\({6 \over 9 p} - {3 \over 9 p} = {3 \over 9 p} = {1 \over 3 p}\)

\({7 \over a} - {9 \over 2 a} = {14 \over 2 a} - {9 \over 2 a} = {5 \over 2 a}\)

\({2 \over 8 a} + {6 \over 3 b} = {6 b \over 24 a b} + {48 a \over 24 a b} = {6 b + 48 a \over 24 a b} = {b + 8 a \over 4 a b}\)

\(8 + {7 \over 3 x} = {8 \over 1} + {7 \over 3 x} = {24 x \over 3 x} + {7 \over 3 x} = {24 x + 7 \over 3 x}\)

\(6 x - {7 \over 9 x} = {6 x \over 1} ⋅ {9 x \over 9 x} - {7 \over 9 x} = {54 x^{2} \over 9 x} - {7 \over 9 x} = {54 x^{2} - 7 \over 9 x}\)

\({7 x \over y} + {2 \over 5 y} = {35 x \over 5 y} + {2 \over 5 y} = {35 x + 2 \over 5 y}\)

\({6 b \over 5 a} - {2 a \over 4 b} = {24 b^{2} \over 20 a b} - {10 a^{2} \over 20 a b} = {-10 a^{2} + 24 b^{2} \over 20 a b} = {-5 a^{2} + 12 b^{2} \over 10 a b}\)

\({5 a \over a} = {5 \over 1} = 5\)

\({p \over 5 p} = {1 \over 5}\)

\({-12 x \over 21 x} = -\frac{4}{7}\)

\({40 p \over 5 p} = 8\)

\({-28 a b \over 36 a c} = -{7 b \over 9 c}\)

\({18 b \over -21 a b} = -{6 \over 7 a}\)

\({35 x y z \over -5 y z} = -7 x\)

\({4 x y \over y} - {3 x z \over z} = 4 x - 3 x = x\)

\({2 \over p} ⋅ {6 \over q} = {12 \over p q}\)

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

\(-{6 \over 7} ⋅ a = -{6 a \over 7}\)

\({7 \over a} : {3 \over b} = {7 \over a} ⋅ {b \over 3} = {7 b \over 3 a}\)

\({8 \over 7} : x = {8 \over 7} : {x \over 1} = {8 \over 7} ⋅ {1 \over x} = {8 \over 7 x}\)

\({3 a \over 5} + {a + 1 \over 7} = {21 a \over 35} + {5 (a + 1) \over 35} = {21 a + 5 (a + 1) \over 35} = {26 a + 5 \over 35}\)

\({9 q \over p} ⋅ {p - 8 \over 5} = {9 q (p - 8) \over 5 p} = {9 p q - 72 q \over 5 p}\)

\(-{8 \over 7} : {x - 6 y \over y} = -{8 \over 7} ⋅ {y \over x - 6 y} = -{8 y \over 7 (x - 6 y)} = -{8 y \over 7 x - 42 y}\)

\({6 a^{2} - 4 a - 20 \over 2 a} = {6 a^{2} \over 2 a} - {4 a \over 2 a} - {20 \over 2 a} = 3 a - 2 - {10 \over a}\)

\({7 a^{2} - a + 8 \over 2 a^{2}} = {7 a^{2} \over 2 a^{2}} - {a \over 2 a^{2}} + {8 \over 2 a^{2}} = 3\frac{1}{2} - {1 \over 2 a} + {4 \over a^{2}}\)