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Suppose x & y are positive integers,x>y, and 3x+4y & 2x+3y when divided by 5,leave remainders 2&3 respectively. It follows that when (x-y) is divided by 5,the remainder necessarily equals
(a) 2 (b)1 (c) 4 (d) none of these

According to the problem,

$(3x+2y)\equiv 2 \bmod{5}$

$(2x+3y)\equiv 3 \bmod{5}$

subtracting the above 2 relations we get,

$(x-y)\equiv (-1) \bmod{5}$

i.e. $(x-y)\equiv 4 \bmod{5}$

Hence the remainder is 4