Q1) \(x + 3\over 2\) - \(x + 9\over 5\) = [ \(3 x -3\over 10\) ]

Q1) \(6\over x+ 2\) + \(5\over x +3\) = [ \(11x + 28\over x^{2}+ 5 x +6 \)]

Q1) \(3\over x+ 2\) + \(8\over x +3\) = [ \(11x + 25\over x^{2}+5x +6 \)]

Q2) \(x + 8\over 4\) - \(x + 8\over 7\) = [ \(3 x + 24\over 28\) ]

Q2) \(10\over x+ 4\) - \(8\over x +3\) = [ \(2 x -2\over x^{2}+ 7 x +12 \)]

Q2) \(5\over x+ 4\) + \(10\over x +8\) = [ \(15 x + 80\over x^{2}+12x +32 \)]

Q3) \(x + 10\over 3\) - \(x + 10\over 9\) = [ \(2 x + 20\over 9\) ]

Q3) \(10\over x+ 8\) - \(7\over x +5\) = [ \(3 x -6\over x^{2}+ 13 x +40 \)]

Q3) \(7\over x+ 5\) + \(5\over x +4\) = [ \(12 x + 53\over x^{2}+9x +20 \)]

Q4) \(x + 9\over 3\) + \(x + 10\over 8\) = [ \(11x + 102\over 24\) ]

Q4) \(9\over x+ 4\) - \(7\over x +6\) = [ \(2 x + 26\over x^{2}+ 10 x +24 \)]

Q4) \(10\over x+ 7\) - \(5\over x +2\) = [ \(5 x -15\over x^{2}+9x +14 \)]

Q5) \(x + 5\over 3\) - \(x + 10\over 9\) = [ \(2 x + 5\over 9\) ]

Q5) \(7\over x+ 5\) + \(8\over x +7\) = [ \(15 x + 89\over x^{2}+ 12 x +35 \)]

Q5) \(8\over x+ 5\) + \(6\over x -5\) = [ \(14 x -10\over x^{2} -25 \)]

Q6) \(x + 10\over 5\) + \(x + 10\over 8\) = [ \(13 x + 130\over 40\) ]

Q6) \(7\over x+ 2\) - \(4\over x +2\) = [ \(3 x + 6\over x^{2}+ 4 x +4 \)]

Q6) \(10\over x+ 4\) + \(6\over x +5\) = [ \(16 x + 74\over x^{2}+9x +20 \)]

Q7) \(x + 4\over 2\) - \(x + 7\over 5\) = [ \(3 x + 6\over 10\) ]

Q7) \(10\over x+ 6\) + \(10\over x +3\) = [ \(20 x + 90\over x^{2}+ 9 x +18 \)]

Q7) \(7\over x+ 2\) - \(3\over x -10\) = [ \(4 x -76\over x^{2}-8x -20 \)]

Q8) \(x + 9\over 8\) + \(x + 6\over 2\) = [ \(5 x + 33\over 8\) ]

Q8) \(7\over x+ 5\) + \(7\over x +2\) = [ \(14 x + 49\over x^{2}+ 7 x +10 \)]

Q8) \(9\over x+ 2\) + \(3\over x -4\) = [ \(12 x -30\over x^{2}-2x -8 \)]

Q9) \(x + 9\over 6\) + \(x + 9\over 8\) = [ \(7 x + 63\over 24\) ]

Q9) \(10\over x+ 2\) + \(5\over x +2\) = [ \(15 x + 30\over x^{2}+ 4 x +4 \)]

Q9) \(9\over x+ 7\) - \(5\over x -3\) = [ \(4 x -62\over x^{2}+4x -21 \)]

Q10) \(x + 8\over 4\) - \(x + 10\over 7\) = [ \(3 x + 16\over 28\) ]

Q10) \(8\over x+ 3\) - \(3\over x +2\) = [ \(5 x + 7\over x^{2}+ 5 x +6 \)]

Q10) \(5\over x+ 2\) - \(3\over x -4\) = [ \(2 x -26\over x^{2}-2x -8 \)]