Respuesta :
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: Â " 2x + y = 11 " ;
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Consider choice [A]: Â "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} ;Â
  →  2x − 6y = 4 ;Â
Then, add this to the second equation:
Â
   2x − 6y = 4Â
 +  2x +  y = 11
_________________ Â
   4x  − 5y = 15 ;   → RULE OUT "Choice [A]" .
________________________________Â
Consider choice [B]: Â "Multiply the first equation by 3 and then add that result to the second equation"Â ;
→ Multiply the first equation by "3" :
 3 * {x − 3y = 2} ;Â
  →  3x − 9y = 6 ;Â
Then, add this to the second equation:
Â
   3x − 9y =  6Â
 +  2x +  y  =  11
 _________________ Â
   5x − 8y = 17  ;   → RULE OUT "Choice [B]" .
____________________________________________________.
Consider choice [C]: Â "Multiply the second equation by 2 and then add that result to the first equation"Â ;
→ Multiply the SECOND equation by "2" :
 2 * (2x + y = 11} ;Â
  →  4x + 2y = 22 ;Â
Then, add this to the first equation:
Â
   4x + 2y  =  22
 +  x  − 3y  =  11
 _________________ Â
   5x − y = 33  ;   → This is a correct answer choice—since we can easily isolate "y" on one side of the equation:
  →  5x − y = 33 ;Â
  ↔  -y + 5x = 33 ;
Subtract "5x" from each side of the equation;
  →  -y + 5x − 5x = 33 − 5x ;Â
  →  -y  = 33 − 5x ;Â
Multiply each side of the equation by "-1" ;Â
  → (-1) * (-y) = (-1) * (33 − 5x) ;Â
  →  y = 5x − 33 ;
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Consider choice [D]: Â "Multiply the second equation by 3 and then add that result to the first equation"Â ;
→ Multiply the SECOND equation by "3" :
Â
    3 (2x + y  = 11) Â
  →  6x + 3y = 33 ;Â
Then, add this to the first equation:
Â
   6x + 3y  =  33Â
 +   x  − 3y  =  2
___________________
  7x       =  35 ;Â
Now, we can solve for "x" ;Â
 7x = 35 ;
Divide EACH SIDE of the equation by "7" ;Â
 to isolate "x" on one side of the equation; & to solve for "x" ;Â
7x / 7 Â = 35/7 ;Â
x = 5 ;   →  Yes; this answer choice, [D]; is a correct step.
Furthermore, we can take:  "y = 5x − 33 " ;  from "choice [D]:Â
 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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__________________________________________________________
 [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: Â " 2x + y = 11 " ;
_____________________________________________________
Consider choice [A]: Â "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} ;Â
  →  2x − 6y = 4 ;Â
Then, add this to the second equation:
Â
   2x − 6y = 4Â
 +  2x +  y = 11
_________________ Â
   4x  − 5y = 15 ;   → RULE OUT "Choice [A]" .
________________________________Â
Consider choice [B]: Â "Multiply the first equation by 3 and then add that result to the second equation"Â ;
→ Multiply the first equation by "3" :
 3 * {x − 3y = 2} ;Â
  →  3x − 9y = 6 ;Â
Then, add this to the second equation:
Â
   3x − 9y =  6Â
 +  2x +  y  =  11
 _________________ Â
   5x − 8y = 17  ;   → RULE OUT "Choice [B]" .
____________________________________________________.
Consider choice [C]: Â "Multiply the second equation by 2 and then add that result to the first equation"Â ;
→ Multiply the SECOND equation by "2" :
 2 * (2x + y = 11} ;Â
  →  4x + 2y = 22 ;Â
Then, add this to the first equation:
Â
   4x + 2y  =  22
 +  x  − 3y  =  11
 _________________ Â
   5x − y = 33  ;   → This is a correct answer choice—since we can easily isolate "y" on one side of the equation:
  →  5x − y = 33 ;Â
  ↔  -y + 5x = 33 ;
Subtract "5x" from each side of the equation;
  →  -y + 5x − 5x = 33 − 5x ;Â
  →  -y  = 33 − 5x ;Â
Multiply each side of the equation by "-1" ;Â
  → (-1) * (-y) = (-1) * (33 − 5x) ;Â
  →  y = 5x − 33 ;
____________________________________________________
Consider choice [D]: Â "Multiply the second equation by 3 and then add that result to the first equation"Â ;
→ Multiply the SECOND equation by "3" :
Â
    3 (2x + y  = 11) Â
  →  6x + 3y = 33 ;Â
Then, add this to the first equation:
Â
   6x + 3y  =  33Â
 +   x  − 3y  =  2
___________________
  7x       =  35 ;Â
Now, we can solve for "x" ;Â
 7x = 35 ;
Divide EACH SIDE of the equation by "7" ;Â
 to isolate "x" on one side of the equation; & to solve for "x" ;Â
7x / 7 Â = 35/7 ;Â
x = 5 ;   →  Yes; this answer choice, [D]; is a correct step.
Furthermore, we can take:  "y = 5x − 33 " ;  from "choice [D]:Â
 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:
________________________________________________
 "x = 5, y = 8 " ;  or, write as:  " [5, 8] " .
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We want to see which should be the first step in solving the given system of equations by elimination method, we will see that the correct option is D: "Multiply the second equation by 3 and then add that result to the first equation"
Elimination method.
The elimination method is a method for solving systems of equations, it needs to add or subtract the equations in such a way that one of the variables is removed.
In this case the system is:
x - 3y = 2
2x + y = 11
Notice that all the options say "add", so we will only work with additions. Because in both equations we have a positive coefficient multiplying the x variable, we can't eliminate it.
But the signs of the coefficients for the y variable are different. So we can just multiply the second equation by 3 so we get:
3*2*x + 3*y = 3*11
6x + 3y = 33
Now we add the two equations:
(x - 3y) + (6x + 3y) = 2 + 33
7x = 35
Now we removed a variable and we can continue solving the system, so the correct option D: "Multiply the second equation by 3 and then add that result to the first equation"
If you want to learn more about systems of equations, you can read:
https://brainly.com/question/13729904