\({2 \over 5 a} - {7 \over 5 a} = -{5 \over 5 a} = -{1 \over a}\)

\({3 \over p} + {7 \over 5 p} = {15 \over 5 p} + {7 \over 5 p} = {22 \over 5 p}\)

\({5 \over 9 a} - {4 \over 3 b} = {5 b \over 9 a b} - {12 a \over 9 a b} = {5 b - 12 a \over 9 a b}\)

\(3 - {4 \over 9 x} = {3 \over 1} - {4 \over 9 x} = {27 x \over 9 x} - {4 \over 9 x} = {27 x - 4 \over 9 x}\)

\({6 x \over y} - {8 \over 2 y} = {12 x \over 2 y} - {8 \over 2 y} = {12 x - 8 \over 2 y} = {6 x - 4 \over y}\)

\({9 p \over p} = {9 \over 1} = 9\)

\({a \over 9 a} = {1 \over 9}\)

\({10 a \over -15 a} = -\frac{2}{3}\)

\({20 x \over 5 x} = 4\)

\({-9 x y \over -21 x z} = {3 y \over 7 z}\)

\({16 b \over -36 a b} = -{4 \over 9 a}\)

\({-24 p q r \over 3 q r} = -8 p\)

\({5 a b \over b} + {2 a c \over c} = 5 a + 2 a = 7 a\)

\(7 a + {6 \over 5 a} = {7 a \over 1} ⋅ {5 a \over 5 a} + {6 \over 5 a} = {35 a^{2} \over 5 a} + {6 \over 5 a} = {35 a^{2} + 6 \over 5 a}\)

\({4 b \over 2 a} + {8 a \over 5 b} = {20 b^{2} \over 10 a b} + {16 a^{2} \over 10 a b} = {16 a^{2} + 20 b^{2} \over 10 a b} = {8 a^{2} + 10 b^{2} \over 5 a b}\)

\({6 \over x} ⋅ {8 \over y} = {48 \over x y}\)

\({p \over 6} ⋅ {4 \over q} = {4 p \over 6 q} = {2 p \over 3 q}\)

\(-{2 \over 9} ⋅ x = -{2 x \over 9}\)

\({8 b \over a} ⋅ {a - 9 \over 2} = {8 b (a - 9) \over 2 a} = {4 b (a - 9) \over a} = {4 a b - 36 b \over a}\)

\({4 \over x} : {8 \over y} = {4 \over x} ⋅ {y \over 8} = {4 y \over 8 x} = {y \over 2 x}\)

\(-{5 \over 9} : p = -{5 \over 9} : {p \over 1} = -{5 \over 9} ⋅ {1 \over p} = -{5 \over 9 p}\)

\({8 \over 3} : {a + 4 b \over b} = {8 \over 3} ⋅ {b \over a + 4 b} = {8 b \over 3 (a + 4 b)} = {8 b \over 3 a + 12 b}\)

\({x \over 4} + {x - 2 \over 7} = {7 x \over 28} + {4 (x - 2) \over 28} = {7 x + 4 (x - 2) \over 28} = {11 x - 8 \over 28}\)

\({-2 p + 4 \over -3 p + 8} - 7 = {-2 p + 4 \over -3 p + 8} + {-7 (-3 p + 8) \over -3 p + 8} = {-2 p + 4 - 7 (-3 p + 8) \over -3 p + 8} = {-2 p + 4 + 21 p - 56 \over -3 p + 8} = {19 p - 52 \over -3 p + 8}\)

\({6 a^{2} + 9 a + 30 \over 3 a} = {6 a^{2} \over 3 a} + {9 a \over 3 a} + {30 \over 3 a} = 2 a + 3 + {10 \over a}\)

\({5 x^{2} - 9 x - 6 \over 3 x^{2}} = {5 x^{2} \over 3 x^{2}} - {9 x \over 3 x^{2}} - {6 \over 3 x^{2}} = 1\frac{2}{3} - {3 \over x} - {2 \over x^{2}}\)