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307 | Solve for $k$, $ \dfrac{8}{2k} = -\dfrac{10}{4k} + \dfrac{5k - 10}{k} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2k$ $4k$ and $k$ The common denominator is $4k$ To get $4k$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{8}{2k} \times \dfrac{2}{2} = \dfrac{16}{4k} $ The denominator of the second term is already $4k$ , so we don't need to change it. To get $4k$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{5k - 10}{k} \times \dfrac{4}{4} = \dfrac{20k - 40}{4k} $ This give us: $ \dfrac{16}{4k} = -\dfrac{10}{4k} + \dfrac{20k - 40}{4k} $ If we multiply both sides of the equation by $4k$ , we get: $ 16 = -10 + 20k - 40$ $ 16 = 20k - 50$ $ 66 = 20k $ $ k = \dfrac{33}{10}$ |
307 | Solve for $x$, $ -\dfrac{5}{8x + 16} = \dfrac{7}{10x + 20} + \dfrac{x + 6}{2x + 4} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8x + 16$ $10x + 20$ and $2x + 4$ The common denominator is $40x + 80$ To get $40x + 80$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{8x + 16} \times \dfrac{5}{5} = -\dfrac{25}{40x + 80} $ To get $40x + 80$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{7}{10x + 20} \times \dfrac{4}{4} = \dfrac{28}{40x + 80} $ To get $40x + 80$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ \dfrac{x + 6}{2x + 4} \times \dfrac{20}{20} = \dfrac{20x + 120}{40x + 80} $ This give us: $ -\dfrac{25}{40x + 80} = \dfrac{28}{40x + 80} + \dfrac{20x + 120}{40x + 80} $ If we multiply both sides of the equation by $40x + 80$ , we get: $ -25 = 28 + 20x + 120$ $ -25 = 20x + 148$ $ -173 = 20x $ $ x = -\dfrac{173}{20}$ |
307 | Solve for $t$, $ -\dfrac{5t + 2}{2t + 5} = -\dfrac{8}{2t + 5} + \dfrac{1}{6t + 15} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2t + 5$ $2t + 5$ and $6t + 15$ The common denominator is $6t + 15$ To get $6t + 15$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{5t + 2}{2t + 5} \times \dfrac{3}{3} = -\dfrac{15t + 6}{6t + 15} $ To get $6t + 15$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{8}{2t + 5} \times \dfrac{3}{3} = -\dfrac{24}{6t + 15} $ The denominator of the third term is already $6t + 15$ , so we don't need to change it. This give us: $ -\dfrac{15t + 6}{6t + 15} = -\dfrac{24}{6t + 15} + \dfrac{1}{6t + 15} $ If we multiply both sides of the equation by $6t + 15$ , we get: $ -15t - 6 = -24 + 1$ $ -15t - 6 = -23$ $ -15t = -17 $ $ t = \dfrac{17}{15}$ |
307 | Solve for $p$, $ \dfrac{p - 1}{16p} = -\dfrac{6}{8p} - \dfrac{7}{8p} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16p$ $8p$ and $8p$ The common denominator is $16p$ The denominator of the first term is already $16p$ , so we don't need to change it. To get $16p$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{6}{8p} \times \dfrac{2}{2} = -\dfrac{12}{16p} $ To get $16p$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{7}{8p} \times \dfrac{2}{2} = -\dfrac{14}{16p} $ This give us: $ \dfrac{p - 1}{16p} = -\dfrac{12}{16p} - \dfrac{14}{16p} $ If we multiply both sides of the equation by $16p$ , we get: $ p - 1 = -12 - 14$ $ p - 1 = -26$ $ p = -25 $ |
307 | Solve for $z$, $ \dfrac{3}{4z^2} = -\dfrac{4}{2z^2} - \dfrac{4z - 6}{2z^2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4z^2$ $2z^2$ and $2z^2$ The common denominator is $4z^2$ The denominator of the first term is already $4z^2$ , so we don't need to change it. To get $4z^2$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{4}{2z^2} \times \dfrac{2}{2} = -\dfrac{8}{4z^2} $ To get $4z^2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{4z - 6}{2z^2} \times \dfrac{2}{2} = -\dfrac{8z - 12}{4z^2} $ This give us: $ \dfrac{3}{4z^2} = -\dfrac{8}{4z^2} - \dfrac{8z - 12}{4z^2} $ If we multiply both sides of the equation by $4z^2$ , we get: $ 3 = -8 - 8z + 12$ $ 3 = -8z + 4$ $ -1 = -8z $ $ z = \dfrac{1}{8}$ |
307 | Solve for $z$, $ \dfrac{10}{8z^2} = \dfrac{z - 3}{8z^2} + \dfrac{10}{16z^2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8z^2$ $8z^2$ and $16z^2$ The common denominator is $16z^2$ To get $16z^2$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{10}{8z^2} \times \dfrac{2}{2} = \dfrac{20}{16z^2} $ To get $16z^2$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{z - 3}{8z^2} \times \dfrac{2}{2} = \dfrac{2z - 6}{16z^2} $ The denominator of the third term is already $16z^2$ , so we don't need to change it. This give us: $ \dfrac{20}{16z^2} = \dfrac{2z - 6}{16z^2} + \dfrac{10}{16z^2} $ If we multiply both sides of the equation by $16z^2$ , we get: $ 20 = 2z - 6 + 10$ $ 20 = 2z + 4$ $ 16 = 2z $ $ z = 8$ |
307 | Solve for $t$, $ -\dfrac{9}{3t} = \dfrac{5}{3t} + \dfrac{3t + 8}{15t} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3t$ $3t$ and $15t$ The common denominator is $15t$ To get $15t$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{9}{3t} \times \dfrac{5}{5} = -\dfrac{45}{15t} $ To get $15t$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{5}{3t} \times \dfrac{5}{5} = \dfrac{25}{15t} $ The denominator of the third term is already $15t$ , so we don't need to change it. This give us: $ -\dfrac{45}{15t} = \dfrac{25}{15t} + \dfrac{3t + 8}{15t} $ If we multiply both sides of the equation by $15t$ , we get: $ -45 = 25 + 3t + 8$ $ -45 = 3t + 33$ $ -78 = 3t $ $ t = -26$ |
307 | Solve for $n$, $ \dfrac{n + 7}{5n - 1} = \dfrac{3}{5n - 1} + \dfrac{8}{25n - 5} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5n - 1$ $5n - 1$ and $25n - 5$ The common denominator is $25n - 5$ To get $25n - 5$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{n + 7}{5n - 1} \times \dfrac{5}{5} = \dfrac{5n + 35}{25n - 5} $ To get $25n - 5$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{3}{5n - 1} \times \dfrac{5}{5} = \dfrac{15}{25n - 5} $ The denominator of the third term is already $25n - 5$ , so we don't need to change it. This give us: $ \dfrac{5n + 35}{25n - 5} = \dfrac{15}{25n - 5} + \dfrac{8}{25n - 5} $ If we multiply both sides of the equation by $25n - 5$ , we get: $ 5n + 35 = 15 + 8$ $ 5n + 35 = 23$ $ 5n = -12 $ $ n = -\dfrac{12}{5}$ |
307 | Solve for $z$, $ \dfrac{4}{4z + 20} = -\dfrac{z - 10}{z + 5} - \dfrac{8}{2z + 10} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4z + 20$ $z + 5$ and $2z + 10$ The common denominator is $4z + 20$ The denominator of the first term is already $4z + 20$ , so we don't need to change it. To get $4z + 20$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{z - 10}{z + 5} \times \dfrac{4}{4} = -\dfrac{4z - 40}{4z + 20} $ To get $4z + 20$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{8}{2z + 10} \times \dfrac{2}{2} = -\dfrac{16}{4z + 20} $ This give us: $ \dfrac{4}{4z + 20} = -\dfrac{4z - 40}{4z + 20} - \dfrac{16}{4z + 20} $ If we multiply both sides of the equation by $4z + 20$ , we get: $ 4 = -4z + 40 - 16$ $ 4 = -4z + 24$ $ -20 = -4z $ $ z = 5$ |
307 | Solve for $a$, $ \dfrac{10}{4a^3} = -\dfrac{5a + 9}{4a^3} - \dfrac{5}{4a^3} $ | If we multiply both sides of the equation by $4a^3$ , we get: $ 10 = -5a - 9 - 5$ $ 10 = -5a - 14$ $ 24 = -5a $ $ a = -\dfrac{24}{5}$ |
307 | Solve for $k$, $ -\dfrac{6}{5k + 5} = \dfrac{5k - 7}{10k + 10} + \dfrac{6}{25k + 25} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5k + 5$ $10k + 10$ and $25k + 25$ The common denominator is $50k + 50$ To get $50k + 50$ in the denominator of the first term, multiply it by $\frac{10}{10}$ $ -\dfrac{6}{5k + 5} \times \dfrac{10}{10} = -\dfrac{60}{50k + 50} $ To get $50k + 50$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{5k - 7}{10k + 10} \times \dfrac{5}{5} = \dfrac{25k - 35}{50k + 50} $ To get $50k + 50$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{6}{25k + 25} \times \dfrac{2}{2} = \dfrac{12}{50k + 50} $ This give us: $ -\dfrac{60}{50k + 50} = \dfrac{25k - 35}{50k + 50} + \dfrac{12}{50k + 50} $ If we multiply both sides of the equation by $50k + 50$ , we get: $ -60 = 25k - 35 + 12$ $ -60 = 25k - 23$ $ -37 = 25k $ $ k = -\dfrac{37}{25}$ |
307 | Solve for $p$, $ -\dfrac{3}{2p} = \dfrac{6}{p} - \dfrac{4p + 5}{p} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2p$ $p$ and $p$ The common denominator is $2p$ The denominator of the first term is already $2p$ , so we don't need to change it. To get $2p$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{6}{p} \times \dfrac{2}{2} = \dfrac{12}{2p} $ To get $2p$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{4p + 5}{p} \times \dfrac{2}{2} = -\dfrac{8p + 10}{2p} $ This give us: $ -\dfrac{3}{2p} = \dfrac{12}{2p} - \dfrac{8p + 10}{2p} $ If we multiply both sides of the equation by $2p$ , we get: $ -3 = 12 - 8p - 10$ $ -3 = -8p + 2$ $ -5 = -8p $ $ p = \dfrac{5}{8}$ |
307 | Solve for $t$, $ -\dfrac{4t - 6}{3t^3} = \dfrac{1}{12t^3} + \dfrac{5}{3t^3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3t^3$ $12t^3$ and $3t^3$ The common denominator is $12t^3$ To get $12t^3$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{4t - 6}{3t^3} \times \dfrac{4}{4} = -\dfrac{16t - 24}{12t^3} $ The denominator of the second term is already $12t^3$ , so we don't need to change it. To get $12t^3$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{5}{3t^3} \times \dfrac{4}{4} = \dfrac{20}{12t^3} $ This give us: $ -\dfrac{16t - 24}{12t^3} = \dfrac{1}{12t^3} + \dfrac{20}{12t^3} $ If we multiply both sides of the equation by $12t^3$ , we get: $ -16t + 24 = 1 + 20$ $ -16t + 24 = 21$ $ -16t = -3 $ $ t = \dfrac{3}{16}$ |
307 | Solve for $q$, $ -\dfrac{9}{5q + 5} = -\dfrac{q - 8}{5q + 5} - \dfrac{7}{q + 1} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5q + 5$ $5q + 5$ and $q + 1$ The common denominator is $5q + 5$ The denominator of the first term is already $5q + 5$ , so we don't need to change it. The denominator of the second term is already $5q + 5$ , so we don't need to change it. To get $5q + 5$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{7}{q + 1} \times \dfrac{5}{5} = -\dfrac{35}{5q + 5} $ This give us: $ -\dfrac{9}{5q + 5} = -\dfrac{q - 8}{5q + 5} - \dfrac{35}{5q + 5} $ If we multiply both sides of the equation by $5q + 5$ , we get: $ -9 = -q + 8 - 35$ $ -9 = -q - 27$ $ 18 = -q $ $ q = -18$ |
307 | Solve for $p$, $ -\dfrac{6}{3p} = \dfrac{p + 4}{3p} + \dfrac{1}{3p} $ | If we multiply both sides of the equation by $3p$ , we get: $ -6 = p + 4 + 1$ $ -6 = p + 5$ $ -11 = p $ $ p = -11$ |
307 | Solve for $y$, $ -\dfrac{8}{y + 5} = -\dfrac{1}{y + 5} - \dfrac{4y + 9}{3y + 15} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $y + 5$ $y + 5$ and $3y + 15$ The common denominator is $3y + 15$ To get $3y + 15$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{8}{y + 5} \times \dfrac{3}{3} = -\dfrac{24}{3y + 15} $ To get $3y + 15$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{1}{y + 5} \times \dfrac{3}{3} = -\dfrac{3}{3y + 15} $ The denominator of the third term is already $3y + 15$ , so we don't need to change it. This give us: $ -\dfrac{24}{3y + 15} = -\dfrac{3}{3y + 15} - \dfrac{4y + 9}{3y + 15} $ If we multiply both sides of the equation by $3y + 15$ , we get: $ -24 = -3 - 4y - 9$ $ -24 = -4y - 12$ $ -12 = -4y $ $ y = 3$ |
307 | Solve for $r$, $ -\dfrac{r - 1}{6r} = \dfrac{8}{6r} + \dfrac{1}{6r} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $6r$ $6r$ and $6r$ The common denominator is $6r$ The denominator of the first term is already $6r$ , so we don't need to change it. The denominator of the second term is already $6r$ , so we don't need to change it. The denominator of the third term is already $6r$ , so we don't need to change it. This give us: $ -\dfrac{r - 1}{6r} = \dfrac{8}{6r} + \dfrac{1}{6r} $ If we multiply both sides of the equation by $6r$ , we get: $ -r + 1 = 8 + 1$ $ -r + 1 = 9$ $ -r = 8 $ $ r = -8$ |
307 | Solve for $t$, $ \dfrac{5}{2t} = -\dfrac{8}{2t} - \dfrac{3t - 6}{2t} $ | If we multiply both sides of the equation by $2t$ , we get: $ 5 = -8 - 3t + 6$ $ 5 = -3t - 2$ $ 7 = -3t $ $ t = -\dfrac{7}{3}$ |
307 | Solve for $n$, $ -\dfrac{n - 2}{n + 3} = \dfrac{1}{3n + 9} - \dfrac{3}{n + 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $n + 3$ $3n + 9$ and $n + 3$ The common denominator is $3n + 9$ To get $3n + 9$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{n - 2}{n + 3} \times \dfrac{3}{3} = -\dfrac{3n - 6}{3n + 9} $ The denominator of the second term is already $3n + 9$ , so we don't need to change it. To get $3n + 9$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{3}{n + 3} \times \dfrac{3}{3} = -\dfrac{9}{3n + 9} $ This give us: $ -\dfrac{3n - 6}{3n + 9} = \dfrac{1}{3n + 9} - \dfrac{9}{3n + 9} $ If we multiply both sides of the equation by $3n + 9$ , we get: $ -3n + 6 = 1 - 9$ $ -3n + 6 = -8$ $ -3n = -14 $ $ n = \dfrac{14}{3}$ |
307 | Solve for $t$, $ \dfrac{10}{9t + 12} = -\dfrac{1}{6t + 8} + \dfrac{2t + 1}{15t + 20} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $9t + 12$ $6t + 8$ and $15t + 20$ The common denominator is $90t + 120$ To get $90t + 120$ in the denominator of the first term, multiply it by $\frac{10}{10}$ $ \dfrac{10}{9t + 12} \times \dfrac{10}{10} = \dfrac{100}{90t + 120} $ To get $90t + 120$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ -\dfrac{1}{6t + 8} \times \dfrac{15}{15} = -\dfrac{15}{90t + 120} $ To get $90t + 120$ in the denominator of the third term, multiply it by $\frac{6}{6}$ $ \dfrac{2t + 1}{15t + 20} \times \dfrac{6}{6} = \dfrac{12t + 6}{90t + 120} $ This give us: $ \dfrac{100}{90t + 120} = -\dfrac{15}{90t + 120} + \dfrac{12t + 6}{90t + 120} $ If we multiply both sides of the equation by $90t + 120$ , we get: $ 100 = -15 + 12t + 6$ $ 100 = 12t - 9$ $ 109 = 12t $ $ t = \dfrac{109}{12}$ |
307 | Solve for $r$, $ \dfrac{8}{5r + 1} = -\dfrac{4}{20r + 4} + \dfrac{r}{15r + 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5r + 1$ $20r + 4$ and $15r + 3$ The common denominator is $60r + 12$ To get $60r + 12$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ \dfrac{8}{5r + 1} \times \dfrac{12}{12} = \dfrac{96}{60r + 12} $ To get $60r + 12$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{4}{20r + 4} \times \dfrac{3}{3} = -\dfrac{12}{60r + 12} $ To get $60r + 12$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{r}{15r + 3} \times \dfrac{4}{4} = \dfrac{4r}{60r + 12} $ This give us: $ \dfrac{96}{60r + 12} = -\dfrac{12}{60r + 12} + \dfrac{4r}{60r + 12} $ If we multiply both sides of the equation by $60r + 12$ , we get: $ 96 = -12 + 4r$ $ 96 = 4r - 12$ $ 108 = 4r $ $ r = 27$ |
307 | Solve for $t$, $ -\dfrac{5}{6t} = \dfrac{t + 5}{2t} + \dfrac{1}{4t} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $6t$ $2t$ and $4t$ The common denominator is $12t$ To get $12t$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{5}{6t} \times \dfrac{2}{2} = -\dfrac{10}{12t} $ To get $12t$ in the denominator of the second term, multiply it by $\frac{6}{6}$ $ \dfrac{t + 5}{2t} \times \dfrac{6}{6} = \dfrac{6t + 30}{12t} $ To get $12t$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{1}{4t} \times \dfrac{3}{3} = \dfrac{3}{12t} $ This give us: $ -\dfrac{10}{12t} = \dfrac{6t + 30}{12t} + \dfrac{3}{12t} $ If we multiply both sides of the equation by $12t$ , we get: $ -10 = 6t + 30 + 3$ $ -10 = 6t + 33$ $ -43 = 6t $ $ t = -\dfrac{43}{6}$ |
307 | Solve for $a$, $ -\dfrac{5a + 5}{20a - 12} = -\dfrac{1}{20a - 12} - \dfrac{6}{10a - 6} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $20a - 12$ $20a - 12$ and $10a - 6$ The common denominator is $20a - 12$ The denominator of the first term is already $20a - 12$ , so we don't need to change it. The denominator of the second term is already $20a - 12$ , so we don't need to change it. To get $20a - 12$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{6}{10a - 6} \times \dfrac{2}{2} = -\dfrac{12}{20a - 12} $ This give us: $ -\dfrac{5a + 5}{20a - 12} = -\dfrac{1}{20a - 12} - \dfrac{12}{20a - 12} $ If we multiply both sides of the equation by $20a - 12$ , we get: $ -5a - 5 = -1 - 12$ $ -5a - 5 = -13$ $ -5a = -8 $ $ a = \dfrac{8}{5}$ |
307 | Solve for $p$, $ -\dfrac{10}{3p - 5} = \dfrac{5p + 1}{9p - 15} - \dfrac{7}{3p - 5} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3p - 5$ $9p - 15$ and $3p - 5$ The common denominator is $9p - 15$ To get $9p - 15$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{10}{3p - 5} \times \dfrac{3}{3} = -\dfrac{30}{9p - 15} $ The denominator of the second term is already $9p - 15$ , so we don't need to change it. To get $9p - 15$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{7}{3p - 5} \times \dfrac{3}{3} = -\dfrac{21}{9p - 15} $ This give us: $ -\dfrac{30}{9p - 15} = \dfrac{5p + 1}{9p - 15} - \dfrac{21}{9p - 15} $ If we multiply both sides of the equation by $9p - 15$ , we get: $ -30 = 5p + 1 - 21$ $ -30 = 5p - 20$ $ -10 = 5p $ $ p = -2$ |
307 | Solve for $x$, $ -\dfrac{7}{2x - 2} = -\dfrac{x}{3x - 3} + \dfrac{3}{4x - 4} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2x - 2$ $3x - 3$ and $4x - 4$ The common denominator is $12x - 12$ To get $12x - 12$ in the denominator of the first term, multiply it by $\frac{6}{6}$ $ -\dfrac{7}{2x - 2} \times \dfrac{6}{6} = -\dfrac{42}{12x - 12} $ To get $12x - 12$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{x}{3x - 3} \times \dfrac{4}{4} = -\dfrac{4x}{12x - 12} $ To get $12x - 12$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{3}{4x - 4} \times \dfrac{3}{3} = \dfrac{9}{12x - 12} $ This give us: $ -\dfrac{42}{12x - 12} = -\dfrac{4x}{12x - 12} + \dfrac{9}{12x - 12} $ If we multiply both sides of the equation by $12x - 12$ , we get: $ -42 = -4x + 9$ $ -42 = -4x + 9$ $ -51 = -4x $ $ x = \dfrac{51}{4}$ |
307 | Solve for $p$, $ -\dfrac{10}{10p^3} = \dfrac{5p - 6}{10p^3} - \dfrac{4}{2p^3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $10p^3$ $10p^3$ and $2p^3$ The common denominator is $10p^3$ The denominator of the first term is already $10p^3$ , so we don't need to change it. The denominator of the second term is already $10p^3$ , so we don't need to change it. To get $10p^3$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{4}{2p^3} \times \dfrac{5}{5} = -\dfrac{20}{10p^3} $ This give us: $ -\dfrac{10}{10p^3} = \dfrac{5p - 6}{10p^3} - \dfrac{20}{10p^3} $ If we multiply both sides of the equation by $10p^3$ , we get: $ -10 = 5p - 6 - 20$ $ -10 = 5p - 26$ $ 16 = 5p $ $ p = \dfrac{16}{5}$ |
307 | Solve for $r$, $ \dfrac{6}{8r} = \dfrac{5r + 8}{20r} - \dfrac{5}{4r} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8r$ $20r$ and $4r$ The common denominator is $40r$ To get $40r$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{6}{8r} \times \dfrac{5}{5} = \dfrac{30}{40r} $ To get $40r$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{5r + 8}{20r} \times \dfrac{2}{2} = \dfrac{10r + 16}{40r} $ To get $40r$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ -\dfrac{5}{4r} \times \dfrac{10}{10} = -\dfrac{50}{40r} $ This give us: $ \dfrac{30}{40r} = \dfrac{10r + 16}{40r} - \dfrac{50}{40r} $ If we multiply both sides of the equation by $40r$ , we get: $ 30 = 10r + 16 - 50$ $ 30 = 10r - 34$ $ 64 = 10r $ $ r = \dfrac{32}{5}$ |
307 | Solve for $r$, $ \dfrac{r - 5}{2r - 2} = -\dfrac{5}{4r - 4} + \dfrac{4}{5r - 5} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2r - 2$ $4r - 4$ and $5r - 5$ The common denominator is $20r - 20$ To get $20r - 20$ in the denominator of the first term, multiply it by $\frac{10}{10}$ $ \dfrac{r - 5}{2r - 2} \times \dfrac{10}{10} = \dfrac{10r - 50}{20r - 20} $ To get $20r - 20$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{4r - 4} \times \dfrac{5}{5} = -\dfrac{25}{20r - 20} $ To get $20r - 20$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{4}{5r - 5} \times \dfrac{4}{4} = \dfrac{16}{20r - 20} $ This give us: $ \dfrac{10r - 50}{20r - 20} = -\dfrac{25}{20r - 20} + \dfrac{16}{20r - 20} $ If we multiply both sides of the equation by $20r - 20$ , we get: $ 10r - 50 = -25 + 16$ $ 10r - 50 = -9$ $ 10r = 41 $ $ r = \dfrac{41}{10}$ |
307 | Solve for $t$, $ \dfrac{10}{3t + 12} = -\dfrac{5t - 5}{t + 4} - \dfrac{4}{5t + 20} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3t + 12$ $t + 4$ and $5t + 20$ The common denominator is $15t + 60$ To get $15t + 60$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{10}{3t + 12} \times \dfrac{5}{5} = \dfrac{50}{15t + 60} $ To get $15t + 60$ in the denominator of the second term, multiply it by $\frac{15}{15}$ $ -\dfrac{5t - 5}{t + 4} \times \dfrac{15}{15} = -\dfrac{75t - 75}{15t + 60} $ To get $15t + 60$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{4}{5t + 20} \times \dfrac{3}{3} = -\dfrac{12}{15t + 60} $ This give us: $ \dfrac{50}{15t + 60} = -\dfrac{75t - 75}{15t + 60} - \dfrac{12}{15t + 60} $ If we multiply both sides of the equation by $15t + 60$ , we get: $ 50 = -75t + 75 - 12$ $ 50 = -75t + 63$ $ -13 = -75t $ $ t = \dfrac{13}{75}$ |
307 | Solve for $k$, $ -\dfrac{5k + 7}{4k - 12} = \dfrac{10}{k - 3} + \dfrac{9}{k - 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4k - 12$ $k - 3$ and $k - 3$ The common denominator is $4k - 12$ The denominator of the first term is already $4k - 12$ , so we don't need to change it. To get $4k - 12$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{10}{k - 3} \times \dfrac{4}{4} = \dfrac{40}{4k - 12} $ To get $4k - 12$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{9}{k - 3} \times \dfrac{4}{4} = \dfrac{36}{4k - 12} $ This give us: $ -\dfrac{5k + 7}{4k - 12} = \dfrac{40}{4k - 12} + \dfrac{36}{4k - 12} $ If we multiply both sides of the equation by $4k - 12$ , we get: $ -5k - 7 = 40 + 36$ $ -5k - 7 = 76$ $ -5k = 83 $ $ k = -\dfrac{83}{5}$ |
307 | Solve for $y$, $ -\dfrac{3y - 3}{y - 4} = -\dfrac{5}{3y - 12} + \dfrac{1}{5y - 20} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $y - 4$ $3y - 12$ and $5y - 20$ The common denominator is $15y - 60$ To get $15y - 60$ in the denominator of the first term, multiply it by $\frac{15}{15}$ $ -\dfrac{3y - 3}{y - 4} \times \dfrac{15}{15} = -\dfrac{45y - 45}{15y - 60} $ To get $15y - 60$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{3y - 12} \times \dfrac{5}{5} = -\dfrac{25}{15y - 60} $ To get $15y - 60$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{1}{5y - 20} \times \dfrac{3}{3} = \dfrac{3}{15y - 60} $ This give us: $ -\dfrac{45y - 45}{15y - 60} = -\dfrac{25}{15y - 60} + \dfrac{3}{15y - 60} $ If we multiply both sides of the equation by $15y - 60$ , we get: $ -45y + 45 = -25 + 3$ $ -45y + 45 = -22$ $ -45y = -67 $ $ y = \dfrac{67}{45}$ |
307 | Solve for $a$, $ \dfrac{8}{a + 5} = -\dfrac{2}{a + 5} + \dfrac{5a - 4}{a + 5} $ | If we multiply both sides of the equation by $a + 5$ , we get: $ 8 = -2 + 5a - 4$ $ 8 = 5a - 6$ $ 14 = 5a $ $ a = \dfrac{14}{5}$ |
307 | Solve for $k$, $ -\dfrac{4}{k + 2} = -\dfrac{5}{k + 2} + \dfrac{k + 1}{k + 2} $ | If we multiply both sides of the equation by $k + 2$ , we get: $ -4 = -5 + k + 1$ $ -4 = k - 4$ $ 0 = k $ $ k = 0$ |
307 | Solve for $z$, $ \dfrac{z - 7}{z} = \dfrac{3}{z} - \dfrac{1}{3z} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $z$ $z$ and $3z$ The common denominator is $3z$ To get $3z$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{z - 7}{z} \times \dfrac{3}{3} = \dfrac{3z - 21}{3z} $ To get $3z$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ \dfrac{3}{z} \times \dfrac{3}{3} = \dfrac{9}{3z} $ The denominator of the third term is already $3z$ , so we don't need to change it. This give us: $ \dfrac{3z - 21}{3z} = \dfrac{9}{3z} - \dfrac{1}{3z} $ If we multiply both sides of the equation by $3z$ , we get: $ 3z - 21 = 9 - 1$ $ 3z - 21 = 8$ $ 3z = 29 $ $ z = \dfrac{29}{3}$ |
307 | Solve for $x$, $ \dfrac{x + 3}{5x + 10} = -\dfrac{6}{4x + 8} + \dfrac{10}{x + 2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5x + 10$ $4x + 8$ and $x + 2$ The common denominator is $20x + 40$ To get $20x + 40$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{x + 3}{5x + 10} \times \dfrac{4}{4} = \dfrac{4x + 12}{20x + 40} $ To get $20x + 40$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{6}{4x + 8} \times \dfrac{5}{5} = -\dfrac{30}{20x + 40} $ To get $20x + 40$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ \dfrac{10}{x + 2} \times \dfrac{20}{20} = \dfrac{200}{20x + 40} $ This give us: $ \dfrac{4x + 12}{20x + 40} = -\dfrac{30}{20x + 40} + \dfrac{200}{20x + 40} $ If we multiply both sides of the equation by $20x + 40$ , we get: $ 4x + 12 = -30 + 200$ $ 4x + 12 = 170$ $ 4x = 158 $ $ x = \dfrac{79}{2}$ |
307 | Solve for $p$, $ -\dfrac{6}{2p + 5} = -\dfrac{p + 2}{8p + 20} + \dfrac{1}{2p + 5} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2p + 5$ $8p + 20$ and $2p + 5$ The common denominator is $8p + 20$ To get $8p + 20$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{6}{2p + 5} \times \dfrac{4}{4} = -\dfrac{24}{8p + 20} $ The denominator of the second term is already $8p + 20$ , so we don't need to change it. To get $8p + 20$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{1}{2p + 5} \times \dfrac{4}{4} = \dfrac{4}{8p + 20} $ This give us: $ -\dfrac{24}{8p + 20} = -\dfrac{p + 2}{8p + 20} + \dfrac{4}{8p + 20} $ If we multiply both sides of the equation by $8p + 20$ , we get: $ -24 = -p - 2 + 4$ $ -24 = -p + 2$ $ -26 = -p $ $ p = 26$ |
307 | Solve for $y$, $ \dfrac{6}{5y + 10} = \dfrac{3y - 6}{3y + 6} - \dfrac{1}{y + 2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5y + 10$ $3y + 6$ and $y + 2$ The common denominator is $15y + 30$ To get $15y + 30$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{6}{5y + 10} \times \dfrac{3}{3} = \dfrac{18}{15y + 30} $ To get $15y + 30$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{3y - 6}{3y + 6} \times \dfrac{5}{5} = \dfrac{15y - 30}{15y + 30} $ To get $15y + 30$ in the denominator of the third term, multiply it by $\frac{15}{15}$ $ -\dfrac{1}{y + 2} \times \dfrac{15}{15} = -\dfrac{15}{15y + 30} $ This give us: $ \dfrac{18}{15y + 30} = \dfrac{15y - 30}{15y + 30} - \dfrac{15}{15y + 30} $ If we multiply both sides of the equation by $15y + 30$ , we get: $ 18 = 15y - 30 - 15$ $ 18 = 15y - 45$ $ 63 = 15y $ $ y = \dfrac{21}{5}$ |
307 | Solve for $p$, $ -\dfrac{10}{3p - 9} = \dfrac{5p - 1}{2p - 6} + \dfrac{3}{p - 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3p - 9$ $2p - 6$ and $p - 3$ The common denominator is $6p - 18$ To get $6p - 18$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{10}{3p - 9} \times \dfrac{2}{2} = -\dfrac{20}{6p - 18} $ To get $6p - 18$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ \dfrac{5p - 1}{2p - 6} \times \dfrac{3}{3} = \dfrac{15p - 3}{6p - 18} $ To get $6p - 18$ in the denominator of the third term, multiply it by $\frac{6}{6}$ $ \dfrac{3}{p - 3} \times \dfrac{6}{6} = \dfrac{18}{6p - 18} $ This give us: $ -\dfrac{20}{6p - 18} = \dfrac{15p - 3}{6p - 18} + \dfrac{18}{6p - 18} $ If we multiply both sides of the equation by $6p - 18$ , we get: $ -20 = 15p - 3 + 18$ $ -20 = 15p + 15$ $ -35 = 15p $ $ p = -\dfrac{7}{3}$ |
307 | Solve for $z$, $ -\dfrac{z + 4}{8z - 20} = \dfrac{10}{2z - 5} + \dfrac{4}{2z - 5} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8z - 20$ $2z - 5$ and $2z - 5$ The common denominator is $8z - 20$ The denominator of the first term is already $8z - 20$ , so we don't need to change it. To get $8z - 20$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{10}{2z - 5} \times \dfrac{4}{4} = \dfrac{40}{8z - 20} $ To get $8z - 20$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{4}{2z - 5} \times \dfrac{4}{4} = \dfrac{16}{8z - 20} $ This give us: $ -\dfrac{z + 4}{8z - 20} = \dfrac{40}{8z - 20} + \dfrac{16}{8z - 20} $ If we multiply both sides of the equation by $8z - 20$ , we get: $ -z - 4 = 40 + 16$ $ -z - 4 = 56$ $ -z = 60 $ $ z = -60$ |
307 | Solve for $r$, $ \dfrac{r + 7}{5r - 20} = \dfrac{5}{3r - 12} + \dfrac{6}{4r - 16} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5r - 20$ $3r - 12$ and $4r - 16$ The common denominator is $60r - 240$ To get $60r - 240$ in the denominator of the first term, multiply it by $\frac{12}{12}$ $ \dfrac{r + 7}{5r - 20} \times \dfrac{12}{12} = \dfrac{12r + 84}{60r - 240} $ To get $60r - 240$ in the denominator of the second term, multiply it by $\frac{20}{20}$ $ \dfrac{5}{3r - 12} \times \dfrac{20}{20} = \dfrac{100}{60r - 240} $ To get $60r - 240$ in the denominator of the third term, multiply it by $\frac{15}{15}$ $ \dfrac{6}{4r - 16} \times \dfrac{15}{15} = \dfrac{90}{60r - 240} $ This give us: $ \dfrac{12r + 84}{60r - 240} = \dfrac{100}{60r - 240} + \dfrac{90}{60r - 240} $ If we multiply both sides of the equation by $60r - 240$ , we get: $ 12r + 84 = 100 + 90$ $ 12r + 84 = 190$ $ 12r = 106 $ $ r = \dfrac{53}{6}$ |
307 | Solve for $n$, $ -\dfrac{3n - 1}{n^2} = -\dfrac{9}{4n^2} - \dfrac{1}{n^2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $n^2$ $4n^2$ and $n^2$ The common denominator is $4n^2$ To get $4n^2$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{3n - 1}{n^2} \times \dfrac{4}{4} = -\dfrac{12n - 4}{4n^2} $ The denominator of the second term is already $4n^2$ , so we don't need to change it. To get $4n^2$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{1}{n^2} \times \dfrac{4}{4} = -\dfrac{4}{4n^2} $ This give us: $ -\dfrac{12n - 4}{4n^2} = -\dfrac{9}{4n^2} - \dfrac{4}{4n^2} $ If we multiply both sides of the equation by $4n^2$ , we get: $ -12n + 4 = -9 - 4$ $ -12n + 4 = -13$ $ -12n = -17 $ $ n = \dfrac{17}{12}$ |
307 | Solve for $t$, $ -\dfrac{1}{5t^3} = \dfrac{t - 9}{5t^3} - \dfrac{5}{5t^3} $ | If we multiply both sides of the equation by $5t^3$ , we get: $ -1 = t - 9 - 5$ $ -1 = t - 14$ $ 13 = t $ $ t = 13$ |
307 | Solve for $n$, $ \dfrac{8}{4n + 4} = -\dfrac{3}{5n + 5} + \dfrac{n - 10}{n + 1} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4n + 4$ $5n + 5$ and $n + 1$ The common denominator is $20n + 20$ To get $20n + 20$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{8}{4n + 4} \times \dfrac{5}{5} = \dfrac{40}{20n + 20} $ To get $20n + 20$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{3}{5n + 5} \times \dfrac{4}{4} = -\dfrac{12}{20n + 20} $ To get $20n + 20$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ \dfrac{n - 10}{n + 1} \times \dfrac{20}{20} = \dfrac{20n - 200}{20n + 20} $ This give us: $ \dfrac{40}{20n + 20} = -\dfrac{12}{20n + 20} + \dfrac{20n - 200}{20n + 20} $ If we multiply both sides of the equation by $20n + 20$ , we get: $ 40 = -12 + 20n - 200$ $ 40 = 20n - 212$ $ 252 = 20n $ $ n = \dfrac{63}{5}$ |
307 | Solve for $x$, $ -\dfrac{6}{5x^2} = -\dfrac{3}{25x^2} - \dfrac{x + 10}{5x^2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5x^2$ $25x^2$ and $5x^2$ The common denominator is $25x^2$ To get $25x^2$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{6}{5x^2} \times \dfrac{5}{5} = -\dfrac{30}{25x^2} $ The denominator of the second term is already $25x^2$ , so we don't need to change it. To get $25x^2$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{x + 10}{5x^2} \times \dfrac{5}{5} = -\dfrac{5x + 50}{25x^2} $ This give us: $ -\dfrac{30}{25x^2} = -\dfrac{3}{25x^2} - \dfrac{5x + 50}{25x^2} $ If we multiply both sides of the equation by $25x^2$ , we get: $ -30 = -3 - 5x - 50$ $ -30 = -5x - 53$ $ 23 = -5x $ $ x = -\dfrac{23}{5}$ |
307 | Solve for $n$, $ -\dfrac{7}{n + 5} = \dfrac{3}{n + 5} + \dfrac{3n + 10}{n + 5} $ | If we multiply both sides of the equation by $n + 5$ , we get: $ -7 = 3 + 3n + 10$ $ -7 = 3n + 13$ $ -20 = 3n $ $ n = -\dfrac{20}{3}$ |
307 | Solve for $r$, $ \dfrac{2}{16r} = -\dfrac{8}{20r} - \dfrac{4r - 1}{4r} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $16r$ $20r$ and $4r$ The common denominator is $80r$ To get $80r$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{2}{16r} \times \dfrac{5}{5} = \dfrac{10}{80r} $ To get $80r$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{20r} \times \dfrac{4}{4} = -\dfrac{32}{80r} $ To get $80r$ in the denominator of the third term, multiply it by $\frac{20}{20}$ $ -\dfrac{4r - 1}{4r} \times \dfrac{20}{20} = -\dfrac{80r - 20}{80r} $ This give us: $ \dfrac{10}{80r} = -\dfrac{32}{80r} - \dfrac{80r - 20}{80r} $ If we multiply both sides of the equation by $80r$ , we get: $ 10 = -32 - 80r + 20$ $ 10 = -80r - 12$ $ 22 = -80r $ $ r = -\dfrac{11}{40}$ |
307 | Solve for $a$, $ -\dfrac{7}{a + 2} = \dfrac{8}{a + 2} - \dfrac{a - 10}{a + 2} $ | If we multiply both sides of the equation by $a + 2$ , we get: $ -7 = 8 - a + 10$ $ -7 = -a + 18$ $ -25 = -a $ $ a = 25$ |
307 | Solve for $z$, $ -\dfrac{4z + 3}{3z + 2} = \dfrac{3}{3z + 2} - \dfrac{4}{12z + 8} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3z + 2$ $3z + 2$ and $12z + 8$ The common denominator is $12z + 8$ To get $12z + 8$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{4z + 3}{3z + 2} \times \dfrac{4}{4} = -\dfrac{16z + 12}{12z + 8} $ To get $12z + 8$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{3}{3z + 2} \times \dfrac{4}{4} = \dfrac{12}{12z + 8} $ The denominator of the third term is already $12z + 8$ , so we don't need to change it. This give us: $ -\dfrac{16z + 12}{12z + 8} = \dfrac{12}{12z + 8} - \dfrac{4}{12z + 8} $ If we multiply both sides of the equation by $12z + 8$ , we get: $ -16z - 12 = 12 - 4$ $ -16z - 12 = 8$ $ -16z = 20 $ $ z = -\dfrac{5}{4}$ |
307 | Solve for $k$, $ \dfrac{6}{6k + 6} = \dfrac{k - 7}{3k + 3} - \dfrac{4}{3k + 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $6k + 6$ $3k + 3$ and $3k + 3$ The common denominator is $6k + 6$ The denominator of the first term is already $6k + 6$ , so we don't need to change it. To get $6k + 6$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{k - 7}{3k + 3} \times \dfrac{2}{2} = \dfrac{2k - 14}{6k + 6} $ To get $6k + 6$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{4}{3k + 3} \times \dfrac{2}{2} = -\dfrac{8}{6k + 6} $ This give us: $ \dfrac{6}{6k + 6} = \dfrac{2k - 14}{6k + 6} - \dfrac{8}{6k + 6} $ If we multiply both sides of the equation by $6k + 6$ , we get: $ 6 = 2k - 14 - 8$ $ 6 = 2k - 22$ $ 28 = 2k $ $ k = 14$ |
307 | Solve for $p$, $ -\dfrac{5p + 5}{2p} = -\dfrac{1}{2p} + \dfrac{7}{2p} $ | If we multiply both sides of the equation by $2p$ , we get: $ -5p - 5 = -1 + 7$ $ -5p - 5 = 6$ $ -5p = 11 $ $ p = -\dfrac{11}{5}$ |
307 | Solve for $y$, $ \dfrac{y - 9}{5y} = -\dfrac{2}{5y} + \dfrac{10}{5y} $ | If we multiply both sides of the equation by $5y$ , we get: $ y - 9 = -2 + 10$ $ y - 9 = 8$ $ y = 17 $ |
307 | Solve for $z$, $ \dfrac{1}{10z - 6} = \dfrac{5}{10z - 6} - \dfrac{3z + 7}{10z - 6} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $10z - 6$ $10z - 6$ and $10z - 6$ The common denominator is $10z - 6$ The denominator of the first term is already $10z - 6$ , so we don't need to change it. The denominator of the second term is already $10z - 6$ , so we don't need to change it. The denominator of the third term is already $10z - 6$ , so we don't need to change it. This give us: $ \dfrac{1}{10z - 6} = \dfrac{5}{10z - 6} - \dfrac{3z + 7}{10z - 6} $ If we multiply both sides of the equation by $10z - 6$ , we get: $ 1 = 5 - 3z - 7$ $ 1 = -3z - 2$ $ 3 = -3z $ $ z = -1$ |
307 | Solve for $r$, $ -\dfrac{1}{5r - 5} = \dfrac{6}{r - 1} + \dfrac{4r - 10}{2r - 2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5r - 5$ $r - 1$ and $2r - 2$ The common denominator is $10r - 10$ To get $10r - 10$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{1}{5r - 5} \times \dfrac{2}{2} = -\dfrac{2}{10r - 10} $ To get $10r - 10$ in the denominator of the second term, multiply it by $\frac{10}{10}$ $ \dfrac{6}{r - 1} \times \dfrac{10}{10} = \dfrac{60}{10r - 10} $ To get $10r - 10$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{4r - 10}{2r - 2} \times \dfrac{5}{5} = \dfrac{20r - 50}{10r - 10} $ This give us: $ -\dfrac{2}{10r - 10} = \dfrac{60}{10r - 10} + \dfrac{20r - 50}{10r - 10} $ If we multiply both sides of the equation by $10r - 10$ , we get: $ -2 = 60 + 20r - 50$ $ -2 = 20r + 10$ $ -12 = 20r $ $ r = -\dfrac{3}{5}$ |
307 | Solve for $n$, $ \dfrac{3n - 3}{n - 2} = -\dfrac{4}{4n - 8} + \dfrac{2}{n - 2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $n - 2$ $4n - 8$ and $n - 2$ The common denominator is $4n - 8$ To get $4n - 8$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{3n - 3}{n - 2} \times \dfrac{4}{4} = \dfrac{12n - 12}{4n - 8} $ The denominator of the second term is already $4n - 8$ , so we don't need to change it. To get $4n - 8$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{2}{n - 2} \times \dfrac{4}{4} = \dfrac{8}{4n - 8} $ This give us: $ \dfrac{12n - 12}{4n - 8} = -\dfrac{4}{4n - 8} + \dfrac{8}{4n - 8} $ If we multiply both sides of the equation by $4n - 8$ , we get: $ 12n - 12 = -4 + 8$ $ 12n - 12 = 4$ $ 12n = 16 $ $ n = \dfrac{4}{3}$ |
307 | Solve for $k$, $ \dfrac{k + 1}{k + 3} = -\dfrac{5}{k + 3} + \dfrac{7}{2k + 6} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $k + 3$ $k + 3$ and $2k + 6$ The common denominator is $2k + 6$ To get $2k + 6$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{k + 1}{k + 3} \times \dfrac{2}{2} = \dfrac{2k + 2}{2k + 6} $ To get $2k + 6$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{5}{k + 3} \times \dfrac{2}{2} = -\dfrac{10}{2k + 6} $ The denominator of the third term is already $2k + 6$ , so we don't need to change it. This give us: $ \dfrac{2k + 2}{2k + 6} = -\dfrac{10}{2k + 6} + \dfrac{7}{2k + 6} $ If we multiply both sides of the equation by $2k + 6$ , we get: $ 2k + 2 = -10 + 7$ $ 2k + 2 = -3$ $ 2k = -5 $ $ k = -\dfrac{5}{2}$ |
307 | Solve for $p$, $ -\dfrac{5}{3p} = -\dfrac{6}{3p} + \dfrac{p + 2}{3p} $ | If we multiply both sides of the equation by $3p$ , we get: $ -5 = -6 + p + 2$ $ -5 = p - 4$ $ -1 = p $ $ p = -1$ |
307 | Solve for $n$, $ -\dfrac{8}{12n - 12} = \dfrac{n - 3}{15n - 15} - \dfrac{2}{15n - 15} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $12n - 12$ $15n - 15$ and $15n - 15$ The common denominator is $60n - 60$ To get $60n - 60$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{8}{12n - 12} \times \dfrac{5}{5} = -\dfrac{40}{60n - 60} $ To get $60n - 60$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{n - 3}{15n - 15} \times \dfrac{4}{4} = \dfrac{4n - 12}{60n - 60} $ To get $60n - 60$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{2}{15n - 15} \times \dfrac{4}{4} = -\dfrac{8}{60n - 60} $ This give us: $ -\dfrac{40}{60n - 60} = \dfrac{4n - 12}{60n - 60} - \dfrac{8}{60n - 60} $ If we multiply both sides of the equation by $60n - 60$ , we get: $ -40 = 4n - 12 - 8$ $ -40 = 4n - 20$ $ -20 = 4n $ $ n = -5$ |
307 | Solve for $t$, $ \dfrac{1}{5t^3} = -\dfrac{2}{t^3} - \dfrac{4t - 9}{2t^3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5t^3$ $t^3$ and $2t^3$ The common denominator is $10t^3$ To get $10t^3$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{1}{5t^3} \times \dfrac{2}{2} = \dfrac{2}{10t^3} $ To get $10t^3$ in the denominator of the second term, multiply it by $\frac{10}{10}$ $ -\dfrac{2}{t^3} \times \dfrac{10}{10} = -\dfrac{20}{10t^3} $ To get $10t^3$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{4t - 9}{2t^3} \times \dfrac{5}{5} = -\dfrac{20t - 45}{10t^3} $ This give us: $ \dfrac{2}{10t^3} = -\dfrac{20}{10t^3} - \dfrac{20t - 45}{10t^3} $ If we multiply both sides of the equation by $10t^3$ , we get: $ 2 = -20 - 20t + 45$ $ 2 = -20t + 25$ $ -23 = -20t $ $ t = \dfrac{23}{20}$ |
307 | Solve for $q$, $ -\dfrac{5q + 5}{25q + 20} = \dfrac{10}{5q + 4} - \dfrac{9}{5q + 4} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25q + 20$ $5q + 4$ and $5q + 4$ The common denominator is $25q + 20$ The denominator of the first term is already $25q + 20$ , so we don't need to change it. To get $25q + 20$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{10}{5q + 4} \times \dfrac{5}{5} = \dfrac{50}{25q + 20} $ To get $25q + 20$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{9}{5q + 4} \times \dfrac{5}{5} = -\dfrac{45}{25q + 20} $ This give us: $ -\dfrac{5q + 5}{25q + 20} = \dfrac{50}{25q + 20} - \dfrac{45}{25q + 20} $ If we multiply both sides of the equation by $25q + 20$ , we get: $ -5q - 5 = 50 - 45$ $ -5q - 5 = 5$ $ -5q = 10 $ $ q = -2$ |
307 | Solve for $k$, $ \dfrac{2k - 9}{k + 3} = -\dfrac{10}{4k + 12} + \dfrac{3}{k + 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $k + 3$ $4k + 12$ and $k + 3$ The common denominator is $4k + 12$ To get $4k + 12$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{2k - 9}{k + 3} \times \dfrac{4}{4} = \dfrac{8k - 36}{4k + 12} $ The denominator of the second term is already $4k + 12$ , so we don't need to change it. To get $4k + 12$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{3}{k + 3} \times \dfrac{4}{4} = \dfrac{12}{4k + 12} $ This give us: $ \dfrac{8k - 36}{4k + 12} = -\dfrac{10}{4k + 12} + \dfrac{12}{4k + 12} $ If we multiply both sides of the equation by $4k + 12$ , we get: $ 8k - 36 = -10 + 12$ $ 8k - 36 = 2$ $ 8k = 38 $ $ k = \dfrac{19}{4}$ |
307 | Solve for $t$, $ -\dfrac{8}{2t + 5} = \dfrac{5}{2t + 5} - \dfrac{2t + 1}{2t + 5} $ | If we multiply both sides of the equation by $2t + 5$ , we get: $ -8 = 5 - 2t - 1$ $ -8 = -2t + 4$ $ -12 = -2t $ $ t = 6$ |
307 | Solve for $n$, $ \dfrac{6}{8n} = -\dfrac{n - 8}{2n} - \dfrac{8}{6n} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8n$ $2n$ and $6n$ The common denominator is $24n$ To get $24n$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{6}{8n} \times \dfrac{3}{3} = \dfrac{18}{24n} $ To get $24n$ in the denominator of the second term, multiply it by $\frac{12}{12}$ $ -\dfrac{n - 8}{2n} \times \dfrac{12}{12} = -\dfrac{12n - 96}{24n} $ To get $24n$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{6n} \times \dfrac{4}{4} = -\dfrac{32}{24n} $ This give us: $ \dfrac{18}{24n} = -\dfrac{12n - 96}{24n} - \dfrac{32}{24n} $ If we multiply both sides of the equation by $24n$ , we get: $ 18 = -12n + 96 - 32$ $ 18 = -12n + 64$ $ -46 = -12n $ $ n = \dfrac{23}{6}$ |
307 | Solve for $r$, $ -\dfrac{4}{4r + 10} = -\dfrac{r - 8}{2r + 5} + \dfrac{1}{2r + 5} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4r + 10$ $2r + 5$ and $2r + 5$ The common denominator is $4r + 10$ The denominator of the first term is already $4r + 10$ , so we don't need to change it. To get $4r + 10$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{r - 8}{2r + 5} \times \dfrac{2}{2} = -\dfrac{2r - 16}{4r + 10} $ To get $4r + 10$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{1}{2r + 5} \times \dfrac{2}{2} = \dfrac{2}{4r + 10} $ This give us: $ -\dfrac{4}{4r + 10} = -\dfrac{2r - 16}{4r + 10} + \dfrac{2}{4r + 10} $ If we multiply both sides of the equation by $4r + 10$ , we get: $ -4 = -2r + 16 + 2$ $ -4 = -2r + 18$ $ -22 = -2r $ $ r = 11$ |
307 | Solve for $a$, $ \dfrac{a + 8}{12a - 8} = \dfrac{6}{9a - 6} + \dfrac{5}{12a - 8} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $12a - 8$ $9a - 6$ and $12a - 8$ The common denominator is $36a - 24$ To get $36a - 24$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{a + 8}{12a - 8} \times \dfrac{3}{3} = \dfrac{3a + 24}{36a - 24} $ To get $36a - 24$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{6}{9a - 6} \times \dfrac{4}{4} = \dfrac{24}{36a - 24} $ To get $36a - 24$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{5}{12a - 8} \times \dfrac{3}{3} = \dfrac{15}{36a - 24} $ This give us: $ \dfrac{3a + 24}{36a - 24} = \dfrac{24}{36a - 24} + \dfrac{15}{36a - 24} $ If we multiply both sides of the equation by $36a - 24$ , we get: $ 3a + 24 = 24 + 15$ $ 3a + 24 = 39$ $ 3a = 15 $ $ a = 5$ |
307 | Solve for $r$, $ \dfrac{r + 7}{2r + 1} = -\dfrac{6}{10r + 5} + \dfrac{8}{6r + 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2r + 1$ $10r + 5$ and $6r + 3$ The common denominator is $30r + 15$ To get $30r + 15$ in the denominator of the first term, multiply it by $\frac{15}{15}$ $ \dfrac{r + 7}{2r + 1} \times \dfrac{15}{15} = \dfrac{15r + 105}{30r + 15} $ To get $30r + 15$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{6}{10r + 5} \times \dfrac{3}{3} = -\dfrac{18}{30r + 15} $ To get $30r + 15$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{8}{6r + 3} \times \dfrac{5}{5} = \dfrac{40}{30r + 15} $ This give us: $ \dfrac{15r + 105}{30r + 15} = -\dfrac{18}{30r + 15} + \dfrac{40}{30r + 15} $ If we multiply both sides of the equation by $30r + 15$ , we get: $ 15r + 105 = -18 + 40$ $ 15r + 105 = 22$ $ 15r = -83 $ $ r = -\dfrac{83}{15}$ |
307 | Solve for $n$, $ \dfrac{5}{2n - 3} = -\dfrac{4n - 10}{2n - 3} + \dfrac{2}{2n - 3} $ | If we multiply both sides of the equation by $2n - 3$ , we get: $ 5 = -4n + 10 + 2$ $ 5 = -4n + 12$ $ -7 = -4n $ $ n = \dfrac{7}{4}$ |
307 | Solve for $p$, $ -\dfrac{5p + 9}{3p^3} = \dfrac{6}{3p^3} - \dfrac{10}{6p^3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3p^3$ $3p^3$ and $6p^3$ The common denominator is $6p^3$ To get $6p^3$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{5p + 9}{3p^3} \times \dfrac{2}{2} = -\dfrac{10p + 18}{6p^3} $ To get $6p^3$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{6}{3p^3} \times \dfrac{2}{2} = \dfrac{12}{6p^3} $ The denominator of the third term is already $6p^3$ , so we don't need to change it. This give us: $ -\dfrac{10p + 18}{6p^3} = \dfrac{12}{6p^3} - \dfrac{10}{6p^3} $ If we multiply both sides of the equation by $6p^3$ , we get: $ -10p - 18 = 12 - 10$ $ -10p - 18 = 2$ $ -10p = 20 $ $ p = -2$ |
307 | Solve for $y$, $ \dfrac{2}{4y} = \dfrac{3}{8y} + \dfrac{2y + 10}{4y} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4y$ $8y$ and $4y$ The common denominator is $8y$ To get $8y$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{2}{4y} \times \dfrac{2}{2} = \dfrac{4}{8y} $ The denominator of the second term is already $8y$ , so we don't need to change it. To get $8y$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{2y + 10}{4y} \times \dfrac{2}{2} = \dfrac{4y + 20}{8y} $ This give us: $ \dfrac{4}{8y} = \dfrac{3}{8y} + \dfrac{4y + 20}{8y} $ If we multiply both sides of the equation by $8y$ , we get: $ 4 = 3 + 4y + 20$ $ 4 = 4y + 23$ $ -19 = 4y $ $ y = -\dfrac{19}{4}$ |
307 | Solve for $a$, $ \dfrac{5a - 1}{a} = \dfrac{1}{3a} + \dfrac{9}{a} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $a$ $3a$ and $a$ The common denominator is $3a$ To get $3a$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{5a - 1}{a} \times \dfrac{3}{3} = \dfrac{15a - 3}{3a} $ The denominator of the second term is already $3a$ , so we don't need to change it. To get $3a$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ \dfrac{9}{a} \times \dfrac{3}{3} = \dfrac{27}{3a} $ This give us: $ \dfrac{15a - 3}{3a} = \dfrac{1}{3a} + \dfrac{27}{3a} $ If we multiply both sides of the equation by $3a$ , we get: $ 15a - 3 = 1 + 27$ $ 15a - 3 = 28$ $ 15a = 31 $ $ a = \dfrac{31}{15}$ |
307 | Solve for $y$, $ -\dfrac{9}{y + 5} = \dfrac{5}{4y + 20} - \dfrac{4y - 3}{5y + 25} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $y + 5$ $4y + 20$ and $5y + 25$ The common denominator is $20y + 100$ To get $20y + 100$ in the denominator of the first term, multiply it by $\frac{20}{20}$ $ -\dfrac{9}{y + 5} \times \dfrac{20}{20} = -\dfrac{180}{20y + 100} $ To get $20y + 100$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{5}{4y + 20} \times \dfrac{5}{5} = \dfrac{25}{20y + 100} $ To get $20y + 100$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{4y - 3}{5y + 25} \times \dfrac{4}{4} = -\dfrac{16y - 12}{20y + 100} $ This give us: $ -\dfrac{180}{20y + 100} = \dfrac{25}{20y + 100} - \dfrac{16y - 12}{20y + 100} $ If we multiply both sides of the equation by $20y + 100$ , we get: $ -180 = 25 - 16y + 12$ $ -180 = -16y + 37$ $ -217 = -16y $ $ y = \dfrac{217}{16}$ |
307 | Solve for $y$, $ -\dfrac{8}{3y^3} = \dfrac{5y - 9}{3y^3} - \dfrac{10}{3y^3} $ | If we multiply both sides of the equation by $3y^3$ , we get: $ -8 = 5y - 9 - 10$ $ -8 = 5y - 19$ $ 11 = 5y $ $ y = \dfrac{11}{5}$ |
307 | Solve for $n$, $ \dfrac{n - 4}{5n^3} = \dfrac{6}{n^3} - \dfrac{5}{n^3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5n^3$ $n^3$ and $n^3$ The common denominator is $5n^3$ The denominator of the first term is already $5n^3$ , so we don't need to change it. To get $5n^3$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{6}{n^3} \times \dfrac{5}{5} = \dfrac{30}{5n^3} $ To get $5n^3$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{n^3} \times \dfrac{5}{5} = -\dfrac{25}{5n^3} $ This give us: $ \dfrac{n - 4}{5n^3} = \dfrac{30}{5n^3} - \dfrac{25}{5n^3} $ If we multiply both sides of the equation by $5n^3$ , we get: $ n - 4 = 30 - 25$ $ n - 4 = 5$ $ n = 9 $ |
307 | Solve for $q$, $ -\dfrac{10}{9q + 15} = -\dfrac{q + 5}{12q + 20} - \dfrac{8}{3q + 5} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $9q + 15$ $12q + 20$ and $3q + 5$ The common denominator is $36q + 60$ To get $36q + 60$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{10}{9q + 15} \times \dfrac{4}{4} = -\dfrac{40}{36q + 60} $ To get $36q + 60$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{q + 5}{12q + 20} \times \dfrac{3}{3} = -\dfrac{3q + 15}{36q + 60} $ To get $36q + 60$ in the denominator of the third term, multiply it by $\frac{12}{12}$ $ -\dfrac{8}{3q + 5} \times \dfrac{12}{12} = -\dfrac{96}{36q + 60} $ This give us: $ -\dfrac{40}{36q + 60} = -\dfrac{3q + 15}{36q + 60} - \dfrac{96}{36q + 60} $ If we multiply both sides of the equation by $36q + 60$ , we get: $ -40 = -3q - 15 - 96$ $ -40 = -3q - 111$ $ 71 = -3q $ $ q = -\dfrac{71}{3}$ |
307 | Solve for $x$, $ \dfrac{9}{15x + 6} = -\dfrac{10}{20x + 8} - \dfrac{3x - 1}{5x + 2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15x + 6$ $20x + 8$ and $5x + 2$ The common denominator is $60x + 24$ To get $60x + 24$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{9}{15x + 6} \times \dfrac{4}{4} = \dfrac{36}{60x + 24} $ To get $60x + 24$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{10}{20x + 8} \times \dfrac{3}{3} = -\dfrac{30}{60x + 24} $ To get $60x + 24$ in the denominator of the third term, multiply it by $\frac{12}{12}$ $ -\dfrac{3x - 1}{5x + 2} \times \dfrac{12}{12} = -\dfrac{36x - 12}{60x + 24} $ This give us: $ \dfrac{36}{60x + 24} = -\dfrac{30}{60x + 24} - \dfrac{36x - 12}{60x + 24} $ If we multiply both sides of the equation by $60x + 24$ , we get: $ 36 = -30 - 36x + 12$ $ 36 = -36x - 18$ $ 54 = -36x $ $ x = -\dfrac{3}{2}$ |
307 | Solve for $r$, $ \dfrac{7}{r} = -\dfrac{9}{3r} - \dfrac{4r - 7}{r} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $r$ $3r$ and $r$ The common denominator is $3r$ To get $3r$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{7}{r} \times \dfrac{3}{3} = \dfrac{21}{3r} $ The denominator of the second term is already $3r$ , so we don't need to change it. To get $3r$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{4r - 7}{r} \times \dfrac{3}{3} = -\dfrac{12r - 21}{3r} $ This give us: $ \dfrac{21}{3r} = -\dfrac{9}{3r} - \dfrac{12r - 21}{3r} $ If we multiply both sides of the equation by $3r$ , we get: $ 21 = -9 - 12r + 21$ $ 21 = -12r + 12$ $ 9 = -12r $ $ r = -\dfrac{3}{4}$ |
307 | Solve for $k$, $ -\dfrac{5}{k^2} = -\dfrac{6}{4k^2} - \dfrac{5k + 2}{k^2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $k^2$ $4k^2$ and $k^2$ The common denominator is $4k^2$ To get $4k^2$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{5}{k^2} \times \dfrac{4}{4} = -\dfrac{20}{4k^2} $ The denominator of the second term is already $4k^2$ , so we don't need to change it. To get $4k^2$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ -\dfrac{5k + 2}{k^2} \times \dfrac{4}{4} = -\dfrac{20k + 8}{4k^2} $ This give us: $ -\dfrac{20}{4k^2} = -\dfrac{6}{4k^2} - \dfrac{20k + 8}{4k^2} $ If we multiply both sides of the equation by $4k^2$ , we get: $ -20 = -6 - 20k - 8$ $ -20 = -20k - 14$ $ -6 = -20k $ $ k = \dfrac{3}{10}$ |
307 | Solve for $q$, $ -\dfrac{5}{15q - 5} = \dfrac{9}{3q - 1} - \dfrac{4q + 5}{3q - 1} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15q - 5$ $3q - 1$ and $3q - 1$ The common denominator is $15q - 5$ The denominator of the first term is already $15q - 5$ , so we don't need to change it. To get $15q - 5$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{9}{3q - 1} \times \dfrac{5}{5} = \dfrac{45}{15q - 5} $ To get $15q - 5$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{4q + 5}{3q - 1} \times \dfrac{5}{5} = -\dfrac{20q + 25}{15q - 5} $ This give us: $ -\dfrac{5}{15q - 5} = \dfrac{45}{15q - 5} - \dfrac{20q + 25}{15q - 5} $ If we multiply both sides of the equation by $15q - 5$ , we get: $ -5 = 45 - 20q - 25$ $ -5 = -20q + 20$ $ -25 = -20q $ $ q = \dfrac{5}{4}$ |
307 | Solve for $r$, $ \dfrac{2}{25r} = \dfrac{3}{15r} + \dfrac{2r + 8}{5r} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25r$ $15r$ and $5r$ The common denominator is $75r$ To get $75r$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{2}{25r} \times \dfrac{3}{3} = \dfrac{6}{75r} $ To get $75r$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{3}{15r} \times \dfrac{5}{5} = \dfrac{15}{75r} $ To get $75r$ in the denominator of the third term, multiply it by $\frac{15}{15}$ $ \dfrac{2r + 8}{5r} \times \dfrac{15}{15} = \dfrac{30r + 120}{75r} $ This give us: $ \dfrac{6}{75r} = \dfrac{15}{75r} + \dfrac{30r + 120}{75r} $ If we multiply both sides of the equation by $75r$ , we get: $ 6 = 15 + 30r + 120$ $ 6 = 30r + 135$ $ -129 = 30r $ $ r = -\dfrac{43}{10}$ |
307 | Solve for $t$, $ -\dfrac{2t}{3t - 9} = \dfrac{1}{t - 3} - \dfrac{4}{t - 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $3t - 9$ $t - 3$ and $t - 3$ The common denominator is $3t - 9$ The denominator of the first term is already $3t - 9$ , so we don't need to change it. To get $3t - 9$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ \dfrac{1}{t - 3} \times \dfrac{3}{3} = \dfrac{3}{3t - 9} $ To get $3t - 9$ in the denominator of the third term, multiply it by $\frac{3}{3}$ $ -\dfrac{4}{t - 3} \times \dfrac{3}{3} = -\dfrac{12}{3t - 9} $ This give us: $ -\dfrac{2t}{3t - 9} = \dfrac{3}{3t - 9} - \dfrac{12}{3t - 9} $ If we multiply both sides of the equation by $3t - 9$ , we get: $ -2t = 3 - 12$ $ -2t = -9$ $ -2t = -9 $ $ t = \dfrac{9}{2}$ |
307 | Solve for $t$, $ \dfrac{3t - 9}{5t - 15} = -\dfrac{6}{t - 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 $5t - 15$ $t - 3$ and $t - 3$ The common denominator is $5t - 15$ The denominator of the first term is already $5t - 15$ , so we don't need to change it. To get $5t - 15$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{6}{t - 3} \times \dfrac{5}{5} = -\dfrac{30}{5t - 15} $ To get $5t - 15$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{7}{t - 3} \times \dfrac{5}{5} = \dfrac{35}{5t - 15} $ This give us: $ \dfrac{3t - 9}{5t - 15} = -\dfrac{30}{5t - 15} + \dfrac{35}{5t - 15} $ If we multiply both sides of the equation by $5t - 15$ , we get: $ 3t - 9 = -30 + 35$ $ 3t - 9 = 5$ $ 3t = 14 $ $ t = \dfrac{14}{3}$ |
307 | Solve for $x$, $ \dfrac{x - 4}{10x} = \dfrac{4}{5x} - \dfrac{3}{20x} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $10x$ $5x$ and $20x$ The common denominator is $20x$ To get $20x$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{x - 4}{10x} \times \dfrac{2}{2} = \dfrac{2x - 8}{20x} $ To get $20x$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{4}{5x} \times \dfrac{4}{4} = \dfrac{16}{20x} $ The denominator of the third term is already $20x$ , so we don't need to change it. This give us: $ \dfrac{2x - 8}{20x} = \dfrac{16}{20x} - \dfrac{3}{20x} $ If we multiply both sides of the equation by $20x$ , we get: $ 2x - 8 = 16 - 3$ $ 2x - 8 = 13$ $ 2x = 21 $ $ x = \dfrac{21}{2}$ |
307 | Solve for $y$, $ \dfrac{5}{25y + 25} = -\dfrac{2y - 7}{5y + 5} - \dfrac{1}{20y + 20} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25y + 25$ $5y + 5$ and $20y + 20$ The common denominator is $100y + 100$ To get $100y + 100$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{5}{25y + 25} \times \dfrac{4}{4} = \dfrac{20}{100y + 100} $ To get $100y + 100$ in the denominator of the second term, multiply it by $\frac{20}{20}$ $ -\dfrac{2y - 7}{5y + 5} \times \dfrac{20}{20} = -\dfrac{40y - 140}{100y + 100} $ To get $100y + 100$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ -\dfrac{1}{20y + 20} \times \dfrac{5}{5} = -\dfrac{5}{100y + 100} $ This give us: $ \dfrac{20}{100y + 100} = -\dfrac{40y - 140}{100y + 100} - \dfrac{5}{100y + 100} $ If we multiply both sides of the equation by $100y + 100$ , we get: $ 20 = -40y + 140 - 5$ $ 20 = -40y + 135$ $ -115 = -40y $ $ y = \dfrac{23}{8}$ |
307 | Solve for $x$, $ -\dfrac{8}{x - 5} = -\dfrac{x - 2}{4x - 20} + \dfrac{10}{x - 5} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $x - 5$ $4x - 20$ and $x - 5$ The common denominator is $4x - 20$ To get $4x - 20$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{x - 5} \times \dfrac{4}{4} = -\dfrac{32}{4x - 20} $ The denominator of the second term is already $4x - 20$ , so we don't need to change it. To get $4x - 20$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{10}{x - 5} \times \dfrac{4}{4} = \dfrac{40}{4x - 20} $ This give us: $ -\dfrac{32}{4x - 20} = -\dfrac{x - 2}{4x - 20} + \dfrac{40}{4x - 20} $ If we multiply both sides of the equation by $4x - 20$ , we get: $ -32 = -x + 2 + 40$ $ -32 = -x + 42$ $ -74 = -x $ $ x = 74$ |
307 | Solve for $q$, $ \dfrac{4}{q + 2} = \dfrac{q - 7}{5q + 10} + \dfrac{2}{q + 2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $q + 2$ $5q + 10$ and $q + 2$ The common denominator is $5q + 10$ To get $5q + 10$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{4}{q + 2} \times \dfrac{5}{5} = \dfrac{20}{5q + 10} $ The denominator of the second term is already $5q + 10$ , so we don't need to change it. To get $5q + 10$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{2}{q + 2} \times \dfrac{5}{5} = \dfrac{10}{5q + 10} $ This give us: $ \dfrac{20}{5q + 10} = \dfrac{q - 7}{5q + 10} + \dfrac{10}{5q + 10} $ If we multiply both sides of the equation by $5q + 10$ , we get: $ 20 = q - 7 + 10$ $ 20 = q + 3$ $ 17 = q $ $ q = 17$ |
307 | Solve for $r$, $ -\dfrac{6}{r} = -\dfrac{r + 6}{r} - \dfrac{8}{r} $ | If we multiply both sides of the equation by $r$ , we get: $ -6 = -r - 6 - 8$ $ -6 = -r - 14$ $ 8 = -r $ $ r = -8$ |
307 | Solve for $q$, $ -\dfrac{6}{25q - 20} = \dfrac{3q + 2}{10q - 8} + \dfrac{8}{25q - 20} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25q - 20$ $10q - 8$ and $25q - 20$ The common denominator is $50q - 40$ To get $50q - 40$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{6}{25q - 20} \times \dfrac{2}{2} = -\dfrac{12}{50q - 40} $ To get $50q - 40$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{3q + 2}{10q - 8} \times \dfrac{5}{5} = \dfrac{15q + 10}{50q - 40} $ To get $50q - 40$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{8}{25q - 20} \times \dfrac{2}{2} = \dfrac{16}{50q - 40} $ This give us: $ -\dfrac{12}{50q - 40} = \dfrac{15q + 10}{50q - 40} + \dfrac{16}{50q - 40} $ If we multiply both sides of the equation by $50q - 40$ , we get: $ -12 = 15q + 10 + 16$ $ -12 = 15q + 26$ $ -38 = 15q $ $ q = -\dfrac{38}{15}$ |
307 | Solve for $a$, $ -\dfrac{6}{2a^3} = \dfrac{2a + 9}{2a^3} + \dfrac{8}{2a^3} $ | If we multiply both sides of the equation by $2a^3$ , we get: $ -6 = 2a + 9 + 8$ $ -6 = 2a + 17$ $ -23 = 2a $ $ a = -\dfrac{23}{2}$ |
307 | Solve for $y$, $ \dfrac{3}{5y} = \dfrac{9}{5y} + \dfrac{y - 9}{5y} $ | If we multiply both sides of the equation by $5y$ , we get: $ 3 = 9 + y - 9$ $ 3 = y$ $ 3 = y $ $ y = 3$ |
307 | Solve for $y$, $ \dfrac{1}{25y} = \dfrac{8}{25y} + \dfrac{y - 6}{15y} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25y$ $25y$ and $15y$ The common denominator is $75y$ To get $75y$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ \dfrac{1}{25y} \times \dfrac{3}{3} = \dfrac{3}{75y} $ To get $75y$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ \dfrac{8}{25y} \times \dfrac{3}{3} = \dfrac{24}{75y} $ To get $75y$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{y - 6}{15y} \times \dfrac{5}{5} = \dfrac{5y - 30}{75y} $ This give us: $ \dfrac{3}{75y} = \dfrac{24}{75y} + \dfrac{5y - 30}{75y} $ If we multiply both sides of the equation by $75y$ , we get: $ 3 = 24 + 5y - 30$ $ 3 = 5y - 6$ $ 9 = 5y $ $ y = \dfrac{9}{5}$ |
307 | Solve for $q$, $ -\dfrac{8}{5q - 3} = \dfrac{3q + 6}{20q - 12} + \dfrac{2}{10q - 6} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $5q - 3$ $20q - 12$ and $10q - 6$ The common denominator is $20q - 12$ To get $20q - 12$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ -\dfrac{8}{5q - 3} \times \dfrac{4}{4} = -\dfrac{32}{20q - 12} $ The denominator of the second term is already $20q - 12$ , so we don't need to change it. To get $20q - 12$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{2}{10q - 6} \times \dfrac{2}{2} = \dfrac{4}{20q - 12} $ This give us: $ -\dfrac{32}{20q - 12} = \dfrac{3q + 6}{20q - 12} + \dfrac{4}{20q - 12} $ If we multiply both sides of the equation by $20q - 12$ , we get: $ -32 = 3q + 6 + 4$ $ -32 = 3q + 10$ $ -42 = 3q $ $ q = -14$ |
307 | Solve for $k$, $ -\dfrac{9}{2k - 1} = -\dfrac{9}{2k - 1} - \dfrac{k + 4}{6k - 3} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2k - 1$ $2k - 1$ and $6k - 3$ The common denominator is $6k - 3$ To get $6k - 3$ in the denominator of the first term, multiply it by $\frac{3}{3}$ $ -\dfrac{9}{2k - 1} \times \dfrac{3}{3} = -\dfrac{27}{6k - 3} $ To get $6k - 3$ in the denominator of the second term, multiply it by $\frac{3}{3}$ $ -\dfrac{9}{2k - 1} \times \dfrac{3}{3} = -\dfrac{27}{6k - 3} $ The denominator of the third term is already $6k - 3$ , so we don't need to change it. This give us: $ -\dfrac{27}{6k - 3} = -\dfrac{27}{6k - 3} - \dfrac{k + 4}{6k - 3} $ If we multiply both sides of the equation by $6k - 3$ , we get: $ -27 = -27 - k - 4$ $ -27 = -k - 31$ $ 4 = -k $ $ k = -4$ |
307 | Solve for $r$, $ \dfrac{9}{15r + 25} = -\dfrac{r - 5}{12r + 20} - \dfrac{3}{6r + 10} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $15r + 25$ $12r + 20$ and $6r + 10$ The common denominator is $60r + 100$ To get $60r + 100$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{9}{15r + 25} \times \dfrac{4}{4} = \dfrac{36}{60r + 100} $ To get $60r + 100$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{r - 5}{12r + 20} \times \dfrac{5}{5} = -\dfrac{5r - 25}{60r + 100} $ To get $60r + 100$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ -\dfrac{3}{6r + 10} \times \dfrac{10}{10} = -\dfrac{30}{60r + 100} $ This give us: $ \dfrac{36}{60r + 100} = -\dfrac{5r - 25}{60r + 100} - \dfrac{30}{60r + 100} $ If we multiply both sides of the equation by $60r + 100$ , we get: $ 36 = -5r + 25 - 30$ $ 36 = -5r - 5$ $ 41 = -5r $ $ r = -\dfrac{41}{5}$ |
307 | Solve for $a$, $ \dfrac{9}{a} = \dfrac{5}{a} - \dfrac{a - 2}{a} $ | If we multiply both sides of the equation by $a$ , we get: $ 9 = 5 - a + 2$ $ 9 = -a + 7$ $ 2 = -a $ $ a = -2$ |
307 | Solve for $q$, $ -\dfrac{4q + 9}{8q} = \dfrac{10}{8q} + \dfrac{8}{2q} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8q$ $8q$ and $2q$ The common denominator is $8q$ The denominator of the first term is already $8q$ , so we don't need to change it. The denominator of the second term is already $8q$ , so we don't need to change it. To get $8q$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{8}{2q} \times \dfrac{4}{4} = \dfrac{32}{8q} $ This give us: $ -\dfrac{4q + 9}{8q} = \dfrac{10}{8q} + \dfrac{32}{8q} $ If we multiply both sides of the equation by $8q$ , we get: $ -4q - 9 = 10 + 32$ $ -4q - 9 = 42$ $ -4q = 51 $ $ q = -\dfrac{51}{4}$ |
307 | Solve for $r$, $ \dfrac{3r - 2}{2r + 4} = \dfrac{1}{4r + 8} + \dfrac{10}{2r + 4} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2r + 4$ $4r + 8$ and $2r + 4$ The common denominator is $4r + 8$ To get $4r + 8$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ \dfrac{3r - 2}{2r + 4} \times \dfrac{2}{2} = \dfrac{6r - 4}{4r + 8} $ The denominator of the second term is already $4r + 8$ , so we don't need to change it. To get $4r + 8$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{10}{2r + 4} \times \dfrac{2}{2} = \dfrac{20}{4r + 8} $ This give us: $ \dfrac{6r - 4}{4r + 8} = \dfrac{1}{4r + 8} + \dfrac{20}{4r + 8} $ If we multiply both sides of the equation by $4r + 8$ , we get: $ 6r - 4 = 1 + 20$ $ 6r - 4 = 21$ $ 6r = 25 $ $ r = \dfrac{25}{6}$ |
307 | Solve for $t$, $ \dfrac{2}{4t} = \dfrac{3t - 10}{t} - \dfrac{8}{2t} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4t$ $t$ and $2t$ The common denominator is $4t$ The denominator of the first term is already $4t$ , so we don't need to change it. To get $4t$ in the denominator of the second term, multiply it by $\frac{4}{4}$ $ \dfrac{3t - 10}{t} \times \dfrac{4}{4} = \dfrac{12t - 40}{4t} $ To get $4t$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ -\dfrac{8}{2t} \times \dfrac{2}{2} = -\dfrac{16}{4t} $ This give us: $ \dfrac{2}{4t} = \dfrac{12t - 40}{4t} - \dfrac{16}{4t} $ If we multiply both sides of the equation by $4t$ , we get: $ 2 = 12t - 40 - 16$ $ 2 = 12t - 56$ $ 58 = 12t $ $ t = \dfrac{29}{6}$ |
307 | Solve for $a$, $ -\dfrac{7}{10a + 10} = \dfrac{5}{2a + 2} + \dfrac{3a - 1}{2a + 2} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $10a + 10$ $2a + 2$ and $2a + 2$ The common denominator is $10a + 10$ The denominator of the first term is already $10a + 10$ , so we don't need to change it. To get $10a + 10$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{5}{2a + 2} \times \dfrac{5}{5} = \dfrac{25}{10a + 10} $ To get $10a + 10$ in the denominator of the third term, multiply it by $\frac{5}{5}$ $ \dfrac{3a - 1}{2a + 2} \times \dfrac{5}{5} = \dfrac{15a - 5}{10a + 10} $ This give us: $ -\dfrac{7}{10a + 10} = \dfrac{25}{10a + 10} + \dfrac{15a - 5}{10a + 10} $ If we multiply both sides of the equation by $10a + 10$ , we get: $ -7 = 25 + 15a - 5$ $ -7 = 15a + 20$ $ -27 = 15a $ $ a = -\dfrac{9}{5}$ |
307 | Solve for $z$, $ -\dfrac{6}{z - 3} = -\dfrac{z - 6}{z - 3} - \dfrac{10}{2z - 6} $ | First we need to find a common denominator for all the expressions. This means finding the least common multiple of $z - 3$ $z - 3$ and $2z - 6$ The common denominator is $2z - 6$ To get $2z - 6$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{6}{z - 3} \times \dfrac{2}{2} = -\dfrac{12}{2z - 6} $ To get $2z - 6$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{z - 6}{z - 3} \times \dfrac{2}{2} = -\dfrac{2z - 12}{2z - 6} $ The denominator of the third term is already $2z - 6$ , so we don't need to change it. This give us: $ -\dfrac{12}{2z - 6} = -\dfrac{2z - 12}{2z - 6} - \dfrac{10}{2z - 6} $ If we multiply both sides of the equation by $2z - 6$ , we get: $ -12 = -2z + 12 - 10$ $ -12 = -2z + 2$ $ -14 = -2z $ $ z = 7$ |
307 | Solve for $y$, $ -\dfrac{2y + 10}{y - 2} = -\dfrac{6}{y - 2} - \dfrac{6}{y - 2} $ | If we multiply both sides of the equation by $y - 2$ , we get: $ -2y - 10 = -6 - 6$ $ -2y - 10 = -12$ $ -2y = -2 $ $ y = 1$ |
307 | Solve for $n$, $ \dfrac{4}{n + 5} = -\dfrac{n - 4}{n + 5} - \dfrac{6}{n + 5} $ | If we multiply both sides of the equation by $n + 5$ , we get: $ 4 = -n + 4 - 6$ $ 4 = -n - 2$ $ 6 = -n $ $ n = -6$ |
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