What is the solution of this system of linear equations? 3y = 3/2x + 6 1/2y –1/4 x = 3 a; (3, 6) b; (2, 1) c; no solution d; infinite number of solutions hurry up plz i'm on a time and need a good grade

The answer is:  [C]:  "no solution" .


Given:  3y = 3/2 x + 6 ; Multiply each side by "2" ; to get rid of the fraction

              2 * { 3y = 3/2 x + 6 } ;  to get:  6y = 3x + 12 

           
Given:  1/2 y –1/4 x = 3 ;  Multiply EACH SIDE by "4" ; to get rid of the fraction ;

            4 * { 1/2 y – 1/4 x = 3 } ;  to get:  2y – x = 12 ;

So we have:

6y = 3x + 12 ;

2y – x = 12 ; 

Multiply BOTH side of the "second equation" by "-1" ;

-1 * { 2y – x = 12 } ; 

to get :   x – 2y = -12  ;  

Now, add "2y" to each side of the equation;  
 to isolate "x" on one side of the equation; & to solve for "x" ; 

  x – 2y + 2y = -12 + 2y ; 

to get:  x = 2y – 12 ; 

Now, consider the "first equation" ; 

6y = 3x + 12 ; 
 
Divide EACH SIDE of this equation by "3" ; 

6y / 3 = (3x + 12) / 3 ; 

to get:  2y = x + 4 ; 

Divide EACH SIDE of the equation by "2" ; 

2y/ 2  = ( x + 4) /2 ; 

to get:  y = (x + 4) / 2 ;
 
Now plug in our calculated value for "x" ; which is:   

        " x = 2y – 12 " ; for the "x" values

 y = [ (2y – 12)] + 4)] / 2 ; 

y = (2y – 8 )/ 2 

2y  = 2y  – 8 ??? ;; 

The answer is:  [C]:  "no solution" .

Just look outside and then you’ll see where the awnser is