  or, write as:  " [5, 8] " .
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