Which of the following statements is always false when x=3 y=2 and z=4

The correct answer is:  [A]:  " 1(x) = 1 " . 

                   →  { that is:  " 1 *(x)  =  1 " . } .

Explanation:

Note:

The question asks:

Which of the following statements is always false when:

                   →    " x = 3 " ,  " y = 2 " ,  and:  " z = 4 " ? 

with the following answer choices:


  [A]:  " 1* (x) = 1 " ;

  [B]:  " 0 * (x) = 0 " .

  [C]:  " x + y = y + x " ;

  [D]:  " x(y + z) = xy + xz " .

  →  Let us consider each answer choice; and plug in the known values:

 
Choice:  [A]:  " 1* (x) = 1 " ;  
Note:  We are given: " x = 3 " ;  so; plug in "3" and solve:

                    →  1(3) = " 3 ";  and " 3 ≠ 1 " .

Furthermore:  Given:   " 1(x) = 1 " ; in the answer choice ;

Note that, on the "left-hand side" of the equation:
       
                   →  1(x) = 1 * x  = x * 1 = "x" ;  
                                     
                   →  since "any value" ; multiplied by "1" ; equals that same value.

→ As such, substitute: "x" (in lieu of "1x") ; on the "left-hand side" of the equation ;  and:
 
→  REWRITE the equation:

               →  " x = 1 " ; 

               →  Since we are given:  "x = 3"  ; substitute "3" for "x" in the equation, as follows:

               →  " 3 = 1 " ;   which is INCORRECT; 

               →  since:  " 3 ≠ 1 " .

→ This answer is ALWAYS false;  so:

Answer choice:  [A]:  " 1(x) = 1 " ; should be CORRECT.


Choice:  [B]:  " 0 * (x) = 0 " ;

Note:  We are given:  " x = 3 " ;  so; plug in "3" and solve:

                    →  0(3) = 0 ;   which is: "true" in this case.

→ As such, this answer is NOT "always false" ;

→ and:  Answer choice:  [B]:  is INCORRECT.


Choice:  [C]:  " x + y = y + x " ; 

Note:  We are given:  " x = 3 " ;  AND:  " y = 2 " ;

So plug in these given values into the equation:

                    →  3 + 2 = 2 + 3 ; 

                    →  5 = 5.  Yes! 
                   
→ As such, this answer is NOT "always false" ; 

→ and:  Answer choice:  [C]:  is INCORRECT.


Choice:  [D]:  " x(y + z) = xy + xz " ; 

Note:  We are given:  " x = 3 " ;  " y = 2 " ; AND:  " z = 4 " ; 

So plug in these given values into the equation:

                    →  3(2 + 4) = (3*2) + (3*4)  ; 

                    →. 3(6) = 6 + 12 ;
                  
                    →  18 = 18.  Yes!  
                   
→ As such, this answer is NOT "always false" ; 

→ and:  Answer choice:  [D]:  is INCORRECT.

So; the correct answer is: 


Answer choice:  [A]:  " 1(x) = 1 " . 

                   →  { that is:  " 1 *(x)  =  1 " . } .

 Hope this answer—and lengthy explanation—helps! 
 Best wishes in the community—and in your academic pursuits!

It is an outlier because idk sorry