\({6 \over 5 p} - {8 \over 5 p} = -{2 \over 5 p}\)

\({9 \over x} + {4 \over 6 x} = {54 \over 6 x} + {4 \over 6 x} = {58 \over 6 x} = {29 \over 3 x}\)

\({9 \over 5 a} - {7 \over 4 b} = {36 b \over 20 a b} - {35 a \over 20 a b} = {36 b - 35 a \over 20 a b}\)

\(4 + {9 \over 5 x} = {4 \over 1} + {9 \over 5 x} = {20 x \over 5 x} + {9 \over 5 x} = {20 x + 9 \over 5 x}\)

\({8 a \over b} + {4 \over 5 b} = {40 a \over 5 b} + {4 \over 5 b} = {40 a + 4 \over 5 b}\)

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

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

\({-16 a \over 18 a} = -\frac{8}{9}\)

\({12 a \over -2 a} = -6\)

\({15 x y \over 21 x z} = {5 y \over 7 z}\)

\({21 y \over 27 x y} = {7 \over 9 x}\)

\({24 a b c \over 3 b c} = 8 a\)

\({6 x y \over y} - {3 x z \over z} = 6 x - 3 x = 3 x\)

\(6 a - {4 \over 5 a} = {6 a \over 1} ⋅ {5 a \over 5 a} - {4 \over 5 a} = {30 a^{2} \over 5 a} - {4 \over 5 a} = {30 a^{2} - 4 \over 5 a}\)

\({2 q \over 6 p} - {9 p \over 8 q} = {8 q^{2} \over 24 p q} - {27 p^{2} \over 24 p q} = {-27 p^{2} + 8 q^{2} \over 24 p q}\)

\({5 \over x} ⋅ {8 \over y} = {40 \over x y}\)

\({a \over 5} ⋅ -{6 \over b} = -{6 a \over 5 b}\)

\(-{6 \over 7} ⋅ x = -{6 x \over 7}\)

\({2 q \over p} ⋅ {p - 6 \over 7} = {2 q (p - 6) \over 7 p} = {2 p q - 12 q \over 7 p}\)

\({3 \over a} : {9 \over b} = {3 \over a} ⋅ {b \over 9} = {3 b \over 9 a} = {b \over 3 a}\)

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

\({8 \over 3} : {x - 4 y \over y} = {8 \over 3} ⋅ {y \over x - 4 y} = {8 y \over 3 (x - 4 y)} = {8 y \over 3 x - 12 y}\)

\({x \over 9} + {x - 6 \over 8} = {8 x \over 72} + {9 (x - 6) \over 72} = {8 x + 9 (x - 6) \over 72} = {17 x - 54 \over 72}\)

\({-8 a + 7 \over 9 a + 4} - 1 = {-8 a + 7 \over 9 a + 4} - {1 (9 a + 4) \over 9 a + 4} = {-8 a + 7 - 1 (9 a + 4) \over 9 a + 4} = {-8 a + 7 - 9 a - 4 \over 9 a + 4} = {-17 a + 3 \over 9 a + 4}\)

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

\({6 a^{2} - 7 a - 3 \over 8 a^{2}} = {6 a^{2} \over 8 a^{2}} - {7 a \over 8 a^{2}} - {3 \over 8 a^{2}} = \frac{3}{4} - {7 \over 8 a} - {3 \over 8 a^{2}}\)