\({5 \over 8 p} + {9 \over 8 p} = {14 \over 8 p} = {7 \over 4 p}\)

\({9 \over a} + {6 \over 2 a} = {18 \over 2 a} + {6 \over 2 a} = {24 \over 2 a} = {12 \over a}\)

\({5 \over 8 x} + {9 \over 4 y} = {5 y \over 8 x y} + {18 x \over 8 x y} = {5 y + 18 x \over 8 x y}\)

\(9 + {8 \over 5 x} = {9 \over 1} + {8 \over 5 x} = {45 x \over 5 x} + {8 \over 5 x} = {45 x + 8 \over 5 x}\)

\({6 a \over b} + {2 \over 8 b} = {48 a \over 8 b} + {2 \over 8 b} = {48 a + 2 \over 8 b} = {24 a + 1 \over 4 b}\)

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

\({a \over 3 a} = {1 \over 3}\)

\({-15 p \over 25 p} = -\frac{3}{5}\)

\({-6 a \over 2 a} = -3\)

\({8 x y \over 36 x z} = {2 y \over 9 z}\)

\({4 b \over -6 a b} = -{2 \over 3 a}\)

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

\({7 x y \over y} - {6 x z \over z} = 7 x - 6 x = x\)

\(4 x - {5 \over 2 x} = {4 x \over 1} ⋅ {2 x \over 2 x} - {5 \over 2 x} = {8 x^{2} \over 2 x} - {5 \over 2 x} = {8 x^{2} - 5 \over 2 x}\)

\({7 b \over 8 a} - {3 a \over 9 b} = {63 b^{2} \over 72 a b} - {24 a^{2} \over 72 a b} = {-24 a^{2} + 63 b^{2} \over 72 a b} = {-8 a^{2} + 21 b^{2} \over 24 a b}\)

\({8 \over p} ⋅ {4 \over q} = {32 \over p q}\)

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

\(-{7 \over 6} ⋅ a = -{7 a \over 6}\)

\({8 b \over a} ⋅ {a + 3 \over 6} = {8 b (a + 3) \over 6 a} = {4 b (a + 3) \over 3 a} = {4 a b + 12 b \over 3 a}\)

\({3 \over x} : {7 \over y} = {3 \over x} ⋅ {y \over 7} = {3 y \over 7 x}\)

\({4 \over 7} : p = {4 \over 7} : {p \over 1} = {4 \over 7} ⋅ {1 \over p} = {4 \over 7 p}\)

\(-{1 \over 3} : {a - 5 b \over b} = -{1 \over 3} ⋅ {b \over a - 5 b} = -{b \over 3 (a - 5 b)} = -{b \over 3 a - 15 b}\)

\({5 x \over 9} + {x + 7 \over 4} = {20 x \over 36} + {9 (x + 7) \over 36} = {20 x + 9 (x + 7) \over 36} = {29 x + 63 \over 36}\)

\({6 a - 2 \over 3 a + 1} + 9 = {6 a - 2 \over 3 a + 1} + {9 (3 a + 1) \over 3 a + 1} = {6 a - 2 + 9 (3 a + 1) \over 3 a + 1} = {6 a - 2 + 27 a + 9 \over 3 a + 1} = {33 a + 7 \over 3 a + 1}\)

\({7 p^{2} - 2 p - 1 \over 4 p^{2}} = {7 p^{2} \over 4 p^{2}} - {2 p \over 4 p^{2}} - {1 \over 4 p^{2}} = 1\frac{3}{4} - {1 \over 2 p} - {1 \over 4 p^{2}}\)