Question

This 1 seems really complicated

Answer 1

The solution to this system set is:  "x = 4" , "y = 0" ;  or write as:  [4, 0] .
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Given: 
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 y = - 4x + 16 ; 

 4y − x + 4 = 0 ;
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"Solve the system using substitution" .
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First, let us simplify the second equation given, to get rid of the "0" ; 

→  4y − x + 4 = 0 ; 

Subtract "4" from each side of the equation ; 

→  4y − x + 4 − 4 = 0 − 4 ;

→  4y − x = -4 ;
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So, we can now rewrite the two (2) equations in the given system:
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y = - 4x + 16 ;   ===> Refer to this as "Equation 1" ; 

4y − x =  -4 ;     ===> Refer to this as "Equation 2" ; 
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Solve for "x" and "y" ;  using "substitution" :
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We are given, as "Equation 1" ;

→  " y = - 4x + 16 " ;
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→  Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

       to solve for "x" ;   as follows:
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Note:  "Equation 2" :

     →  " 4y − x =  - 4 " ; 
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Substitute the value for "y" {i.e., the value provided for "y";  in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→   as follows:  
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→   " 4 (-4x + 16) − x = -4 " ;
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Note the "distributive property" of multiplication :
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   a(b + c)  = ab + ac ;   AND: 

   a(b − c) = ab − ac .
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As such:

We have:  

→   " 4 (-4x + 16) − x = - 4 " ;
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AND:

→    "4 (-4x + 16) "  =  (4* -4x) + (4 *16)  =  " -16x + 64 " ;
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Now, we can write the entire equation:

→  " -16x + 64 − x = - 4 " ; 

Note:  " - 16x − x =  -16x − 1x = -17x " ; 

→  " -17x + 64 = - 4 " ;   Solve for "x" ; 

Subtract "64" from EACH SIDE of the equation:

→  " -17x + 64 − 64 = - 4 − 64 " ;   

to get:  

→  " -17x = -68 " ;

Divide EACH side of the equation by "-17" ; 
   to isolate "x" on one side of the equation; & to solve for "x" ; 

→  -17x / -17 = -68/ -17 ; 

to get:  

→  x = 4  ;
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Now, Plug this value for "x" ; into "{Equation 1"} ; 

which is:  " y = -4x + 16" ; to solve for "y".
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→  y = -4(4) + 16 ; 

        = -16 + 16 ; 

→ y = 0 .
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The solution to this system set is:  "x = 4" , "y = 0" ;  or write as:  [4, 0] .
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Now, let us check our answers—as directed in this very question itself ; 
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→  Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten; 

→  Let us check;  

→  For EACH of these 2 (TWO) equations;  do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ; 

→ Consider the first equation given in our problem, as originally written in the system of equations:

→  " y = - 4x + 16 " ;    

→ Substitute:  "4" for "x" and "0" for "y" ;  When done, are both sides equal?

→  "0 = ?  -4(4) + 16 " ?? ;   →  "0 = ? -16 + 16 ?? " ;  →  Yes!  ;

 {Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→  " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

 {that is:  "4" for the "x-value" ; & "0" for the "y-value" ;  

→  to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→    " 4(0)  −  4 + 4 = ? 0 ?? " ;

      →  " 0  −  4  + 4 = ? 0 ?? " ;

      →  " - 4  + 4 = ? 0 ?? " ;  Yes!
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→  As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
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→   "x = 4" and "y = 0" ;  or; write as:  [0, 4]  ;  are correct.
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Hope this lenghty explanation is of help!  Best wishes!
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