\({4 \over 7 p} - {8 \over 7 p} = -{4 \over 7 p}\)

\({8 \over a} - {4 \over 5 a} = {40 \over 5 a} - {4 \over 5 a} = {36 \over 5 a}\)

\({5 \over 3 x} - {7 \over 2 y} = {10 y \over 6 x y} - {21 x \over 6 x y} = {10 y - 21 x \over 6 x y}\)

\(9 + {5 \over 3 a} = {9 \over 1} + {5 \over 3 a} = {27 a \over 3 a} + {5 \over 3 a} = {27 a + 5 \over 3 a}\)

\({8 x \over y} + {5 \over 2 y} = {16 x \over 2 y} + {5 \over 2 y} = {16 x + 5 \over 2 y}\)

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

\({a \over 2 a} = {1 \over 2}\)

\({20 a \over 36 a} = \frac{5}{9}\)

\({30 p \over 5 p} = 6\)

\({-16 x y \over -36 x z} = {4 y \over 9 z}\)

\({4 q \over 14 p q} = {2 \over 7 p}\)

\({-24 x y z \over -3 y z} = 8 x\)

\({7 a b \over b} - {3 a c \over c} = 7 a - 3 a = 4 a\)

\(7 x + {4 \over 9 x} = {7 x \over 1} ⋅ {9 x \over 9 x} + {4 \over 9 x} = {63 x^{2} \over 9 x} + {4 \over 9 x} = {63 x^{2} + 4 \over 9 x}\)

\({9 b \over 4 a} - {3 a \over 7 b} = {63 b^{2} \over 28 a b} - {12 a^{2} \over 28 a b} = {-12 a^{2} + 63 b^{2} \over 28 a b}\)

\({5 \over p} ⋅ {6 \over q} = {30 \over p q}\)

\({x \over 9} ⋅ {6 \over y} = {6 x \over 9 y} = {2 x \over 3 y}\)

\({1 \over 5} ⋅ a = {a \over 5}\)

\({4 y \over x} ⋅ {x - 9 \over 3} = {4 y (x - 9) \over 3 x} = {4 x y - 36 y \over 3 x}\)

\({7 \over p} : {4 \over q} = {7 \over p} ⋅ {q \over 4} = {7 q \over 4 p}\)

\(-{5 \over 3} : x = -{5 \over 3} : {x \over 1} = -{5 \over 3} ⋅ {1 \over x} = -{5 \over 3 x}\)

\(-{8 \over 5} : {a - 9 b \over b} = -{8 \over 5} ⋅ {b \over a - 9 b} = -{8 b \over 5 (a - 9 b)} = -{8 b \over 5 a - 45 b}\)

\({5 a \over 9} + {a + 8 \over 7} = {35 a \over 63} + {9 (a + 8) \over 63} = {35 a + 9 (a + 8) \over 63} = {44 a + 72 \over 63}\)

\({-7 x + 3 \over 9 x + 4} - 8 = {-7 x + 3 \over 9 x + 4} - {8 (9 x + 4) \over 9 x + 4} = {-7 x + 3 - 8 (9 x + 4) \over 9 x + 4} = {-7 x + 3 - 72 x - 32 \over 9 x + 4} = {-79 x - 29 \over 9 x + 4}\)

\({6 a^{2} + 2 a - 40 \over 2 a} = {6 a^{2} \over 2 a} + {2 a \over 2 a} - {40 \over 2 a} = 3 a + 1 - {20 \over a}\)

\({6 a^{2} - 9 a - 7 \over 5 a^{2}} = {6 a^{2} \over 5 a^{2}} - {9 a \over 5 a^{2}} - {7 \over 5 a^{2}} = 1\frac{1}{5} - {9 \over 5 a} - {7 \over 5 a^{2}}\)