\({5 \over 8 x} + {7 \over 8 x} = {12 \over 8 x} = {3 \over 2 x}\)

\({8 \over a} + {7 \over 5 a} = {40 \over 5 a} + {7 \over 5 a} = {47 \over 5 a}\)

\({8 \over 4 p} + {2 \over 7 q} = {56 q \over 28 p q} + {8 p \over 28 p q} = {56 q + 8 p \over 28 p q} = {14 q + 2 p \over 7 p q}\)

\(8 - {5 \over 2 a} = {8 \over 1} - {5 \over 2 a} = {16 a \over 2 a} - {5 \over 2 a} = {16 a - 5 \over 2 a}\)

\({6 x \over y} + {5 \over 3 y} = {18 x \over 3 y} + {5 \over 3 y} = {18 x + 5 \over 3 y}\)

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

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

\({-15 a \over -18 a} = \frac{5}{6}\)

\({-20 p \over -4 p} = 5\)

\({-6 a b \over 9 a c} = -{2 b \over 3 c}\)

\({-10 y \over -15 x y} = {2 \over 3 x}\)

\({-20 x y z \over 5 y z} = -4 x\)

\({6 a b \over b} + {3 a c \over c} = 6 a + 3 a = 9 a\)

\(8 x - {7 \over 6 x} = {8 x \over 1} ⋅ {6 x \over 6 x} - {7 \over 6 x} = {48 x^{2} \over 6 x} - {7 \over 6 x} = {48 x^{2} - 7 \over 6 x}\)

\({5 q \over 7 p} - {6 p \over 4 q} = {20 q^{2} \over 28 p q} - {42 p^{2} \over 28 p q} = {-42 p^{2} + 20 q^{2} \over 28 p q} = {-21 p^{2} + 10 q^{2} \over 14 p q}\)

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

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

\(-{1 \over 8} ⋅ a = -{a \over 8}\)

\({7 b \over a} ⋅ {a + 2 \over 9} = {7 b (a + 2) \over 9 a} = {7 a b + 14 b \over 9 a}\)

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

\({9 \over 7} : a = {9 \over 7} : {a \over 1} = {9 \over 7} ⋅ {1 \over a} = {9 \over 7 a}\)

\(-{4 \over 3} : {x + 7 y \over y} = -{4 \over 3} ⋅ {y \over x + 7 y} = -{4 y \over 3 (x + 7 y)} = -{4 y \over 3 x + 21 y}\)

\({2 x \over 7} + {x + 5 \over 6} = {12 x \over 42} + {7 (x + 5) \over 42} = {12 x + 7 (x + 5) \over 42} = {19 x + 35 \over 42}\)

\({3 x - 8 \over 2 x - 4} - 9 = {3 x - 8 \over 2 x - 4} - {9 (2 x - 4) \over 2 x - 4} = {3 x - 8 - 9 (2 x - 4) \over 2 x - 4} = {3 x - 8 - 18 x + 36 \over 2 x - 4} = {-15 x + 28 \over 2 x - 4}\)

\({x^{2} + 3 x - 20 \over x} = {x^{2} \over x} + {3 x \over x} - {20 \over x} = x + 3 - {20 \over x}\)

\({p^{2} + 2 p + 4 \over 5 p^{2}} = {p^{2} \over 5 p^{2}} + {2 p \over 5 p^{2}} + {4 \over 5 p^{2}} = \frac{1}{5} + {2 \over 5 p} + {4 \over 5 p^{2}}\)