Q1) \(x + 8\over 2\) - \(x + 7\over 6\) = [ \(2 x + 17\over 6\) ]

Q1) \(5\over x+ 3\) + \(10\over x +2\) = [ \(15 x + 40\over x^{2}+ 5 x +6 \)]

Q1) \(10\over x+ 5\) + \(4\over x +2\) = [ \(14 x + 40\over x^{2}+7x +10 \)]

Q2) \(x + 10\over 6\) + \(x + 6\over 3\) = [ \(3 x + 22\over 6\) ]

Q2) \(10\over x+ 5\) - \(5\over x +4\) = [ \(5 x + 15\over x^{2}+ 9 x +20 \)]

Q2) \(10\over x+ 2\) - \(8\over x -9\) = [ \(2 x -106\over x^{2}-7x -18 \)]

Q3) \(x + 5\over 2\) - \(x + 6\over 5\) = [ \(3 x + 13\over 10\) ]

Q3) \(10\over x+ 7\) + \(10\over x +9\) = [ \(20 x + 160\over x^{2}+ 16 x +63 \)]

Q3) \(9\over x+ 3\) + \(5\over x -3\) = [ \(14 x -12\over x^{2} -9 \)]

Q4) \(x + 7\over 4\) + \(x + 9\over 3\) = [ \(7 x + 57\over 12\) ]

Q4) \(9\over x+ 4\) - \(4\over x +3\) = [ \(5 x + 11\over x^{2}+ 7 x +12 \)]

Q4) \(9\over x+ 2\) + \(7\over x -4\) = [ \(16 x -22\over x^{2}-2x -8 \)]

Q5) \(x + 8\over 6\) + \(x + 7\over 6\) = [ \(2 x + 15\over 6\) ]

Q5) \(10\over x+ 7\) - \(6\over x +5\) = [ \(4 x + 8\over x^{2}+ 12 x +35 \)]

Q5) \(9\over x+ 5\) - \(5\over x -9\) = [ \(4 x -106\over x^{2}-4x -45 \)]

Q6) \(x + 8\over 3\) + \(x + 7\over 3\) = [ \(2 x + 15\over 3\) ]

Q6) \(8\over x+ 2\) - \(6\over x +5\) = [ \(2 x + 28\over x^{2}+ 7 x +10 \)]

Q6) \(9\over x+ 5\) - \(6\over x +5\) = [ \(3 x + 15\over x^{2}+10x +25 \)]

Q7) \(x + 10\over 2\) + \(x + 10\over 6\) = [ \(2 x + 20\over 3\) ]

Q7) \(8\over x+ 4\) + \(10\over x +5\) = [ \(18 x + 80\over x^{2}+ 9 x +20 \)]

Q7) \(9\over x+ 5\) - \(5\over x -2\) = [ \(4 x -43\over x^{2}+3x -10 \)]

Q8) \(x + 5\over 2\) - \(x + 10\over 5\) = [ \(3 x + 5\over 10\) ]

Q8) \(7\over x+ 5\) + \(10\over x +5\) = [ \(17 x + 85\over x^{2}+ 10 x +25 \)]

Q8) \(9\over x+ 4\) - \(4\over x -9\) = [ \(5 x -97\over x^{2}-5x -36 \)]

Q9) \(x + 6\over 5\) - \(x + 9\over 7\) = [ \(2 x -3\over 35\) ]

Q9) \(5\over x+ 3\) + \(5\over x +2\) = [ \(10 x + 25\over x^{2}+ 5 x +6 \)]

Q9) \(4\over x+ 3\) + \(2\over x -9\) = [ \(6 x -30\over x^{2}-6x -27 \)]

Q10) \(x + 9\over 2\) - \(x + 8\over 5\) = [ \(3 x + 29\over 10\) ]

Q10) \(10\over x+ 9\) - \(6\over x +2\) = [ \(4 x -34\over x^{2}+ 11x +18 \)]

Q10) \(9\over x+ 3\) + \(10\over x -4\) = [ \(19 x -6\over x^{2}-x -12 \)]