\({7 \over 9 a} + {4 \over 9 a} = {11 \over 9 a}\)

\({6 \over p} - {4 \over 5 p} = {30 \over 5 p} - {4 \over 5 p} = {26 \over 5 p}\)

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

\(8 - {4 \over 7 x} = {8 \over 1} - {4 \over 7 x} = {56 x \over 7 x} - {4 \over 7 x} = {56 x - 4 \over 7 x}\)

\({4 x \over y} + {3 \over 5 y} = {20 x \over 5 y} + {3 \over 5 y} = {20 x + 3 \over 5 y}\)

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

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

\({-12 a \over 15 a} = -\frac{4}{5}\)

\({25 x \over 5 x} = 5\)

\({-40 p q \over 45 p r} = -{8 q \over 9 r}\)

\({-25 y \over -45 x y} = {5 \over 9 x}\)

\({12 a b c \over 4 b c} = 3 a\)

\({5 p q \over q} - {3 p r \over r} = 5 p - 3 p = 2 p\)

\(4 x - {5 \over 6 x} = {4 x \over 1} ⋅ {6 x \over 6 x} - {5 \over 6 x} = {24 x^{2} \over 6 x} - {5 \over 6 x} = {24 x^{2} - 5 \over 6 x}\)

\({4 b \over 6 a} - {8 a \over 9 b} = {12 b^{2} \over 18 a b} - {16 a^{2} \over 18 a b} = {-16 a^{2} + 12 b^{2} \over 18 a b} = {-8 a^{2} + 6 b^{2} \over 9 a b}\)

\({7 \over a} ⋅ -{5 \over b} = -{35 \over a b}\)

\({x \over 2} ⋅ {5 \over y} = {5 x \over 2 y}\)

\(-{2 \over 3} ⋅ p = -{2 p \over 3}\)

\({3 q \over p} ⋅ {p + 8 \over 5} = {3 q (p + 8) \over 5 p} = {3 p q + 24 q \over 5 p}\)

\({9 \over a} : {2 \over b} = {9 \over a} ⋅ {b \over 2} = {9 b \over 2 a}\)

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

\({5 \over 4} : {x + y \over y} = {5 \over 4} ⋅ {y \over x + y} = {5 y \over 4 (x + y)} = {5 y \over 4 x + 4 y}\)

\({3 x \over 2} + {x - 5 \over 9} = {27 x \over 18} + {2 (x - 5) \over 18} = {27 x + 2 (x - 5) \over 18} = {29 x - 10 \over 18}\)

\({-6 x - 5 \over 7 x - 8} + 2 = {-6 x - 5 \over 7 x - 8} + {2 (7 x - 8) \over 7 x - 8} = {-6 x - 5 + 2 (7 x - 8) \over 7 x - 8} = {-6 x - 5 + 14 x - 16 \over 7 x - 8} = {8 x - 21 \over 7 x - 8}\)

\({3 p^{2} - p - 20 \over p} = {3 p^{2} \over p} - {p \over p} - {20 \over p} = 3 p - 1 - {20 \over p}\)

\({7 a^{2} + 6 a + 5 \over 9 a^{2}} = {7 a^{2} \over 9 a^{2}} + {6 a \over 9 a^{2}} + {5 \over 9 a^{2}} = \frac{7}{9} + {2 \over 3 a} + {5 \over 9 a^{2}}\)