In order to solve the system of equations x-3y=2 and 2x+y=11 using the elimination method, which of the following steps could be used

The correct answers are:

 [C]:  "Multiply the second equation by 2 and then add that result to the first equation" ;  AND:

 [D]:  "Multiply the second equation by 3 and then add that result to the first equation" .

Explanation:

Given:

The first equation:  " x − y = 2 "  ;

The second equation:  " 2x + y = 11 " ;

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

    4x  − 5y = 15 ;    → RULE OUT "Choice [A]" .

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

    5x − 8y = 17  ;    → RULE OUT "Choice [B]" .

.

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

    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 ;

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

  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.

So, the answer to this system of equations is:

 "x = 5, y = 8 " ;  or, write as:  " [5, 8] " .

lol.

it's ok we all do this sometimes.

-aparri