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307

Solve for $a$, $ \dfrac{10}{3a} = -\dfrac{2a - 3}{15a} - \dfrac{10}{3a} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3a$ $15a$ and $3a$ The common denominator is $15a$ To get $15a$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{10}{3a} \times \dfrac{5}{5} = \dfrac{50}{15a} $ The denominator of the second term is already $15a$ , so we don't need to change it. To get $15a$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{10}{3a} \times \dfrac{5}{5} = -\dfrac{50}{15a} $ This give us: $ \dfrac{50}{15a} = -\dfrac{2a - 3}{15a} - \dfrac{50}{15a} $ If we multiply both sides of the equation by $15a$ , we get: $ 50 = -2a + 3 - 50$ $ 50 = -2a - 47$ $ 97 = -2a $ $ a = -\dfrac{97}{2}$

307

Solve for $y$, $ \dfrac{5}{3y} = -\dfrac{4}{9y} - \dfrac{y - 1}{3y} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3y$ $9y$ and $3y$ The common denominator is $9y$ To get $9y$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{5}{3y} \times \dfrac{3}{3} = \dfrac{15}{9y} $ The denominator of the second term is already $9y$ , so we don't need to change it. To get $9y$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{y - 1}{3y} \times \dfrac{3}{3} = -\dfrac{3y - 3}{9y} $ This give us: $ \dfrac{15}{9y} = -\dfrac{4}{9y} - \dfrac{3y - 3}{9y} $ If we multiply both sides of the equation by $9y$ , we get: $ 15 = -4 - 3y + 3$ $ 15 = -3y - 1$ $ 16 = -3y $ $ y = -\dfrac{16}{3}$

307

Solve for $r$, $ \dfrac{7}{25r} = -\dfrac{8}{10r} + \dfrac{5r + 2}{5r} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25r$ $10r$ and $5r$ The common denominator is $50r$ To get $50r$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{7}{25r} \times \dfrac{2}{2} = \dfrac{14}{50r} $ To get $50r$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{8}{10r} \times \dfrac{5}{5} = -\dfrac{40}{50r} $ To get $50r$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ \dfrac{5r + 2}{5r} \times \dfrac{10}{10} = \dfrac{50r + 20}{50r} $ This give us: $ \dfrac{14}{50r} = -\dfrac{40}{50r} + \dfrac{50r + 20}{50r} $ If we multiply both sides of the equation by $50r$ , we get: $ 14 = -40 + 50r + 20$ $ 14 = 50r - 20$ $ 34 = 50r $ $ r = \dfrac{17}{25}$

307

Solve for $q$, $ \dfrac{4q + 4}{4q - 1} = \dfrac{7}{20q - 5} - \dfrac{4}{20q - 5} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4q - 1$ $20q - 5$ and $20q - 5$ The common denominator is $20q - 5$ To get $20q - 5$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{4q + 4}{4q - 1} \times \dfrac{5}{5} = \dfrac{20q + 20}{20q - 5} $ The denominator of the second term is already $20q - 5$ , so we don't need to change it. The denominator of the third term is already $20q - 5$ , so we don't need to change it. This give us: $ \dfrac{20q + 20}{20q - 5} = \dfrac{7}{20q - 5} - \dfrac{4}{20q - 5} $ If we multiply both sides of the equation by $20q - 5$ , we get: $ 20q + 20 = 7 - 4$ $ 20q + 20 = 3$ $ 20q = -17 $ $ q = -\dfrac{17}{20}$

307

Solve for $z$, $ -\dfrac{1}{5z + 5} = -\dfrac{z - 7}{5z + 5} + \dfrac{5}{5z + 5} $

If we multiply both sides of the equation by $5z + 5$ , we get: $ -1 = -z + 7 + 5$ $ -1 = -z + 12$ $ -13 = -z $ $ z = 13$

307

Solve for $a$, $ -\dfrac{3}{a} = \dfrac{5}{4a} + \dfrac{4a - 5}{a} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $a$ $4a$ and $a$ The common denominator is $4a$ To get $4a$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{3}{a} \times \dfrac{4}{4} = -\dfrac{12}{4a} $ The denominator of the second term is already $4a$ , so we don't need to change it. To get $4a$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{4a - 5}{a} \times \dfrac{4}{4} = \dfrac{16a - 20}{4a} $ This give us: $ -\dfrac{12}{4a} = \dfrac{5}{4a} + \dfrac{16a - 20}{4a} $ If we multiply both sides of the equation by $4a$ , we get: $ -12 = 5 + 16a - 20$ $ -12 = 16a - 15$ $ 3 = 16a $ $ a = \dfrac{3}{16}$

307

Solve for $x$, $ -\dfrac{2x - 4}{2x - 4} = \dfrac{8}{2x - 4} - \dfrac{2}{8x - 16} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2x - 4$ $2x - 4$ and $8x - 16$ The common denominator is $8x - 16$ To get $8x - 16$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{2x - 4}{2x - 4} \times \dfrac{4}{4} = -\dfrac{8x - 16}{8x - 16} $ To get $8x - 16$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{8}{2x - 4} \times \dfrac{4}{4} = \dfrac{32}{8x - 16} $ The denominator of the third term is already $8x - 16$ , so we don't need to change it. This give us: $ -\dfrac{8x - 16}{8x - 16} = \dfrac{32}{8x - 16} - \dfrac{2}{8x - 16} $ If we multiply both sides of the equation by $8x - 16$ , we get: $ -8x + 16 = 32 - 2$ $ -8x + 16 = 30$ $ -8x = 14 $ $ x = -\dfrac{7}{4}$

307

Solve for $x$, $ -\dfrac{4x}{2x} = \dfrac{4}{10x} + \dfrac{10}{4x} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2x$ $10x$ and $4x$ The common denominator is $20x$ To get $20x$ in the denominator of the first term, multiply it by $\frac{10}{10}$ $ -\dfrac{4x}{2x} \times \dfrac{10}{10} = -\dfrac{40x}{20x} $ To get $20x$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{4}{10x} \times \dfrac{2}{2} = \dfrac{8}{20x} $ To get $20x$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{10}{4x} \times \dfrac{5}{5} = \dfrac{50}{20x} $ This give us: $ -\dfrac{40x}{20x} = \dfrac{8}{20x} + \dfrac{50}{20x} $ If we multiply both sides of the equation by $20x$ , we get: $ -40x = 8 + 50$ $ -40x = 58$ $ -40x = 58 $ $ x = -\dfrac{29}{20}$

307

Solve for $q$, $ -\dfrac{1}{20q^2} = -\dfrac{4q + 6}{5q^2} + \dfrac{7}{15q^2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $20q^2$ $5q^2$ and $15q^2$ The common denominator is $60q^2$ To get $60q^2$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{1}{20q^2} \times \dfrac{3}{3} = -\dfrac{3}{60q^2} $ To get $60q^2$ in the denominator of the second term, multiply it by $\frac{12}{12}$ $ -\dfrac{4q + 6}{5q^2} \times \dfrac{12}{12} = -\dfrac{48q + 72}{60q^2} $ To get $60q^2$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{7}{15q^2} \times \dfrac{4}{4} = \dfrac{28}{60q^2} $ This give us: $ -\dfrac{3}{60q^2} = -\dfrac{48q + 72}{60q^2} + \dfrac{28}{60q^2} $ If we multiply both sides of the equation by $60q^2$ , we get: $ -3 = -48q - 72 + 28$ $ -3 = -48q - 44$ $ 41 = -48q $ $ q = -\dfrac{41}{48}$

307

Solve for $a$, $ \dfrac{2}{a^2} = \dfrac{9}{3a^2} + \dfrac{3a + 6}{4a^2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $a^2$ $3a^2$ and $4a^2$ The common denominator is $12a^2$ To get $12a^2$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ \dfrac{2}{a^2} \times \dfrac{12}{12} = \dfrac{24}{12a^2} $ To get $12a^2$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{9}{3a^2} \times \dfrac{4}{4} = \dfrac{36}{12a^2} $ To get $12a^2$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{3a + 6}{4a^2} \times \dfrac{3}{3} = \dfrac{9a + 18}{12a^2} $ This give us: $ \dfrac{24}{12a^2} = \dfrac{36}{12a^2} + \dfrac{9a + 18}{12a^2} $ If we multiply both sides of the equation by $12a^2$ , we get: $ 24 = 36 + 9a + 18$ $ 24 = 9a + 54$ $ -30 = 9a $ $ a = -\dfrac{10}{3}$

307

Solve for $p$, $ \dfrac{2}{4p - 5} = -\dfrac{5p + 3}{4p - 5} + \dfrac{9}{4p - 5} $

If we multiply both sides of the equation by $4p - 5$ , we get: $ 2 = -5p - 3 + 9$ $ 2 = -5p + 6$ $ -4 = -5p $ $ p = \dfrac{4}{5}$

307

Solve for $a$, $ -\dfrac{3a + 10}{2a + 2} = -\dfrac{2}{2a + 2} + \dfrac{1}{10a + 10} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2a + 2$ $2a + 2$ and $10a + 10$ The common denominator is $10a + 10$ To get $10a + 10$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{3a + 10}{2a + 2} \times \dfrac{5}{5} = -\dfrac{15a + 50}{10a + 10} $ To get $10a + 10$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{2}{2a + 2} \times \dfrac{5}{5} = -\dfrac{10}{10a + 10} $ The denominator of the third term is already $10a + 10$ , so we don't need to change it. This give us: $ -\dfrac{15a + 50}{10a + 10} = -\dfrac{10}{10a + 10} + \dfrac{1}{10a + 10} $ If we multiply both sides of the equation by $10a + 10$ , we get: $ -15a - 50 = -10 + 1$ $ -15a - 50 = -9$ $ -15a = 41 $ $ a = -\dfrac{41}{15}$

307

Solve for $x$, $ -\dfrac{9}{x + 3} = -\dfrac{5}{x + 3} - \dfrac{x + 9}{x + 3} $

If we multiply both sides of the equation by $x + 3$ , we get: $ -9 = -5 - x - 9$ $ -9 = -x - 14$ $ 5 = -x $ $ x = -5$

307

Solve for $q$, $ \dfrac{5}{2q^2} = -\dfrac{1}{q^2} - \dfrac{q - 7}{5q^2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2q^2$ $q^2$ and $5q^2$ The common denominator is $10q^2$ To get $10q^2$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{5}{2q^2} \times \dfrac{5}{5} = \dfrac{25}{10q^2} $ To get $10q^2$ in the denominator of the second term, multiply it by $\frac{10}{10}$ $ -\dfrac{1}{q^2} \times \dfrac{10}{10} = -\dfrac{10}{10q^2} $ To get $10q^2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{q - 7}{5q^2} \times \dfrac{2}{2} = -\dfrac{2q - 14}{10q^2} $ This give us: $ \dfrac{25}{10q^2} = -\dfrac{10}{10q^2} - \dfrac{2q - 14}{10q^2} $ If we multiply both sides of the equation by $10q^2$ , we get: $ 25 = -10 - 2q + 14$ $ 25 = -2q + 4$ $ 21 = -2q $ $ q = -\dfrac{21}{2}$

307

Solve for $q$, $ -\dfrac{3q - 8}{q - 4} = -\dfrac{7}{3q - 12} - \dfrac{6}{4q - 16} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $q - 4$ $3q - 12$ and $4q - 16$ The common denominator is $12q - 48$ To get $12q - 48$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ -\dfrac{3q - 8}{q - 4} \times \dfrac{12}{12} = -\dfrac{36q - 96}{12q - 48} $ To get $12q - 48$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{7}{3q - 12} \times \dfrac{4}{4} = -\dfrac{28}{12q - 48} $ To get $12q - 48$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{6}{4q - 16} \times \dfrac{3}{3} = -\dfrac{18}{12q - 48} $ This give us: $ -\dfrac{36q - 96}{12q - 48} = -\dfrac{28}{12q - 48} - \dfrac{18}{12q - 48} $ If we multiply both sides of the equation by $12q - 48$ , we get: $ -36q + 96 = -28 - 18$ $ -36q + 96 = -46$ $ -36q = -142 $ $ q = \dfrac{71}{18}$

307

Solve for $x$, $ -\dfrac{8}{x} = -\dfrac{7}{x} + \dfrac{5x + 2}{2x} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $x$ $x$ and $2x$ The common denominator is $2x$ To get $2x$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{8}{x} \times \dfrac{2}{2} = -\dfrac{16}{2x} $ To get $2x$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{7}{x} \times \dfrac{2}{2} = -\dfrac{14}{2x} $ The denominator of the third term is already $2x$ , so we don't need to change it. This give us: $ -\dfrac{16}{2x} = -\dfrac{14}{2x} + \dfrac{5x + 2}{2x} $ If we multiply both sides of the equation by $2x$ , we get: $ -16 = -14 + 5x + 2$ $ -16 = 5x - 12$ $ -4 = 5x $ $ x = -\dfrac{4}{5}$

307

Solve for $p$, $ \dfrac{6}{5p + 15} = \dfrac{9}{p + 3} + \dfrac{p - 4}{3p + 9} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5p + 15$ $p + 3$ and $3p + 9$ The common denominator is $15p + 45$ To get $15p + 45$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{6}{5p + 15} \times \dfrac{3}{3} = \dfrac{18}{15p + 45} $ To get $15p + 45$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ \dfrac{9}{p + 3} \times \dfrac{15}{15} = \dfrac{135}{15p + 45} $ To get $15p + 45$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{p - 4}{3p + 9} \times \dfrac{5}{5} = \dfrac{5p - 20}{15p + 45} $ This give us: $ \dfrac{18}{15p + 45} = \dfrac{135}{15p + 45} + \dfrac{5p - 20}{15p + 45} $ If we multiply both sides of the equation by $15p + 45$ , we get: $ 18 = 135 + 5p - 20$ $ 18 = 5p + 115$ $ -97 = 5p $ $ p = -\dfrac{97}{5}$

307

Solve for $z$, $ -\dfrac{7}{15z + 12} = -\dfrac{3z - 6}{5z + 4} + \dfrac{10}{5z + 4} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15z + 12$ $5z + 4$ and $5z + 4$ The common denominator is $15z + 12$ The denominator of the first term is already $15z + 12$ , so we don't need to change it. To get $15z + 12$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{3z - 6}{5z + 4} \times \dfrac{3}{3} = -\dfrac{9z - 18}{15z + 12} $ To get $15z + 12$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{10}{5z + 4} \times \dfrac{3}{3} = \dfrac{30}{15z + 12} $ This give us: $ -\dfrac{7}{15z + 12} = -\dfrac{9z - 18}{15z + 12} + \dfrac{30}{15z + 12} $ If we multiply both sides of the equation by $15z + 12$ , we get: $ -7 = -9z + 18 + 30$ $ -7 = -9z + 48$ $ -55 = -9z $ $ z = \dfrac{55}{9}$

307

Solve for $t$, $ -\dfrac{5}{t - 5} = -\dfrac{3}{t - 5} - \dfrac{3t}{t - 5} $

If we multiply both sides of the equation by $t - 5$ , we get: $ -5 = -3 - 3t$ $ -5 = -3t - 3$ $ -2 = -3t $ $ t = \dfrac{2}{3}$

307

Solve for $r$, $ -\dfrac{4r - 7}{r + 3} = \dfrac{4}{r + 3} - \dfrac{1}{r + 3} $

If we multiply both sides of the equation by $r + 3$ , we get: $ -4r + 7 = 4 - 1$ $ -4r + 7 = 3$ $ -4r = -4 $ $ r = 1$

307

Solve for $y$, $ -\dfrac{3}{25y + 10} = -\dfrac{10}{10y + 4} + \dfrac{4y - 1}{5y + 2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25y + 10$ $10y + 4$ and $5y + 2$ The common denominator is $50y + 20$ To get $50y + 20$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{3}{25y + 10} \times \dfrac{2}{2} = -\dfrac{6}{50y + 20} $ To get $50y + 20$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{10}{10y + 4} \times \dfrac{5}{5} = -\dfrac{50}{50y + 20} $ To get $50y + 20$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ \dfrac{4y - 1}{5y + 2} \times \dfrac{10}{10} = \dfrac{40y - 10}{50y + 20} $ This give us: $ -\dfrac{6}{50y + 20} = -\dfrac{50}{50y + 20} + \dfrac{40y - 10}{50y + 20} $ If we multiply both sides of the equation by $50y + 20$ , we get: $ -6 = -50 + 40y - 10$ $ -6 = 40y - 60$ $ 54 = 40y $ $ y = \dfrac{27}{20}$

307

Solve for $a$, $ \dfrac{10}{a^2} = -\dfrac{6}{a^2} + \dfrac{5a - 7}{a^2} $

If we multiply both sides of the equation by $a^2$ , we get: $ 10 = -6 + 5a - 7$ $ 10 = 5a - 13$ $ 23 = 5a $ $ a = \dfrac{23}{5}$

307

Solve for $t$, $ -\dfrac{t}{t^3} = -\dfrac{2}{3t^3} + \dfrac{7}{t^3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $t^3$ $3t^3$ and $t^3$ The common denominator is $3t^3$ To get $3t^3$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{t}{t^3} \times \dfrac{3}{3} = -\dfrac{3t}{3t^3} $ The denominator of the second term is already $3t^3$ , so we don't need to change it. To get $3t^3$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{7}{t^3} \times \dfrac{3}{3} = \dfrac{21}{3t^3} $ This give us: $ -\dfrac{3t}{3t^3} = -\dfrac{2}{3t^3} + \dfrac{21}{3t^3} $ If we multiply both sides of the equation by $3t^3$ , we get: $ -3t = -2 + 21$ $ -3t = 19$ $ -3t = 19 $ $ t = -\dfrac{19}{3}$

307

Solve for $k$, $ -\dfrac{k + 1}{25k^3} = -\dfrac{3}{5k^3} - \dfrac{8}{20k^3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25k^3$ $5k^3$ and $20k^3$ The common denominator is $100k^3$ To get $100k^3$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{k + 1}{25k^3} \times \dfrac{4}{4} = -\dfrac{4k + 4}{100k^3} $ To get $100k^3$ in the denominator of the second term, multiply it by $\frac{20}{20}$ $ -\dfrac{3}{5k^3} \times \dfrac{20}{20} = -\dfrac{60}{100k^3} $ To get $100k^3$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{8}{20k^3} \times \dfrac{5}{5} = -\dfrac{40}{100k^3} $ This give us: $ -\dfrac{4k + 4}{100k^3} = -\dfrac{60}{100k^3} - \dfrac{40}{100k^3} $ If we multiply both sides of the equation by $100k^3$ , we get: $ -4k - 4 = -60 - 40$ $ -4k - 4 = -100$ $ -4k = -96 $ $ k = 24$

307

Solve for $t$, $ -\dfrac{10}{2t - 3} = -\dfrac{t - 5}{10t - 15} - \dfrac{3}{2t - 3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2t - 3$ $10t - 15$ and $2t - 3$ The common denominator is $10t - 15$ To get $10t - 15$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{10}{2t - 3} \times \dfrac{5}{5} = -\dfrac{50}{10t - 15} $ The denominator of the second term is already $10t - 15$ , so we don't need to change it. To get $10t - 15$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{3}{2t - 3} \times \dfrac{5}{5} = -\dfrac{15}{10t - 15} $ This give us: $ -\dfrac{50}{10t - 15} = -\dfrac{t - 5}{10t - 15} - \dfrac{15}{10t - 15} $ If we multiply both sides of the equation by $10t - 15$ , we get: $ -50 = -t + 5 - 15$ $ -50 = -t - 10$ $ -40 = -t $ $ t = 40$

307

Solve for $a$, $ \dfrac{a - 7}{2a} = -\dfrac{2}{2a} - \dfrac{9}{2a} $

If we multiply both sides of the equation by $2a$ , we get: $ a - 7 = -2 - 9$ $ a - 7 = -11$ $ a = -4 $

307

Solve for $x$, $ -\dfrac{6}{3x - 5} = \dfrac{x + 10}{6x - 10} + \dfrac{1}{3x - 5} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3x - 5$ $6x - 10$ and $3x - 5$ The common denominator is $6x - 10$ To get $6x - 10$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{6}{3x - 5} \times \dfrac{2}{2} = -\dfrac{12}{6x - 10} $ The denominator of the second term is already $6x - 10$ , so we don't need to change it. To get $6x - 10$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{1}{3x - 5} \times \dfrac{2}{2} = \dfrac{2}{6x - 10} $ This give us: $ -\dfrac{12}{6x - 10} = \dfrac{x + 10}{6x - 10} + \dfrac{2}{6x - 10} $ If we multiply both sides of the equation by $6x - 10$ , we get: $ -12 = x + 10 + 2$ $ -12 = x + 12$ $ -24 = x $ $ x = -24$

307

Solve for $k$, $ -\dfrac{2}{12k - 4} = -\dfrac{k - 5}{6k - 2} + \dfrac{9}{3k - 1} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $12k - 4$ $6k - 2$ and $3k - 1$ The common denominator is $12k - 4$ The denominator of the first term is already $12k - 4$ , so we don't need to change it. To get $12k - 4$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{k - 5}{6k - 2} \times \dfrac{2}{2} = -\dfrac{2k - 10}{12k - 4} $ To get $12k - 4$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{9}{3k - 1} \times \dfrac{4}{4} = \dfrac{36}{12k - 4} $ This give us: $ -\dfrac{2}{12k - 4} = -\dfrac{2k - 10}{12k - 4} + \dfrac{36}{12k - 4} $ If we multiply both sides of the equation by $12k - 4$ , we get: $ -2 = -2k + 10 + 36$ $ -2 = -2k + 46$ $ -48 = -2k $ $ k = 24$

307

Solve for $n$, $ -\dfrac{3}{n} = -\dfrac{n - 3}{5n} - \dfrac{4}{n} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $n$ $5n$ and $n$ The common denominator is $5n$ To get $5n$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{3}{n} \times \dfrac{5}{5} = -\dfrac{15}{5n} $ The denominator of the second term is already $5n$ , so we don't need to change it. To get $5n$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{4}{n} \times \dfrac{5}{5} = -\dfrac{20}{5n} $ This give us: $ -\dfrac{15}{5n} = -\dfrac{n - 3}{5n} - \dfrac{20}{5n} $ If we multiply both sides of the equation by $5n$ , we get: $ -15 = -n + 3 - 20$ $ -15 = -n - 17$ $ 2 = -n $ $ n = -2$

307

Solve for $k$, $ -\dfrac{6}{15k} = -\dfrac{5k - 9}{20k} - \dfrac{6}{5k} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15k$ $20k$ and $5k$ The common denominator is $60k$ To get $60k$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{6}{15k} \times \dfrac{4}{4} = -\dfrac{24}{60k} $ To get $60k$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{5k - 9}{20k} \times \dfrac{3}{3} = -\dfrac{15k - 27}{60k} $ To get $60k$ in the denominator of the third term, multiply it by $\frac{12}{12}$ $ -\dfrac{6}{5k} \times \dfrac{12}{12} = -\dfrac{72}{60k} $ This give us: $ -\dfrac{24}{60k} = -\dfrac{15k - 27}{60k} - \dfrac{72}{60k} $ If we multiply both sides of the equation by $60k$ , we get: $ -24 = -15k + 27 - 72$ $ -24 = -15k - 45$ $ 21 = -15k $ $ k = -\dfrac{7}{5}$

307

Solve for $y$, $ -\dfrac{9}{9y^2} = -\dfrac{5y - 4}{9y^2} - \dfrac{7}{3y^2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $9y^2$ $9y^2$ and $3y^2$ The common denominator is $9y^2$ The denominator of the first term is already $9y^2$ , so we don't need to change it. The denominator of the second term is already $9y^2$ , so we don't need to change it. To get $9y^2$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{7}{3y^2} \times \dfrac{3}{3} = -\dfrac{21}{9y^2} $ This give us: $ -\dfrac{9}{9y^2} = -\dfrac{5y - 4}{9y^2} - \dfrac{21}{9y^2} $ If we multiply both sides of the equation by $9y^2$ , we get: $ -9 = -5y + 4 - 21$ $ -9 = -5y - 17$ $ 8 = -5y $ $ y = -\dfrac{8}{5}$

307

Solve for $t$, $ \dfrac{7}{t + 1} = \dfrac{2}{t + 1} - \dfrac{5t - 6}{t + 1} $

If we multiply both sides of the equation by $t + 1$ , we get: $ 7 = 2 - 5t + 6$ $ 7 = -5t + 8$ $ -1 = -5t $ $ t = \dfrac{1}{5}$

307

Solve for $n$, $ -\dfrac{5}{4n} = -\dfrac{4}{20n} - \dfrac{n + 6}{4n} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4n$ $20n$ and $4n$ The common denominator is $20n$ To get $20n$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{4n} \times \dfrac{5}{5} = -\dfrac{25}{20n} $ The denominator of the second term is already $20n$ , so we don't need to change it. To get $20n$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{n + 6}{4n} \times \dfrac{5}{5} = -\dfrac{5n + 30}{20n} $ This give us: $ -\dfrac{25}{20n} = -\dfrac{4}{20n} - \dfrac{5n + 30}{20n} $ If we multiply both sides of the equation by $20n$ , we get: $ -25 = -4 - 5n - 30$ $ -25 = -5n - 34$ $ 9 = -5n $ $ n = -\dfrac{9}{5}$

307

Solve for $a$, $ -\dfrac{8}{8a^2} = \dfrac{4a - 3}{4a^2} + \dfrac{3}{20a^2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8a^2$ $4a^2$ and $20a^2$ The common denominator is $40a^2$ To get $40a^2$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{8}{8a^2} \times \dfrac{5}{5} = -\dfrac{40}{40a^2} $ To get $40a^2$ in the denominator of the second term, multiply it by $\frac{10}{10}$ $ \dfrac{4a - 3}{4a^2} \times \dfrac{10}{10} = \dfrac{40a - 30}{40a^2} $ To get $40a^2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{3}{20a^2} \times \dfrac{2}{2} = \dfrac{6}{40a^2} $ This give us: $ -\dfrac{40}{40a^2} = \dfrac{40a - 30}{40a^2} + \dfrac{6}{40a^2} $ If we multiply both sides of the equation by $40a^2$ , we get: $ -40 = 40a - 30 + 6$ $ -40 = 40a - 24$ $ -16 = 40a $ $ a = -\dfrac{2}{5}$

307

Solve for $k$, $ \dfrac{4}{4k} = -\dfrac{2}{4k} + \dfrac{4k - 6}{4k} $

If we multiply both sides of the equation by $4k$ , we get: $ 4 = -2 + 4k - 6$ $ 4 = 4k - 8$ $ 12 = 4k $ $ k = 3$

307

Solve for $n$, $ -\dfrac{3}{4n - 3} = \dfrac{n + 6}{4n - 3} + \dfrac{6}{20n - 15} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4n - 3$ $4n - 3$ and $20n - 15$ The common denominator is $20n - 15$ To get $20n - 15$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{3}{4n - 3} \times \dfrac{5}{5} = -\dfrac{15}{20n - 15} $ To get $20n - 15$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{n + 6}{4n - 3} \times \dfrac{5}{5} = \dfrac{5n + 30}{20n - 15} $ The denominator of the third term is already $20n - 15$ , so we don't need to change it. This give us: $ -\dfrac{15}{20n - 15} = \dfrac{5n + 30}{20n - 15} + \dfrac{6}{20n - 15} $ If we multiply both sides of the equation by $20n - 15$ , we get: $ -15 = 5n + 30 + 6$ $ -15 = 5n + 36$ $ -51 = 5n $ $ n = -\dfrac{51}{5}$

307

Solve for $y$, $ \dfrac{3}{16y + 8} = -\dfrac{y + 8}{4y + 2} + \dfrac{5}{16y + 8} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16y + 8$ $4y + 2$ and $16y + 8$ The common denominator is $16y + 8$ The denominator of the first term is already $16y + 8$ , so we don't need to change it. To get $16y + 8$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{y + 8}{4y + 2} \times \dfrac{4}{4} = -\dfrac{4y + 32}{16y + 8} $ The denominator of the third term is already $16y + 8$ , so we don't need to change it. This give us: $ \dfrac{3}{16y + 8} = -\dfrac{4y + 32}{16y + 8} + \dfrac{5}{16y + 8} $ If we multiply both sides of the equation by $16y + 8$ , we get: $ 3 = -4y - 32 + 5$ $ 3 = -4y - 27$ $ 30 = -4y $ $ y = -\dfrac{15}{2}$

307

Solve for $z$, $ -\dfrac{9}{3z - 9} = -\dfrac{8}{z - 3} - \dfrac{5z + 3}{z - 3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3z - 9$ $z - 3$ and $z - 3$ The common denominator is $3z - 9$ The denominator of the first term is already $3z - 9$ , so we don't need to change it. To get $3z - 9$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{8}{z - 3} \times \dfrac{3}{3} = -\dfrac{24}{3z - 9} $ To get $3z - 9$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{5z + 3}{z - 3} \times \dfrac{3}{3} = -\dfrac{15z + 9}{3z - 9} $ This give us: $ -\dfrac{9}{3z - 9} = -\dfrac{24}{3z - 9} - \dfrac{15z + 9}{3z - 9} $ If we multiply both sides of the equation by $3z - 9$ , we get: $ -9 = -24 - 15z - 9$ $ -9 = -15z - 33$ $ 24 = -15z $ $ z = -\dfrac{8}{5}$

307

Solve for $y$, $ \dfrac{5}{12y} = \dfrac{8}{3y} - \dfrac{5y - 6}{9y} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $12y$ $3y$ and $9y$ The common denominator is $36y$ To get $36y$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{5}{12y} \times \dfrac{3}{3} = \dfrac{15}{36y} $ To get $36y$ in the denominator of the second term, multiply it by $\frac{12}{12}$ $ \dfrac{8}{3y} \times \dfrac{12}{12} = \dfrac{96}{36y} $ To get $36y$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{5y - 6}{9y} \times \dfrac{4}{4} = -\dfrac{20y - 24}{36y} $ This give us: $ \dfrac{15}{36y} = \dfrac{96}{36y} - \dfrac{20y - 24}{36y} $ If we multiply both sides of the equation by $36y$ , we get: $ 15 = 96 - 20y + 24$ $ 15 = -20y + 120$ $ -105 = -20y $ $ y = \dfrac{21}{4}$

307

Solve for $p$, $ -\dfrac{1}{p + 2} = -\dfrac{7}{4p + 8} + \dfrac{5p - 6}{4p + 8} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $p + 2$ $4p + 8$ and $4p + 8$ The common denominator is $4p + 8$ To get $4p + 8$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{1}{p + 2} \times \dfrac{4}{4} = -\dfrac{4}{4p + 8} $ The denominator of the second term is already $4p + 8$ , so we don't need to change it. The denominator of the third term is already $4p + 8$ , so we don't need to change it. This give us: $ -\dfrac{4}{4p + 8} = -\dfrac{7}{4p + 8} + \dfrac{5p - 6}{4p + 8} $ If we multiply both sides of the equation by $4p + 8$ , we get: $ -4 = -7 + 5p - 6$ $ -4 = 5p - 13$ $ 9 = 5p $ $ p = \dfrac{9}{5}$

307

Solve for $p$, $ \dfrac{3p - 2}{8p - 4} = \dfrac{1}{12p - 6} + \dfrac{1}{4p - 2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8p - 4$ $12p - 6$ and $4p - 2$ The common denominator is $24p - 12$ To get $24p - 12$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{3p - 2}{8p - 4} \times \dfrac{3}{3} = \dfrac{9p - 6}{24p - 12} $ To get $24p - 12$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{1}{12p - 6} \times \dfrac{2}{2} = \dfrac{2}{24p - 12} $ To get $24p - 12$ in the denominator of the third term, multiply it by $\frac{6}{6}$ $ \dfrac{1}{4p - 2} \times \dfrac{6}{6} = \dfrac{6}{24p - 12} $ This give us: $ \dfrac{9p - 6}{24p - 12} = \dfrac{2}{24p - 12} + \dfrac{6}{24p - 12} $ If we multiply both sides of the equation by $24p - 12$ , we get: $ 9p - 6 = 2 + 6$ $ 9p - 6 = 8$ $ 9p = 14 $ $ p = \dfrac{14}{9}$

307

Solve for $x$, $ \dfrac{x + 10}{4x + 16} = \dfrac{7}{x + 4} + \dfrac{1}{x + 4} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4x + 16$ $x + 4$ and $x + 4$ The common denominator is $4x + 16$ The denominator of the first term is already $4x + 16$ , so we don't need to change it. To get $4x + 16$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{7}{x + 4} \times \dfrac{4}{4} = \dfrac{28}{4x + 16} $ To get $4x + 16$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{1}{x + 4} \times \dfrac{4}{4} = \dfrac{4}{4x + 16} $ This give us: $ \dfrac{x + 10}{4x + 16} = \dfrac{28}{4x + 16} + \dfrac{4}{4x + 16} $ If we multiply both sides of the equation by $4x + 16$ , we get: $ x + 10 = 28 + 4$ $ x + 10 = 32$ $ x = 22 $

307

Solve for $t$, $ \dfrac{1}{5t - 20} = \dfrac{2t - 2}{t - 4} + \dfrac{3}{3t - 12} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5t - 20$ $t - 4$ and $3t - 12$ The common denominator is $15t - 60$ To get $15t - 60$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{1}{5t - 20} \times \dfrac{3}{3} = \dfrac{3}{15t - 60} $ To get $15t - 60$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ \dfrac{2t - 2}{t - 4} \times \dfrac{15}{15} = \dfrac{30t - 30}{15t - 60} $ To get $15t - 60$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{3}{3t - 12} \times \dfrac{5}{5} = \dfrac{15}{15t - 60} $ This give us: $ \dfrac{3}{15t - 60} = \dfrac{30t - 30}{15t - 60} + \dfrac{15}{15t - 60} $ If we multiply both sides of the equation by $15t - 60$ , we get: $ 3 = 30t - 30 + 15$ $ 3 = 30t - 15$ $ 18 = 30t $ $ t = \dfrac{3}{5}$

307

Solve for $k$, $ \dfrac{9}{8k} = \dfrac{4}{20k} - \dfrac{5k + 3}{4k} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8k$ $20k$ and $4k$ The common denominator is $40k$ To get $40k$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{9}{8k} \times \dfrac{5}{5} = \dfrac{45}{40k} $ To get $40k$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{4}{20k} \times \dfrac{2}{2} = \dfrac{8}{40k} $ To get $40k$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ -\dfrac{5k + 3}{4k} \times \dfrac{10}{10} = -\dfrac{50k + 30}{40k} $ This give us: $ \dfrac{45}{40k} = \dfrac{8}{40k} - \dfrac{50k + 30}{40k} $ If we multiply both sides of the equation by $40k$ , we get: $ 45 = 8 - 50k - 30$ $ 45 = -50k - 22$ $ 67 = -50k $ $ k = -\dfrac{67}{50}$

307

Solve for $r$, $ \dfrac{r - 8}{r} = \dfrac{3}{4r} + \dfrac{3}{r} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $r$ $4r$ and $r$ The common denominator is $4r$ To get $4r$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{r - 8}{r} \times \dfrac{4}{4} = \dfrac{4r - 32}{4r} $ The denominator of the second term is already $4r$ , so we don't need to change it. To get $4r$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{3}{r} \times \dfrac{4}{4} = \dfrac{12}{4r} $ This give us: $ \dfrac{4r - 32}{4r} = \dfrac{3}{4r} + \dfrac{12}{4r} $ If we multiply both sides of the equation by $4r$ , we get: $ 4r - 32 = 3 + 12$ $ 4r - 32 = 15$ $ 4r = 47 $ $ r = \dfrac{47}{4}$

307

Solve for $r$, $ -\dfrac{1}{12r - 8} = \dfrac{r + 10}{15r - 10} + \dfrac{5}{3r - 2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $12r - 8$ $15r - 10$ and $3r - 2$ The common denominator is $60r - 40$ To get $60r - 40$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{1}{12r - 8} \times \dfrac{5}{5} = -\dfrac{5}{60r - 40} $ To get $60r - 40$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{r + 10}{15r - 10} \times \dfrac{4}{4} = \dfrac{4r + 40}{60r - 40} $ To get $60r - 40$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ \dfrac{5}{3r - 2} \times \dfrac{20}{20} = \dfrac{100}{60r - 40} $ This give us: $ -\dfrac{5}{60r - 40} = \dfrac{4r + 40}{60r - 40} + \dfrac{100}{60r - 40} $ If we multiply both sides of the equation by $60r - 40$ , we get: $ -5 = 4r + 40 + 100$ $ -5 = 4r + 140$ $ -145 = 4r $ $ r = -\dfrac{145}{4}$

307

Solve for $q$, $ \dfrac{q + 1}{3q} = \dfrac{3}{5q} - \dfrac{1}{q} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3q$ $5q$ and $q$ The common denominator is $15q$ To get $15q$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{q + 1}{3q} \times \dfrac{5}{5} = \dfrac{5q + 5}{15q} $ To get $15q$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ \dfrac{3}{5q} \times \dfrac{3}{3} = \dfrac{9}{15q} $ To get $15q$ in the denominator of the third term, multiply it by $\frac{15}{15}$ $ -\dfrac{1}{q} \times \dfrac{15}{15} = -\dfrac{15}{15q} $ This give us: $ \dfrac{5q + 5}{15q} = \dfrac{9}{15q} - \dfrac{15}{15q} $ If we multiply both sides of the equation by $15q$ , we get: $ 5q + 5 = 9 - 15$ $ 5q + 5 = -6$ $ 5q = -11 $ $ q = -\dfrac{11}{5}$

307

Solve for $x$, $ \dfrac{3x + 4}{4x - 2} = \dfrac{4}{4x - 2} - \dfrac{10}{8x - 4} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4x - 2$ $4x - 2$ and $8x - 4$ The common denominator is $8x - 4$ To get $8x - 4$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{3x + 4}{4x - 2} \times \dfrac{2}{2} = \dfrac{6x + 8}{8x - 4} $ To get $8x - 4$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{4}{4x - 2} \times \dfrac{2}{2} = \dfrac{8}{8x - 4} $ The denominator of the third term is already $8x - 4$ , so we don't need to change it. This give us: $ \dfrac{6x + 8}{8x - 4} = \dfrac{8}{8x - 4} - \dfrac{10}{8x - 4} $ If we multiply both sides of the equation by $8x - 4$ , we get: $ 6x + 8 = 8 - 10$ $ 6x + 8 = -2$ $ 6x = -10 $ $ x = -\dfrac{5}{3}$

307

Solve for $r$, $ \dfrac{r + 5}{8r^3} = -\dfrac{10}{10r^3} - \dfrac{6}{4r^3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8r^3$ $10r^3$ and $4r^3$ The common denominator is $40r^3$ To get $40r^3$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{r + 5}{8r^3} \times \dfrac{5}{5} = \dfrac{5r + 25}{40r^3} $ To get $40r^3$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{10}{10r^3} \times \dfrac{4}{4} = -\dfrac{40}{40r^3} $ To get $40r^3$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ -\dfrac{6}{4r^3} \times \dfrac{10}{10} = -\dfrac{60}{40r^3} $ This give us: $ \dfrac{5r + 25}{40r^3} = -\dfrac{40}{40r^3} - \dfrac{60}{40r^3} $ If we multiply both sides of the equation by $40r^3$ , we get: $ 5r + 25 = -40 - 60$ $ 5r + 25 = -100$ $ 5r = -125 $ $ r = -25$

307

Solve for $q$, $ \dfrac{q + 3}{4q - 5} = -\dfrac{1}{4q - 5} + \dfrac{9}{4q - 5} $

If we multiply both sides of the equation by $4q - 5$ , we get: $ q + 3 = -1 + 9$ $ q + 3 = 8$ $ q = 5 $

307

Solve for $y$, $ \dfrac{9}{10y + 6} = \dfrac{2y - 9}{15y + 9} - \dfrac{3}{5y + 3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $10y + 6$ $15y + 9$ and $5y + 3$ The common denominator is $30y + 18$ To get $30y + 18$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{9}{10y + 6} \times \dfrac{3}{3} = \dfrac{27}{30y + 18} $ To get $30y + 18$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{2y - 9}{15y + 9} \times \dfrac{2}{2} = \dfrac{4y - 18}{30y + 18} $ To get $30y + 18$ in the denominator of the third term, multiply it by $\frac{6}{6}$ $ -\dfrac{3}{5y + 3} \times \dfrac{6}{6} = -\dfrac{18}{30y + 18} $ This give us: $ \dfrac{27}{30y + 18} = \dfrac{4y - 18}{30y + 18} - \dfrac{18}{30y + 18} $ If we multiply both sides of the equation by $30y + 18$ , we get: $ 27 = 4y - 18 - 18$ $ 27 = 4y - 36$ $ 63 = 4y $ $ y = \dfrac{63}{4}$

307

Solve for $k$, $ \dfrac{6}{k} = -\dfrac{3k - 1}{k} + \dfrac{9}{k} $

If we multiply both sides of the equation by $k$ , we get: $ 6 = -3k + 1 + 9$ $ 6 = -3k + 10$ $ -4 = -3k $ $ k = \dfrac{4}{3}$

307

Solve for $z$, $ -\dfrac{z - 8}{20z^3} = -\dfrac{6}{16z^3} + \dfrac{8}{16z^3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $20z^3$ $16z^3$ and $16z^3$ The common denominator is $80z^3$ To get $80z^3$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{z - 8}{20z^3} \times \dfrac{4}{4} = -\dfrac{4z - 32}{80z^3} $ To get $80z^3$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{6}{16z^3} \times \dfrac{5}{5} = -\dfrac{30}{80z^3} $ To get $80z^3$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{8}{16z^3} \times \dfrac{5}{5} = \dfrac{40}{80z^3} $ This give us: $ -\dfrac{4z - 32}{80z^3} = -\dfrac{30}{80z^3} + \dfrac{40}{80z^3} $ If we multiply both sides of the equation by $80z^3$ , we get: $ -4z + 32 = -30 + 40$ $ -4z + 32 = 10$ $ -4z = -22 $ $ z = \dfrac{11}{2}$

307

Solve for $q$, $ \dfrac{7}{8q + 8} = \dfrac{3q + 7}{6q + 6} + \dfrac{10}{2q + 2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8q + 8$ $6q + 6$ and $2q + 2$ The common denominator is $24q + 24$ To get $24q + 24$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{7}{8q + 8} \times \dfrac{3}{3} = \dfrac{21}{24q + 24} $ To get $24q + 24$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{3q + 7}{6q + 6} \times \dfrac{4}{4} = \dfrac{12q + 28}{24q + 24} $ To get $24q + 24$ in the denominator of the third term, multiply it by $\frac{12}{12}$ $ \dfrac{10}{2q + 2} \times \dfrac{12}{12} = \dfrac{120}{24q + 24} $ This give us: $ \dfrac{21}{24q + 24} = \dfrac{12q + 28}{24q + 24} + \dfrac{120}{24q + 24} $ If we multiply both sides of the equation by $24q + 24$ , we get: $ 21 = 12q + 28 + 120$ $ 21 = 12q + 148$ $ -127 = 12q $ $ q = -\dfrac{127}{12}$

307

Solve for $q$, $ -\dfrac{5q + 4}{3q} = -\dfrac{1}{15q} + \dfrac{4}{12q} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3q$ $15q$ and $12q$ The common denominator is $60q$ To get $60q$ in the denominator of the first term, multiply it by $\frac{20}{20}$ $ -\dfrac{5q + 4}{3q} \times \dfrac{20}{20} = -\dfrac{100q + 80}{60q} $ To get $60q$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{1}{15q} \times \dfrac{4}{4} = -\dfrac{4}{60q} $ To get $60q$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{4}{12q} \times \dfrac{5}{5} = \dfrac{20}{60q} $ This give us: $ -\dfrac{100q + 80}{60q} = -\dfrac{4}{60q} + \dfrac{20}{60q} $ If we multiply both sides of the equation by $60q$ , we get: $ -100q - 80 = -4 + 20$ $ -100q - 80 = 16$ $ -100q = 96 $ $ q = -\dfrac{24}{25}$

307

Solve for $y$, $ -\dfrac{8}{3y - 2} = \dfrac{4y + 5}{12y - 8} + \dfrac{5}{3y - 2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3y - 2$ $12y - 8$ and $3y - 2$ The common denominator is $12y - 8$ To get $12y - 8$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{3y - 2} \times \dfrac{4}{4} = -\dfrac{32}{12y - 8} $ The denominator of the second term is already $12y - 8$ , so we don't need to change it. To get $12y - 8$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{5}{3y - 2} \times \dfrac{4}{4} = \dfrac{20}{12y - 8} $ This give us: $ -\dfrac{32}{12y - 8} = \dfrac{4y + 5}{12y - 8} + \dfrac{20}{12y - 8} $ If we multiply both sides of the equation by $12y - 8$ , we get: $ -32 = 4y + 5 + 20$ $ -32 = 4y + 25$ $ -57 = 4y $ $ y = -\dfrac{57}{4}$

307

Solve for $n$, $ \dfrac{2}{8n} = \dfrac{4n - 4}{4n} + \dfrac{1}{16n} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8n$ $4n$ and $16n$ The common denominator is $16n$ To get $16n$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{2}{8n} \times \dfrac{2}{2} = \dfrac{4}{16n} $ To get $16n$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{4n - 4}{4n} \times \dfrac{4}{4} = \dfrac{16n - 16}{16n} $ The denominator of the third term is already $16n$ , so we don't need to change it. This give us: $ \dfrac{4}{16n} = \dfrac{16n - 16}{16n} + \dfrac{1}{16n} $ If we multiply both sides of the equation by $16n$ , we get: $ 4 = 16n - 16 + 1$ $ 4 = 16n - 15$ $ 19 = 16n $ $ n = \dfrac{19}{16}$

307

Solve for $k$, $ \dfrac{8}{5k - 10} = -\dfrac{4}{k - 2} + \dfrac{k + 1}{4k - 8} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5k - 10$ $k - 2$ and $4k - 8$ The common denominator is $20k - 40$ To get $20k - 40$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{8}{5k - 10} \times \dfrac{4}{4} = \dfrac{32}{20k - 40} $ To get $20k - 40$ in the denominator of the second term, multiply it by $\frac{20}{20}$ $ -\dfrac{4}{k - 2} \times \dfrac{20}{20} = -\dfrac{80}{20k - 40} $ To get $20k - 40$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{k + 1}{4k - 8} \times \dfrac{5}{5} = \dfrac{5k + 5}{20k - 40} $ This give us: $ \dfrac{32}{20k - 40} = -\dfrac{80}{20k - 40} + \dfrac{5k + 5}{20k - 40} $ If we multiply both sides of the equation by $20k - 40$ , we get: $ 32 = -80 + 5k + 5$ $ 32 = 5k - 75$ $ 107 = 5k $ $ k = \dfrac{107}{5}$

307

Solve for $r$, $ -\dfrac{r - 6}{15r^2} = -\dfrac{2}{3r^2} + \dfrac{3}{3r^2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15r^2$ $3r^2$ and $3r^2$ The common denominator is $15r^2$ The denominator of the first term is already $15r^2$ , so we don't need to change it. To get $15r^2$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{2}{3r^2} \times \dfrac{5}{5} = -\dfrac{10}{15r^2} $ To get $15r^2$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{3}{3r^2} \times \dfrac{5}{5} = \dfrac{15}{15r^2} $ This give us: $ -\dfrac{r - 6}{15r^2} = -\dfrac{10}{15r^2} + \dfrac{15}{15r^2} $ If we multiply both sides of the equation by $15r^2$ , we get: $ -r + 6 = -10 + 15$ $ -r + 6 = 5$ $ -r = -1 $ $ r = 1$

307

Solve for $k$, $ \dfrac{10}{k - 4} = \dfrac{4}{k - 4} - \dfrac{2k + 6}{k - 4} $

If we multiply both sides of the equation by $k - 4$ , we get: $ 10 = 4 - 2k - 6$ $ 10 = -2k - 2$ $ 12 = -2k $ $ k = -6$

307

Solve for $n$, $ \dfrac{2}{9n - 3} = -\dfrac{n + 1}{9n - 3} + \dfrac{5}{12n - 4} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $9n - 3$ $9n - 3$ and $12n - 4$ The common denominator is $36n - 12$ To get $36n - 12$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{2}{9n - 3} \times \dfrac{4}{4} = \dfrac{8}{36n - 12} $ To get $36n - 12$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{n + 1}{9n - 3} \times \dfrac{4}{4} = -\dfrac{4n + 4}{36n - 12} $ To get $36n - 12$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{5}{12n - 4} \times \dfrac{3}{3} = \dfrac{15}{36n - 12} $ This give us: $ \dfrac{8}{36n - 12} = -\dfrac{4n + 4}{36n - 12} + \dfrac{15}{36n - 12} $ If we multiply both sides of the equation by $36n - 12$ , we get: $ 8 = -4n - 4 + 15$ $ 8 = -4n + 11$ $ -3 = -4n $ $ n = \dfrac{3}{4}$

307

Solve for $z$, $ -\dfrac{4z + 7}{5z + 1} = -\dfrac{6}{5z + 1} + \dfrac{6}{25z + 5} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5z + 1$ $5z + 1$ and $25z + 5$ The common denominator is $25z + 5$ To get $25z + 5$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{4z + 7}{5z + 1} \times \dfrac{5}{5} = -\dfrac{20z + 35}{25z + 5} $ To get $25z + 5$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{6}{5z + 1} \times \dfrac{5}{5} = -\dfrac{30}{25z + 5} $ The denominator of the third term is already $25z + 5$ , so we don't need to change it. This give us: $ -\dfrac{20z + 35}{25z + 5} = -\dfrac{30}{25z + 5} + \dfrac{6}{25z + 5} $ If we multiply both sides of the equation by $25z + 5$ , we get: $ -20z - 35 = -30 + 6$ $ -20z - 35 = -24$ $ -20z = 11 $ $ z = -\dfrac{11}{20}$

307

Solve for $z$, $ \dfrac{4z + 7}{3z + 6} = -\dfrac{9}{z + 2} - \dfrac{4}{5z + 10} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3z + 6$ $z + 2$ and $5z + 10$ The common denominator is $15z + 30$ To get $15z + 30$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{4z + 7}{3z + 6} \times \dfrac{5}{5} = \dfrac{20z + 35}{15z + 30} $ To get $15z + 30$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ -\dfrac{9}{z + 2} \times \dfrac{15}{15} = -\dfrac{135}{15z + 30} $ To get $15z + 30$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{4}{5z + 10} \times \dfrac{3}{3} = -\dfrac{12}{15z + 30} $ This give us: $ \dfrac{20z + 35}{15z + 30} = -\dfrac{135}{15z + 30} - \dfrac{12}{15z + 30} $ If we multiply both sides of the equation by $15z + 30$ , we get: $ 20z + 35 = -135 - 12$ $ 20z + 35 = -147$ $ 20z = -182 $ $ z = -\dfrac{91}{10}$

307

Solve for $z$, $ \dfrac{3}{3z + 9} = \dfrac{10}{4z + 12} - \dfrac{z + 5}{5z + 15} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3z + 9$ $4z + 12$ and $5z + 15$ The common denominator is $60z + 180$ To get $60z + 180$ in the denominator of the first term, multiply it by $\frac{20}{20}$ $ \dfrac{3}{3z + 9} \times \dfrac{20}{20} = \dfrac{60}{60z + 180} $ To get $60z + 180$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ \dfrac{10}{4z + 12} \times \dfrac{15}{15} = \dfrac{150}{60z + 180} $ To get $60z + 180$ in the denominator of the third term, multiply it by $\frac{12}{12}$ $ -\dfrac{z + 5}{5z + 15} \times \dfrac{12}{12} = -\dfrac{12z + 60}{60z + 180} $ This give us: $ \dfrac{60}{60z + 180} = \dfrac{150}{60z + 180} - \dfrac{12z + 60}{60z + 180} $ If we multiply both sides of the equation by $60z + 180$ , we get: $ 60 = 150 - 12z - 60$ $ 60 = -12z + 90$ $ -30 = -12z $ $ z = \dfrac{5}{2}$

307

Solve for $p$, $ \dfrac{8}{16p - 20} = -\dfrac{5p - 6}{4p - 5} - \dfrac{1}{20p - 25} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16p - 20$ $4p - 5$ and $20p - 25$ The common denominator is $80p - 100$ To get $80p - 100$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{8}{16p - 20} \times \dfrac{5}{5} = \dfrac{40}{80p - 100} $ To get $80p - 100$ in the denominator of the second term, multiply it by $\frac{20}{20}$ $ -\dfrac{5p - 6}{4p - 5} \times \dfrac{20}{20} = -\dfrac{100p - 120}{80p - 100} $ To get $80p - 100$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{1}{20p - 25} \times \dfrac{4}{4} = -\dfrac{4}{80p - 100} $ This give us: $ \dfrac{40}{80p - 100} = -\dfrac{100p - 120}{80p - 100} - \dfrac{4}{80p - 100} $ If we multiply both sides of the equation by $80p - 100$ , we get: $ 40 = -100p + 120 - 4$ $ 40 = -100p + 116$ $ -76 = -100p $ $ p = \dfrac{19}{25}$

307

Solve for $x$, $ -\dfrac{8}{20x} = \dfrac{x - 6}{5x} + \dfrac{4}{20x} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $20x$ $5x$ and $20x$ The common denominator is $20x$ The denominator of the first term is already $20x$ , so we don't need to change it. To get $20x$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{x - 6}{5x} \times \dfrac{4}{4} = \dfrac{4x - 24}{20x} $ The denominator of the third term is already $20x$ , so we don't need to change it. This give us: $ -\dfrac{8}{20x} = \dfrac{4x - 24}{20x} + \dfrac{4}{20x} $ If we multiply both sides of the equation by $20x$ , we get: $ -8 = 4x - 24 + 4$ $ -8 = 4x - 20$ $ 12 = 4x $ $ x = 3$

307

Solve for $t$, $ \dfrac{1}{16t^3} = \dfrac{t + 1}{4t^3} - \dfrac{5}{4t^3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16t^3$ $4t^3$ and $4t^3$ The common denominator is $16t^3$ The denominator of the first term is already $16t^3$ , so we don't need to change it. To get $16t^3$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{t + 1}{4t^3} \times \dfrac{4}{4} = \dfrac{4t + 4}{16t^3} $ To get $16t^3$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{5}{4t^3} \times \dfrac{4}{4} = -\dfrac{20}{16t^3} $ This give us: $ \dfrac{1}{16t^3} = \dfrac{4t + 4}{16t^3} - \dfrac{20}{16t^3} $ If we multiply both sides of the equation by $16t^3$ , we get: $ 1 = 4t + 4 - 20$ $ 1 = 4t - 16$ $ 17 = 4t $ $ t = \dfrac{17}{4}$

307

Solve for $r$, $ \dfrac{4}{4r + 3} = -\dfrac{7}{16r + 12} - \dfrac{4r - 3}{16r + 12} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4r + 3$ $16r + 12$ and $16r + 12$ The common denominator is $16r + 12$ To get $16r + 12$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{4}{4r + 3} \times \dfrac{4}{4} = \dfrac{16}{16r + 12} $ The denominator of the second term is already $16r + 12$ , so we don't need to change it. The denominator of the third term is already $16r + 12$ , so we don't need to change it. This give us: $ \dfrac{16}{16r + 12} = -\dfrac{7}{16r + 12} - \dfrac{4r - 3}{16r + 12} $ If we multiply both sides of the equation by $16r + 12$ , we get: $ 16 = -7 - 4r + 3$ $ 16 = -4r - 4$ $ 20 = -4r $ $ r = -5$

307

Solve for $n$, $ -\dfrac{10}{2n} = -\dfrac{10}{2n} - \dfrac{3n - 8}{2n} $

If we multiply both sides of the equation by $2n$ , we get: $ -10 = -10 - 3n + 8$ $ -10 = -3n - 2$ $ -8 = -3n $ $ n = \dfrac{8}{3}$

307

Solve for $p$, $ -\dfrac{7}{5p + 5} = -\dfrac{4p}{3p + 3} - \dfrac{3}{5p + 5} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5p + 5$ $3p + 3$ and $5p + 5$ The common denominator is $15p + 15$ To get $15p + 15$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{7}{5p + 5} \times \dfrac{3}{3} = -\dfrac{21}{15p + 15} $ To get $15p + 15$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{4p}{3p + 3} \times \dfrac{5}{5} = -\dfrac{20p}{15p + 15} $ To get $15p + 15$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{3}{5p + 5} \times \dfrac{3}{3} = -\dfrac{9}{15p + 15} $ This give us: $ -\dfrac{21}{15p + 15} = -\dfrac{20p}{15p + 15} - \dfrac{9}{15p + 15} $ If we multiply both sides of the equation by $15p + 15$ , we get: $ -21 = -20p - 9$ $ -21 = -20p - 9$ $ -12 = -20p $ $ p = \dfrac{3}{5}$

307

Solve for $a$, $ -\dfrac{7}{25a^3} = -\dfrac{2a + 3}{20a^3} - \dfrac{3}{15a^3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25a^3$ $20a^3$ and $15a^3$ The common denominator is $300a^3$ To get $300a^3$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ -\dfrac{7}{25a^3} \times \dfrac{12}{12} = -\dfrac{84}{300a^3} $ To get $300a^3$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ -\dfrac{2a + 3}{20a^3} \times \dfrac{15}{15} = -\dfrac{30a + 45}{300a^3} $ To get $300a^3$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ -\dfrac{3}{15a^3} \times \dfrac{20}{20} = -\dfrac{60}{300a^3} $ This give us: $ -\dfrac{84}{300a^3} = -\dfrac{30a + 45}{300a^3} - \dfrac{60}{300a^3} $ If we multiply both sides of the equation by $300a^3$ , we get: $ -84 = -30a - 45 - 60$ $ -84 = -30a - 105$ $ 21 = -30a $ $ a = -\dfrac{7}{10}$

307

Solve for $y$, $ \dfrac{8}{y - 1} = \dfrac{6}{4y - 4} + \dfrac{4y}{y - 1} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $y - 1$ $4y - 4$ and $y - 1$ The common denominator is $4y - 4$ To get $4y - 4$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{8}{y - 1} \times \dfrac{4}{4} = \dfrac{32}{4y - 4} $ The denominator of the second term is already $4y - 4$ , so we don't need to change it. To get $4y - 4$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{4y}{y - 1} \times \dfrac{4}{4} = \dfrac{16y}{4y - 4} $ This give us: $ \dfrac{32}{4y - 4} = \dfrac{6}{4y - 4} + \dfrac{16y}{4y - 4} $ If we multiply both sides of the equation by $4y - 4$ , we get: $ 32 = 6 + 16y$ $ 32 = 16y + 6$ $ 26 = 16y $ $ y = \dfrac{13}{8}$

307

Solve for $q$, $ \dfrac{4q - 3}{q^2} = \dfrac{10}{3q^2} + \dfrac{3}{3q^2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $q^2$ $3q^2$ and $3q^2$ The common denominator is $3q^2$ To get $3q^2$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{4q - 3}{q^2} \times \dfrac{3}{3} = \dfrac{12q - 9}{3q^2} $ The denominator of the second term is already $3q^2$ , so we don't need to change it. The denominator of the third term is already $3q^2$ , so we don't need to change it. This give us: $ \dfrac{12q - 9}{3q^2} = \dfrac{10}{3q^2} + \dfrac{3}{3q^2} $ If we multiply both sides of the equation by $3q^2$ , we get: $ 12q - 9 = 10 + 3$ $ 12q - 9 = 13$ $ 12q = 22 $ $ q = \dfrac{11}{6}$

307

Solve for $z$, $ \dfrac{8}{z - 4} = \dfrac{3z + 6}{3z - 12} - \dfrac{10}{5z - 20} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $z - 4$ $3z - 12$ and $5z - 20$ The common denominator is $15z - 60$ To get $15z - 60$ in the denominator of the first term, multiply it by $\frac{15}{15}$ $ \dfrac{8}{z - 4} \times \dfrac{15}{15} = \dfrac{120}{15z - 60} $ To get $15z - 60$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{3z + 6}{3z - 12} \times \dfrac{5}{5} = \dfrac{15z + 30}{15z - 60} $ To get $15z - 60$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{10}{5z - 20} \times \dfrac{3}{3} = -\dfrac{30}{15z - 60} $ This give us: $ \dfrac{120}{15z - 60} = \dfrac{15z + 30}{15z - 60} - \dfrac{30}{15z - 60} $ If we multiply both sides of the equation by $15z - 60$ , we get: $ 120 = 15z + 30 - 30$ $ 120 = 15z$ $ 120 = 15z $ $ z = 8$

307

Solve for $q$, $ -\dfrac{5}{4q - 2} = \dfrac{7}{4q - 2} + \dfrac{q + 1}{20q - 10} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4q - 2$ $4q - 2$ and $20q - 10$ The common denominator is $20q - 10$ To get $20q - 10$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{4q - 2} \times \dfrac{5}{5} = -\dfrac{25}{20q - 10} $ To get $20q - 10$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{7}{4q - 2} \times \dfrac{5}{5} = \dfrac{35}{20q - 10} $ The denominator of the third term is already $20q - 10$ , so we don't need to change it. This give us: $ -\dfrac{25}{20q - 10} = \dfrac{35}{20q - 10} + \dfrac{q + 1}{20q - 10} $ If we multiply both sides of the equation by $20q - 10$ , we get: $ -25 = 35 + q + 1$ $ -25 = q + 36$ $ -61 = q $ $ q = -61$

307

Solve for $k$, $ \dfrac{1}{15k - 15} = -\dfrac{1}{6k - 6} - \dfrac{4k + 4}{3k - 3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15k - 15$ $6k - 6$ and $3k - 3$ The common denominator is $30k - 30$ To get $30k - 30$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{1}{15k - 15} \times \dfrac{2}{2} = \dfrac{2}{30k - 30} $ To get $30k - 30$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{1}{6k - 6} \times \dfrac{5}{5} = -\dfrac{5}{30k - 30} $ To get $30k - 30$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ -\dfrac{4k + 4}{3k - 3} \times \dfrac{10}{10} = -\dfrac{40k + 40}{30k - 30} $ This give us: $ \dfrac{2}{30k - 30} = -\dfrac{5}{30k - 30} - \dfrac{40k + 40}{30k - 30} $ If we multiply both sides of the equation by $30k - 30$ , we get: $ 2 = -5 - 40k - 40$ $ 2 = -40k - 45$ $ 47 = -40k $ $ k = -\dfrac{47}{40}$

307

Solve for $a$, $ \dfrac{3}{16a + 16} = -\dfrac{3}{4a + 4} - \dfrac{4a - 5}{12a + 12} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16a + 16$ $4a + 4$ and $12a + 12$ The common denominator is $48a + 48$ To get $48a + 48$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{3}{16a + 16} \times \dfrac{3}{3} = \dfrac{9}{48a + 48} $ To get $48a + 48$ in the denominator of the second term, multiply it by $\frac{12}{12}$ $ -\dfrac{3}{4a + 4} \times \dfrac{12}{12} = -\dfrac{36}{48a + 48} $ To get $48a + 48$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{4a - 5}{12a + 12} \times \dfrac{4}{4} = -\dfrac{16a - 20}{48a + 48} $ This give us: $ \dfrac{9}{48a + 48} = -\dfrac{36}{48a + 48} - \dfrac{16a - 20}{48a + 48} $ If we multiply both sides of the equation by $48a + 48$ , we get: $ 9 = -36 - 16a + 20$ $ 9 = -16a - 16$ $ 25 = -16a $ $ a = -\dfrac{25}{16}$

307

Solve for $k$, $ -\dfrac{8}{5k} = -\dfrac{5k + 10}{5k} - \dfrac{5}{4k} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5k$ $5k$ and $4k$ The common denominator is $20k$ To get $20k$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{5k} \times \dfrac{4}{4} = -\dfrac{32}{20k} $ To get $20k$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{5k + 10}{5k} \times \dfrac{4}{4} = -\dfrac{20k + 40}{20k} $ To get $20k$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{4k} \times \dfrac{5}{5} = -\dfrac{25}{20k} $ This give us: $ -\dfrac{32}{20k} = -\dfrac{20k + 40}{20k} - \dfrac{25}{20k} $ If we multiply both sides of the equation by $20k$ , we get: $ -32 = -20k - 40 - 25$ $ -32 = -20k - 65$ $ 33 = -20k $ $ k = -\dfrac{33}{20}$

307

Solve for $z$, $ \dfrac{4z + 7}{5z - 2} = \dfrac{4}{5z - 2} + \dfrac{7}{5z - 2} $

If we multiply both sides of the equation by $5z - 2$ , we get: $ 4z + 7 = 4 + 7$ $ 4z + 7 = 11$ $ 4z = 4 $ $ z = 1$

307

Solve for $n$, $ \dfrac{8}{5n + 1} = \dfrac{1}{15n + 3} + \dfrac{2n - 7}{25n + 5} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5n + 1$ $15n + 3$ and $25n + 5$ The common denominator is $75n + 15$ To get $75n + 15$ in the denominator of the first term, multiply it by $\frac{15}{15}$ $ \dfrac{8}{5n + 1} \times \dfrac{15}{15} = \dfrac{120}{75n + 15} $ To get $75n + 15$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{1}{15n + 3} \times \dfrac{5}{5} = \dfrac{5}{75n + 15} $ To get $75n + 15$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{2n - 7}{25n + 5} \times \dfrac{3}{3} = \dfrac{6n - 21}{75n + 15} $ This give us: $ \dfrac{120}{75n + 15} = \dfrac{5}{75n + 15} + \dfrac{6n - 21}{75n + 15} $ If we multiply both sides of the equation by $75n + 15$ , we get: $ 120 = 5 + 6n - 21$ $ 120 = 6n - 16$ $ 136 = 6n $ $ n = \dfrac{68}{3}$

307

Solve for $x$, $ \dfrac{x + 8}{15x + 5} = -\dfrac{4}{3x + 1} - \dfrac{4}{15x + 5} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15x + 5$ $3x + 1$ and $15x + 5$ The common denominator is $15x + 5$ The denominator of the first term is already $15x + 5$ , so we don't need to change it. To get $15x + 5$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{4}{3x + 1} \times \dfrac{5}{5} = -\dfrac{20}{15x + 5} $ The denominator of the third term is already $15x + 5$ , so we don't need to change it. This give us: $ \dfrac{x + 8}{15x + 5} = -\dfrac{20}{15x + 5} - \dfrac{4}{15x + 5} $ If we multiply both sides of the equation by $15x + 5$ , we get: $ x + 8 = -20 - 4$ $ x + 8 = -24$ $ x = -32 $

307

Solve for $r$, $ -\dfrac{3}{r - 4} = \dfrac{7}{r - 4} + \dfrac{2r + 9}{5r - 20} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $r - 4$ $r - 4$ and $5r - 20$ The common denominator is $5r - 20$ To get $5r - 20$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{3}{r - 4} \times \dfrac{5}{5} = -\dfrac{15}{5r - 20} $ To get $5r - 20$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{7}{r - 4} \times \dfrac{5}{5} = \dfrac{35}{5r - 20} $ The denominator of the third term is already $5r - 20$ , so we don't need to change it. This give us: $ -\dfrac{15}{5r - 20} = \dfrac{35}{5r - 20} + \dfrac{2r + 9}{5r - 20} $ If we multiply both sides of the equation by $5r - 20$ , we get: $ -15 = 35 + 2r + 9$ $ -15 = 2r + 44$ $ -59 = 2r $ $ r = -\dfrac{59}{2}$

307

Solve for $q$, $ \dfrac{5}{4q^2} = \dfrac{10}{4q^2} + \dfrac{q - 8}{2q^2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4q^2$ $4q^2$ and $2q^2$ The common denominator is $4q^2$ The denominator of the first term is already $4q^2$ , so we don't need to change it. The denominator of the second term is already $4q^2$ , so we don't need to change it. To get $4q^2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{q - 8}{2q^2} \times \dfrac{2}{2} = \dfrac{2q - 16}{4q^2} $ This give us: $ \dfrac{5}{4q^2} = \dfrac{10}{4q^2} + \dfrac{2q - 16}{4q^2} $ If we multiply both sides of the equation by $4q^2$ , we get: $ 5 = 10 + 2q - 16$ $ 5 = 2q - 6$ $ 11 = 2q $ $ q = \dfrac{11}{2}$

307

Solve for $n$, $ -\dfrac{4n + 5}{15n + 10} = \dfrac{9}{9n + 6} - \dfrac{6}{12n + 8} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15n + 10$ $9n + 6$ and $12n + 8$ The common denominator is $180n + 120$ To get $180n + 120$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ -\dfrac{4n + 5}{15n + 10} \times \dfrac{12}{12} = -\dfrac{48n + 60}{180n + 120} $ To get $180n + 120$ in the denominator of the second term, multiply it by $\frac{20}{20}$ $ \dfrac{9}{9n + 6} \times \dfrac{20}{20} = \dfrac{180}{180n + 120} $ To get $180n + 120$ in the denominator of the third term, multiply it by $\frac{15}{15}$ $ -\dfrac{6}{12n + 8} \times \dfrac{15}{15} = -\dfrac{90}{180n + 120} $ This give us: $ -\dfrac{48n + 60}{180n + 120} = \dfrac{180}{180n + 120} - \dfrac{90}{180n + 120} $ If we multiply both sides of the equation by $180n + 120$ , we get: $ -48n - 60 = 180 - 90$ $ -48n - 60 = 90$ $ -48n = 150 $ $ n = -\dfrac{25}{8}$

307

Solve for $a$, $ -\dfrac{10}{2a - 4} = \dfrac{4}{10a - 20} - \dfrac{5a - 2}{2a - 4} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2a - 4$ $10a - 20$ and $2a - 4$ The common denominator is $10a - 20$ To get $10a - 20$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{10}{2a - 4} \times \dfrac{5}{5} = -\dfrac{50}{10a - 20} $ The denominator of the second term is already $10a - 20$ , so we don't need to change it. To get $10a - 20$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{5a - 2}{2a - 4} \times \dfrac{5}{5} = -\dfrac{25a - 10}{10a - 20} $ This give us: $ -\dfrac{50}{10a - 20} = \dfrac{4}{10a - 20} - \dfrac{25a - 10}{10a - 20} $ If we multiply both sides of the equation by $10a - 20$ , we get: $ -50 = 4 - 25a + 10$ $ -50 = -25a + 14$ $ -64 = -25a $ $ a = \dfrac{64}{25}$

307

Solve for $y$, $ \dfrac{10}{y - 3} = -\dfrac{9}{5y - 15} + \dfrac{2y - 4}{5y - 15} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $y - 3$ $5y - 15$ and $5y - 15$ The common denominator is $5y - 15$ To get $5y - 15$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{10}{y - 3} \times \dfrac{5}{5} = \dfrac{50}{5y - 15} $ The denominator of the second term is already $5y - 15$ , so we don't need to change it. The denominator of the third term is already $5y - 15$ , so we don't need to change it. This give us: $ \dfrac{50}{5y - 15} = -\dfrac{9}{5y - 15} + \dfrac{2y - 4}{5y - 15} $ If we multiply both sides of the equation by $5y - 15$ , we get: $ 50 = -9 + 2y - 4$ $ 50 = 2y - 13$ $ 63 = 2y $ $ y = \dfrac{63}{2}$

307

Solve for $p$, $ \dfrac{3p - 10}{10p} = \dfrac{6}{25p} - \dfrac{3}{5p} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $10p$ $25p$ and $5p$ The common denominator is $50p$ To get $50p$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{3p - 10}{10p} \times \dfrac{5}{5} = \dfrac{15p - 50}{50p} $ To get $50p$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{6}{25p} \times \dfrac{2}{2} = \dfrac{12}{50p} $ To get $50p$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ -\dfrac{3}{5p} \times \dfrac{10}{10} = -\dfrac{30}{50p} $ This give us: $ \dfrac{15p - 50}{50p} = \dfrac{12}{50p} - \dfrac{30}{50p} $ If we multiply both sides of the equation by $50p$ , we get: $ 15p - 50 = 12 - 30$ $ 15p - 50 = -18$ $ 15p = 32 $ $ p = \dfrac{32}{15}$

307

Solve for $z$, $ \dfrac{4z + 1}{z} = -\dfrac{6}{3z} + \dfrac{6}{z} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $z$ $3z$ and $z$ The common denominator is $3z$ To get $3z$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{4z + 1}{z} \times \dfrac{3}{3} = \dfrac{12z + 3}{3z} $ The denominator of the second term is already $3z$ , so we don't need to change it. To get $3z$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{6}{z} \times \dfrac{3}{3} = \dfrac{18}{3z} $ This give us: $ \dfrac{12z + 3}{3z} = -\dfrac{6}{3z} + \dfrac{18}{3z} $ If we multiply both sides of the equation by $3z$ , we get: $ 12z + 3 = -6 + 18$ $ 12z + 3 = 12$ $ 12z = 9 $ $ z = \dfrac{3}{4}$

307

Solve for $k$, $ \dfrac{6}{10k - 15} = -\dfrac{2}{8k - 12} - \dfrac{k - 9}{2k - 3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $10k - 15$ $8k - 12$ and $2k - 3$ The common denominator is $40k - 60$ To get $40k - 60$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{6}{10k - 15} \times \dfrac{4}{4} = \dfrac{24}{40k - 60} $ To get $40k - 60$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{2}{8k - 12} \times \dfrac{5}{5} = -\dfrac{10}{40k - 60} $ To get $40k - 60$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ -\dfrac{k - 9}{2k - 3} \times \dfrac{20}{20} = -\dfrac{20k - 180}{40k - 60} $ This give us: $ \dfrac{24}{40k - 60} = -\dfrac{10}{40k - 60} - \dfrac{20k - 180}{40k - 60} $ If we multiply both sides of the equation by $40k - 60$ , we get: $ 24 = -10 - 20k + 180$ $ 24 = -20k + 170$ $ -146 = -20k $ $ k = \dfrac{73}{10}$

307

Solve for $q$, $ \dfrac{10}{2q + 1} = -\dfrac{4}{2q + 1} - \dfrac{2q - 10}{2q + 1} $

If we multiply both sides of the equation by $2q + 1$ , we get: $ 10 = -4 - 2q + 10$ $ 10 = -2q + 6$ $ 4 = -2q $ $ q = -2$

307

Solve for $t$, $ -\dfrac{6}{25t + 15} = -\dfrac{4t + 10}{5t + 3} + \dfrac{5}{10t + 6} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25t + 15$ $5t + 3$ and $10t + 6$ The common denominator is $50t + 30$ To get $50t + 30$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{6}{25t + 15} \times \dfrac{2}{2} = -\dfrac{12}{50t + 30} $ To get $50t + 30$ in the denominator of the second term, multiply it by $\frac{10}{10}$ $ -\dfrac{4t + 10}{5t + 3} \times \dfrac{10}{10} = -\dfrac{40t + 100}{50t + 30} $ To get $50t + 30$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{5}{10t + 6} \times \dfrac{5}{5} = \dfrac{25}{50t + 30} $ This give us: $ -\dfrac{12}{50t + 30} = -\dfrac{40t + 100}{50t + 30} + \dfrac{25}{50t + 30} $ If we multiply both sides of the equation by $50t + 30$ , we get: $ -12 = -40t - 100 + 25$ $ -12 = -40t - 75$ $ 63 = -40t $ $ t = -\dfrac{63}{40}$

307

Solve for $n$, $ \dfrac{1}{4n - 4} = \dfrac{3}{4n - 4} - \dfrac{2n - 5}{4n - 4} $

If we multiply both sides of the equation by $4n - 4$ , we get: $ 1 = 3 - 2n + 5$ $ 1 = -2n + 8$ $ -7 = -2n $ $ n = \dfrac{7}{2}$

307

Solve for $n$, $ \dfrac{3n + 1}{2n} = \dfrac{10}{10n} + \dfrac{6}{10n} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2n$ $10n$ and $10n$ The common denominator is $10n$ To get $10n$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{3n + 1}{2n} \times \dfrac{5}{5} = \dfrac{15n + 5}{10n} $ The denominator of the second term is already $10n$ , so we don't need to change it. The denominator of the third term is already $10n$ , so we don't need to change it. This give us: $ \dfrac{15n + 5}{10n} = \dfrac{10}{10n} + \dfrac{6}{10n} $ If we multiply both sides of the equation by $10n$ , we get: $ 15n + 5 = 10 + 6$ $ 15n + 5 = 16$ $ 15n = 11 $ $ n = \dfrac{11}{15}$

307

Solve for $n$, $ \dfrac{6}{2n} = -\dfrac{5n + 1}{2n} - \dfrac{10}{2n} $

If we multiply both sides of the equation by $2n$ , we get: $ 6 = -5n - 1 - 10$ $ 6 = -5n - 11$ $ 17 = -5n $ $ n = -\dfrac{17}{5}$

307

Solve for $q$, $ -\dfrac{1}{3q - 3} = -\dfrac{10}{15q - 15} - \dfrac{q - 5}{3q - 3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3q - 3$ $15q - 15$ and $3q - 3$ The common denominator is $15q - 15$ To get $15q - 15$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{1}{3q - 3} \times \dfrac{5}{5} = -\dfrac{5}{15q - 15} $ The denominator of the second term is already $15q - 15$ , so we don't need to change it. To get $15q - 15$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{q - 5}{3q - 3} \times \dfrac{5}{5} = -\dfrac{5q - 25}{15q - 15} $ This give us: $ -\dfrac{5}{15q - 15} = -\dfrac{10}{15q - 15} - \dfrac{5q - 25}{15q - 15} $ If we multiply both sides of the equation by $15q - 15$ , we get: $ -5 = -10 - 5q + 25$ $ -5 = -5q + 15$ $ -20 = -5q $ $ q = 4$

307

Solve for $n$, $ \dfrac{3n}{9n - 6} = \dfrac{7}{9n - 6} - \dfrac{1}{3n - 2} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $9n - 6$ $9n - 6$ and $3n - 2$ The common denominator is $9n - 6$ The denominator of the first term is already $9n - 6$ , so we don't need to change it. The denominator of the second term is already $9n - 6$ , so we don't need to change it. To get $9n - 6$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{1}{3n - 2} \times \dfrac{3}{3} = -\dfrac{3}{9n - 6} $ This give us: $ \dfrac{3n}{9n - 6} = \dfrac{7}{9n - 6} - \dfrac{3}{9n - 6} $ If we multiply both sides of the equation by $9n - 6$ , we get: $ 3n = 7 - 3$ $ 3n = 4$ $ 3n = 4 $ $ n = \dfrac{4}{3}$

307

Solve for $q$, $ \dfrac{9}{4q^3} = \dfrac{2q + 3}{q^3} - \dfrac{7}{q^3} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4q^3$ $q^3$ and $q^3$ The common denominator is $4q^3$ The denominator of the first term is already $4q^3$ , so we don't need to change it. To get $4q^3$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{2q + 3}{q^3} \times \dfrac{4}{4} = \dfrac{8q + 12}{4q^3} $ To get $4q^3$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{7}{q^3} \times \dfrac{4}{4} = -\dfrac{28}{4q^3} $ This give us: $ \dfrac{9}{4q^3} = \dfrac{8q + 12}{4q^3} - \dfrac{28}{4q^3} $ If we multiply both sides of the equation by $4q^3$ , we get: $ 9 = 8q + 12 - 28$ $ 9 = 8q - 16$ $ 25 = 8q $ $ q = \dfrac{25}{8}$

307

Solve for $z$, $ \dfrac{5z - 5}{8z + 12} = \dfrac{10}{2z + 3} + \dfrac{8}{6z + 9} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8z + 12$ $2z + 3$ and $6z + 9$ The common denominator is $24z + 36$ To get $24z + 36$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{5z - 5}{8z + 12} \times \dfrac{3}{3} = \dfrac{15z - 15}{24z + 36} $ To get $24z + 36$ in the denominator of the second term, multiply it by $\frac{12}{12}$ $ \dfrac{10}{2z + 3} \times \dfrac{12}{12} = \dfrac{120}{24z + 36} $ To get $24z + 36$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{8}{6z + 9} \times \dfrac{4}{4} = \dfrac{32}{24z + 36} $ This give us: $ \dfrac{15z - 15}{24z + 36} = \dfrac{120}{24z + 36} + \dfrac{32}{24z + 36} $ If we multiply both sides of the equation by $24z + 36$ , we get: $ 15z - 15 = 120 + 32$ $ 15z - 15 = 152$ $ 15z = 167 $ $ z = \dfrac{167}{15}$

307

Solve for $n$, $ -\dfrac{6}{3n} = -\dfrac{9}{12n} - \dfrac{4n - 3}{9n} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3n$ $12n$ and $9n$ The common denominator is $36n$ To get $36n$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ -\dfrac{6}{3n} \times \dfrac{12}{12} = -\dfrac{72}{36n} $ To get $36n$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{9}{12n} \times \dfrac{3}{3} = -\dfrac{27}{36n} $ To get $36n$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{4n - 3}{9n} \times \dfrac{4}{4} = -\dfrac{16n - 12}{36n} $ This give us: $ -\dfrac{72}{36n} = -\dfrac{27}{36n} - \dfrac{16n - 12}{36n} $ If we multiply both sides of the equation by $36n$ , we get: $ -72 = -27 - 16n + 12$ $ -72 = -16n - 15$ $ -57 = -16n $ $ n = \dfrac{57}{16}$

307

Solve for $x$, $ \dfrac{1}{2x - 2} = \dfrac{2}{x - 1} + \dfrac{2x + 10}{x - 1} $

First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2x - 2$ $x - 1$ and $x - 1$ The common denominator is $2x - 2$ The denominator of the first term is already $2x - 2$ , so we don't need to change it. To get $2x - 2$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{2}{x - 1} \times \dfrac{2}{2} = \dfrac{4}{2x - 2} $ To get $2x - 2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{2x + 10}{x - 1} \times \dfrac{2}{2} = \dfrac{4x + 20}{2x - 2} $ This give us: $ \dfrac{1}{2x - 2} = \dfrac{4}{2x - 2} + \dfrac{4x + 20}{2x - 2} $ If we multiply both sides of the equation by $2x - 2$ , we get: $ 1 = 4 + 4x + 20$ $ 1 = 4x + 24$ $ -23 = 4x $ $ x = -\dfrac{23}{4}$