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kodGreya [7K]
3 years ago
15

Please help! Thank you!

Mathematics
1 answer:
KIM [24]3 years ago
3 0
The correct answer is:  [B]:  " (2, 5) ".
__________________________________________
Given:
__________________________________________
  -5x + y = -5 ;
  -4x + 2y = 2 .
___________________________________________
  Consider the first equation:
___________________________
-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".
_____________________________________________
   y - 5x + 5x = -5 + 5x  

   y = -5 + 5x ;  ↔ y = 5x - 5 ;
____________________________________________
 Now, take our second equation:
______________________________
     -4x + 2y = 2 ; and plug in "(5x - 5)" for "y" ;  and solve for "x" :
_____________________________________________________
         -4x + 2(5x - 5) = 2 ; 
______________________________________________________
Note, 2(5x - 5) = 2(5x) - 2(5)  = 10x - 10 ;
__________________________________________
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 ;
_________________________________
Now, take our first given equation; and plug our solved value for "x" ; which is "2" ; and solve for "y" ;
_____________________________________
-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 .
_____________________________
So, we have, x = 2 ; and y = 5 .
____________________________
Now, let us check our work by plugging in "2" for "x" and "5" for "y" in BOTH the original first and second equations:
______________________________
first equation: 

-5x + y = -5 ;

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

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

-4x + 2y = 2 ;

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

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



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\large \bigstar \frak{ } \large\underline{\sf{Solution-}}

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\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered}  \\  \\  \text{So, using this identity, we get} \\  \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

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<h2>Hence,</h2>

\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}

\rule{190pt}{2pt}

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