In the following equations, verify whether the given value of the variable is a solution of  the equation:  
(i) x + 4 = 2x ; x = 4                       
(ii) y – 7 = 3y + 8 ; y = 3       
(iii) 3u + 2 = 2u + 7 ; u = 5
(iv) 2x – 3 = x /2 – 2 ; x = √2        
(v) (5/2)x + 3 = 21/2 ; x = 3
(vi) 24 – 3(u – 2) = u + 8 ; u = -1  
(vii) (x – 2) + (x + 3 ) = x + 8 ; x = 0.

(i) If we substitute x = 4, we get
           L.H.S = x + 4 = 4 + 4 = 8    
    and R.H.S = 2x = 2 x 4 = 8
       .^.. L.H.S = R.H.S.   
      Hence, 4 is a solution of x + 4 = 2x.

(ii) If we substitute y = 3, we get  
             L.H.S = y – 7 = 3 – 7 = -4  
      and R.H.S = 3(3) + 8 = 9 + 8 = 17
         .^.. L.H.S ≠ R.H.S.   
     Hence, 3 is not a solution of y – 7 = 3y + 8

(iii) If we substitute u = 5, we get
            L.H.S = 3u + 2 = 3(5) + 2 = 15 + 2 = 17
      and R.H.S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17      
         .^.. L.H.S = R.H.S.
      Hence, 5 is a solution of 3u + 2 = 2u + 7

(iv) If we substitute x = √2, we get
            L.H.S = 2 x – 3 = 2(√2) – 3 = 2√2 – 3
      and R.H.S = x/2 – 2 = √2/2 - 2
         .^.. L.H.S ≠ R.H.S.
      Hence, √2 is not a solution of 2x – 3 = x/2 – 2
 
(v) If we substitute x = 3, we get
           L.H.S = (5/2)x + 3 = (5/2)(3) + 3 = 15/2 + 3 = (15 + 6)/2 = 21/2  
     and R.H.S = 21/2
     .^.. L.H.S = R.H.S.
     Hence, 3 is a solution of (5/2)x + 3 = 21/2.

(vi) If we substitute u = -1, we get
            L.H.S = 24 – 3{(-1) – 2}= 24 – 3(-1 –2) = 24 – (-9) = 24 + 9 = 33
      and R.H.S = u + 8 = -1 + 8 = 7
         .^.. L.H.S ≠ R.H.S.
      Hence, -1 is not a solution of 24 – 3(u – 2) = u + 8.

(vii) If we substitute x = 0, we get
              L.H.S = (x – 2) + (x + 3) = (0 – 2) + (0 + 3) = -2 + 3 = 1  
       and R.H.S = x + 8 = 0 + 8 = 8
           .^.. L.H.S ≠ R.H.S.
       Hence, 0 is not a solution of (x – 2) + (x + 3) = x + 8.