The correct answer is: [B]: " (2, 5) ".
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Given:
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-5x + y = -5 ;
  -4x + 2y = 2 .
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Consider the first equation:
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-5x + y = -5 ; ↔ y + (-5x) = -5 ;

↔ y - 5x = -5 ; Add "5x" to each side of the equation; to isolate "y" on one side of the equation; and to solve in terms of "y".
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   y - 5x + 5x = -5 + 5x

   y = -5 + 5x ; ↔ y = 5x - 5 ;
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Now, take our second equation:
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     -4x + 2y = 2 ; and plug in "(5x - 5)" for "y" ; and solve for "x" :
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         -4x + 2(5x - 5) = 2 ;
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Note, 2(5x - 5) = 2(5x) - 2(5) = 10x - 10 ;
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So:   -4x + 10x - 10 = 2 ;

On the left-hand side of the equation, combine the "like terms" ;

-4x +10x = 6x ; and rewrite:

6x - 10 = 2 ;

Now, add "10" to each side of the equation:

6x - 10 + 10 = 2 + 10 ;

to get:

           6x = 12 ; Now, divide EACH side of the equation by "6" ; to isolate "x" on one side of the equation; and to solve for "x" ;

           6x/6 = 12 / 6 ;

                 x = 2 ;
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Now, take our first given equation; and plug our solved value for "x" ; which is "2" ; and solve for "y" ;
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-5x + y = -5 ;

-5(2) + y = -5 ;

-10 + y = -5 ; ↔
                              y - 10 = -5 ;

Add "10" to each side of the equation; to isolate "y" on one side of the equation; and to solve for "y" ;

         y - 10 + 10 = -5 + 10 ;

                   y = 5 .
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So, we have, x = 2 ; and y = 5 .
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Now, let us check our work by plugging in "2" for "x" and "5" for "y" in BOTH the original first and second equations:
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first equation:

-5x + y = -5 ;

-5(2) + 5 =? -5?

-10 + 5 =? -5 ? YES!
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second equation:

-4x + 2y = 2 ;

-4(2) + 2(5) =? 2 ?

-8 + 10 =? 2 ? Yes!
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So, the answer is:
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          x = 2 , y = 5 ; or, "(2, 5)" ; which is: "Answer choice: [B] " .
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