Given (x/2) = (x/3) + 1
Transposing (x/3) to LHS we get
(x/2) – (x/3) = 1
(3x – 2x)/6 = 1 [LCM of 3 and 2 is 6]
x/6 = 1
Multiplying 6 to both sides we get,
x = 6
Verification:
Substituting x = 6 in given equation we get
(6/2) = (6/3) + 1
3 = 2 + 1
3 = 3
Thus LHS = RHS
Hence, verified.