Solution

\[ \begin{aligned} \frac{p + 7}{p - 6} & = \frac{1}{3} \\[6pt] 3 \times (p+7) &= 1 \times (p - 6) \\[6pt] 3p + 21 &= p - 6 \\[6pt] 3p - p &= -21 - 6 \\[6pt] 2p &= -27 \\[6pt] \color{green} p &= \color{green} \frac{-27}{2} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } & = \frac{p + 7}{p - 6} \\[6pt] & = \frac{\frac{-27}{2} + 7}{\frac{-27}{2} - 6} \\[6pt] & = \frac{\frac{-27 + 14}{2}}{\frac{-27 - 12}{2} } \\[6pt] & = \frac{\frac{-13}{2}}{\frac{-39}{2} } \\[6pt] & = \frac{\cancel{13}^1}{\cancel2} \times \frac{\cancel 2}{\cancel{39}_3} \\[6pt] \textbf{L.H.S } & = \frac{1}{3} \\[6pt] \textbf{R.H.S } & = \frac{1}{3} \\ \\ L.H.S & = R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} p = \dfrac{-27}{2}\)

Solution

\[ \begin{aligned} \frac{x - 2}{6x + 1} &= 1 \\[6pt] x - 2 &= 1 \times (6x + 1) \\[6pt] x - 2 &= 6x + 1 \\[6pt] x - 6x &= 1 + 2 \\[6pt] -5x &= 3 \\[6pt] \color{green} x &= \color{green} \frac{-3}{5} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{x - 2}{6x + 1} \\[6pt] &= \frac{\frac{-3}{5} - 2}{6(\frac{-3}{5}) + 1} \\[6pt] &= \frac{\frac{-3 - 10}{5}}{\frac{-18}{5} + 1} \\[6pt] &= \frac{\frac{-13}{5}}{\frac{-18 + 5}{5}} \\[6pt] &= \frac{\frac{-13}{5}}{\frac{-13}{5}} \\[6pt] &= \frac{\cancel{13}}{\cancel5} \times \frac{\cancel5}{\cancel{13}} \\[6pt] \textbf{L.H.S } &= 1 \\[6pt] \textbf{R.H.S } &= 1 \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{-3}{5}\)

Solution

\[ \begin{aligned} \frac{3x}{5x - 5} &= -1 \\[6pt] 3x &= -1 \times (5x - 5) \\[6pt] 3x &= -5x + 5 \\[6pt] 3x + 5x &= 5 \\[6pt] 8x &= 5 \\[6pt] \color{green} x &= \color{green} \frac{5}{8} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{3x}{5x - 5} \\[6pt] &= \frac{3(\frac{5}{8})}{5(\frac{5}{8}) - 5} \\[6pt] &= \frac{\frac{15}{8}}{\frac{25}{8} - 5} \\[6pt] &= \frac{\frac{15}{8}}{\frac{25 - 40}{8}} \\[6pt] &= \frac{\frac{15}{8}}{\frac{25 - 40}{8}} \\[6pt] &= \frac{\cancel{15}}{\cancel8} \times \frac{-\cancel{8}^{\ 1}}{\cancel{15}} \\[6pt] \textbf{L.H.S } &= -1 \\[6pt] \textbf{R.H.S } &= -1 \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{5}{8}\)

Solution

\[ \begin{aligned} \frac{2x - 1}{5x} &= \frac{-1}{6} \\[6pt] 6 \times (2x - 1) &= -1 \times 5x \\[6pt] 12x - 6 &= -5x \\[6pt] 12x + 5x &= 6 \\[6pt] 17x &= 6 \\[6pt] \color{green} x &= \color{green} \frac{6}{17} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{2x - 1}{5x} \\[6pt] &= \frac{2(\frac{6}{17}) - 1}{5(\frac{6}{17})} \\[6pt] &= \frac{\frac{12}{17} - 1}{\frac{30}{17}} \\[6pt] &= \frac{\frac{12 - 17}{17}}{\frac{30}{17}} \\[6pt] &= \frac{\frac{-5}{17}}{\frac{30}{17}} \\[6pt] &= \frac{-\cancel5^1}{\cancel{17}} \times \frac{\cancel{17}}{\cancel{30}_6} \\[6pt] &= \frac{-1}{6} \\[6pt] \textbf{L.H.S } &= -\frac{1}{6} \\[6pt] \textbf{R.H.S } &= -\frac{1}{6} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{6}{17}\)

Solution

\[ \begin{aligned} \frac{4x + 1}{3x - 1} &= -2 \\[6pt] 4x + 1 &= -2 \times (3x - 1) \\[6pt] 4x + 1 &= -6x + 2 \\[6pt] 4x + 6x &= 2 - 1 \\[6pt] 10x &= 1 \\[6pt] \color{green} x &= \color{green} \frac{1}{10} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{4x + 1}{3x - 1} \\[6pt] &= \frac{4(\frac{1}{10}) + 1}{3(\frac{1}{10}) - 1} \\[6pt] &= \frac{\frac{2}{5} + 1}{\frac{3}{10} - 1} \\[6pt] &= \frac{\frac{2 + 5}{5}}{\frac{3 - 10}{10}} \\[6pt] &= \frac{\frac{7}{5}}{\frac{-7}{10}} \\[6pt] &= \frac{\cancel7}{\cancel5} \times \frac{-\cancel{10}^2}{\cancel7} \\[6pt] &= -2 \\[6pt] \textbf{L.H.S } &= -2 \\[6pt] \textbf{R.H.S } &= -2 \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{1}{10}\)

Solution

\[ \begin{aligned} \frac{4z - 3}{2z + 1} &= \frac{5}{7} \\[6pt] 7 \times (4z - 3) &= 5 \times (2z + 1) \\[6pt] 28z - 21 &= 10z + 5 \\[6pt] 28z - 10z &= 5 + 21 \\[6pt] 18z &= 26 \\[6pt] z &= \frac{26}{18} \\[6pt] \color{green} z &= \color{green} \frac{13}{9} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{4z - 3}{2z + 1} \\[6pt] &= \frac{4(\frac{13}{9}) - 3}{2(\frac{13}{9}) + 1} \\[6pt] &= \frac{\frac{52}{9} - 3}{\frac{26}{9} + 1} \\[6pt] &= \frac{\frac{52 - 27}{9}}{\frac{26 + 9}{9}} \\[6pt] &= \frac{\frac{25}{9}}{\frac{35}{9}} \\[6pt] &= \frac{\cancel{25}^{5}}{\cancel9} \times \frac{\cancel9}{\cancel{35}_7} \\[6pt] \textbf{L.H.S } &= \frac{5}{7} \\[6pt] \textbf{R.H.S } &= \frac{5}{7} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} z = \dfrac{13}{9}\)

Solution

\[ \begin{aligned} \frac{\frac{2}{5}x + 3}{\frac{1}{3}x - 1} &= \frac{3}{5} \\[6pt] 5 \left(\frac{2}{5}x + 3 \right) &= 3 \left(\frac{1}{3}x - 1 \right) \\[6pt] 2x + 15 &= x - 3 \\[6pt] 2x - x &= -3 - 15 \\[6pt] \color{green} x &= \color{green} -18 \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{\frac{2}{5}x + 3}{\frac{1}{3}x - 1} \\[6pt] &= \frac{\frac{2}{5}(-18) + 3}{\frac{1}{3}(-18) - 1} \\[6pt] &= \frac{\frac{-36}{5} + 3}{-6 - 1} \\[6pt] &= \frac{\frac{-36 + 15}{5}}{-7} \\[6pt] &= \frac{\frac{-21}{5}}{-7} \\[6pt] &= \frac{\cancel{21}^3}{5} \times \frac{1}{\cancel7_1} \\[6pt] \textbf{L.H.S } &= \frac{3}{5} \\[6pt] \textbf{R.H.S } &= \frac{3}{5} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = -18\)

Solution

\[ \begin{aligned} \frac{2x - \frac{3}{4}}{3x + \frac{4}{5}} &= \frac{1}{5} \\[6pt] 5 \left(2x - \frac{3}{4} \right) &= 1 \left(3x + \frac{4}{5} \right) \\[6pt] 10x - \frac{15}{4} &= 3x + \frac{4}{5} \\[6pt] 10x - 3x &= \frac{4}{5} + \frac{15}{4} \\[6pt] 7x &= \frac{16 + 75}{20} \\[6pt] 7x &= \frac{91}{20} \\[6pt] x &= \frac{\cancel{91}^{13}}{20 \times \cancel7_1} \\[6pt] \color{green} x &= \color{green} \frac{13}{20} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{2x - \frac{3}{4}}{3x + \frac{4}{5}} \\[6pt] &= \frac{2(\frac{13}{20}) - \frac{3}{4}}{3(\frac{13}{20}) + \frac{4}{5}} \\[6pt] &= \frac{\frac{13}{10} - \frac{3}{4}}{\frac{39}{20} + \frac{4}{5}} \\[6pt] &= \frac{\frac{26 - 15}{20}}{\frac{39 + 16}{20}} \\[6pt] &= \frac{\frac{11}{20}}{\frac{55}{20}} \\[6pt] &= \frac{\cancel{11}^1}{\cancel{20}} \times \frac{\cancel{20}}{\cancel{55}_5} \\[6pt] \textbf{L.H.S } &= \frac{1}{5} \\[6pt] \textbf{R.H.S } &= \frac{1}{5} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{13}{20}\)

Solution

\[ \begin{aligned} \frac{\frac{3}{4}x + 1}{x + \frac{1}{5}} &= \frac{7}{4} \\[6pt] 4 \left(\frac{3}{4}x + 1 \right) &= 7 \left(x + \frac{1}{5} \right) \\[6pt] 3x + 4 &= 7x + \frac{7}{5} \\[6pt] 3x - 7x &= \frac{7}{5} - 4 \\[6pt] -4x &= \frac{7 - 20}{5} \\[6pt] -4x &= \frac{-13}{5} \\[6pt] x &= \frac{-13}{5 \times (-4)} \\[6pt] \color{green} x &= \color{green} \frac{13}{20} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{\frac{3}{4}x + 1}{x + \frac{1}{5}} \\[6pt] &= \frac{\frac{3}{4}(\frac{13}{20}) + 1}{\frac{13}{20} + \frac{1}{5}} \\[6pt] &= \frac{\frac{39}{80} + 1}{\frac{13 + 4}{20}} \\[6pt] &= \frac{\frac{39 + 80}{80}}{\frac{17}{20}} \\[6pt] &= \frac{\frac{119}{80}}{\frac{17}{20}} \\[6pt] &= \frac{\cancel{119}^{7}}{\cancel{80}_4} \times \frac{\cancel{20}^1}{\cancel{17}_1} \\[6pt] \textbf{L.H.S } &= \frac{7}{4} \\[6pt] \textbf{R.H.S } &= \frac{7}{4} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{13}{20}\)

Solution

\[ \begin{aligned} \frac{3k + 5}{4k - 3} &= \frac{4}{9} \\[6pt] 9 \times (3k + 5) &= 4 \times (4k - 3) \\[6pt] 27k + 45 &= 16k - 12 \\[6pt] 27k - 16k &= -12 - 45 \\[6pt] 11k &= -57 \\[6pt] \color{green} k &= \color{green} \frac{-57}{11} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{3k + 5}{4k - 3} \\[6pt] &= \frac{3(\frac{-57}{11}) + 5}{4(\frac{-57}{11}) - 3} \\[6pt] &= \frac{\frac{-171}{11} + 5}{\frac{-228}{11} - 3} \\[6pt] &= \frac{\frac{-171 + 55}{11}}{\frac{-228 - 33}{11}} \\[6pt] &= \frac{\frac{-116}{11}}{\frac{-261}{11}} \\[6pt] &= \frac{116}{\cancel{11}} \times \frac{\cancel{11}}{261} \\[6pt] &= \frac{\cancel{116}^4}{\cancel{261}_9} \\[6pt] \textbf{L.H.S } &= \frac{4}{9} \\[6pt] \textbf{R.H.S } &= \frac{4}{9} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} k = \dfrac{-57}{11}\)

Solution

\[ \begin{aligned} \frac{0.5x - 4}{2.4x + 6} &= \frac{-5}{3} \\[6pt] 3 \times (0.5x - 4) &= -5 \times (2.4x + 6) \\[6pt] 1.5x - 12 &= -12x - 30 \\[6pt] 1.5x + 12x &= -30 + 12 \\[6pt] 13.5x &= -18 \\[6pt] x &= \frac{-18}{13.5} \\[6pt] x &= \frac{-180}{135} \\[6pt] \color{green} x &= \color{green} \frac{-4}{3} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{0.5x - 4}{2.4x + 6} \\[6pt] &= \frac{0.5(\frac{-4}{3}) - 4}{2.4(\frac{-4}{3}) + 6} \\[6pt] &= \frac{\frac{-2}{3} - 4}{\frac{-9.6}{3} + 6} \\[6pt] &= \frac{\frac{-2 - 12}{3}}{\frac{-9.6 + 18}{3}} \\[6pt] &= \frac{\frac{-14}{3}}{\frac{8.4}{3}} \\[6pt] &= \frac{-14}{\cancel3} \times \frac{\cancel3}{8.4} \\[6pt] &= \frac{-14}{8.4} \\[6pt] &= \frac{-140}{84} \\[6pt] \textbf{L.H.S } &= \frac{-5}{3} \\[6pt] \textbf{R.H.S } &= \frac{-5}{3} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{-4}{3}\)

Solution

\[ \begin{aligned} \frac{\frac{x}{3} - \frac{2}{5}}{\frac{3}{4} - 2x} &= \frac{16}{15} \\[6pt] 15 \times \left( \frac{x}{3} - \frac{2}{5} \right) &= 16 \times \left( \frac{3}{4} - 2x \right) \\[6pt] 5x - 6 &= 12 - 32x \\[6pt] 5x + 32x &= 12 + 6 \\[6pt] 37x &= 18 \\[6pt] \color{green} x &= \color{green} \frac{18}{37} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{\frac{x}{3} - \frac{2}{5}}{\frac{3}{4} - 2x} \\[6pt] &= \frac{\frac{1}{3}(\frac{18}{37}) - \frac{2}{5}}{\frac{3}{4} - 2(\frac{18}{37})} \\[6pt] &= \frac{\frac{6}{37} - \frac{2}{5}}{\frac{3}{4} - \frac{36}{37}} \\[6pt] &= \frac{\frac{30 - 74}{185}}{\frac{111 - 144}{148}} \\[6pt] &= \frac{\frac{-44}{185}}{\frac{-33}{148}} \\[6pt] &= \frac{\cancel{44}^4}{\cancel{185}_5} \times \frac{\cancel{148}^4}{\cancel{33}_3} \\[6pt] &= \frac{4}{5} \times \frac{4}{3} \\[6pt] \textbf{L.H.S } &= \frac{16}{15} \\[6pt] \textbf{R.H.S } &= \frac{16}{15} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{18}{37}\)

Solution

\[ \begin{aligned} \frac{(1 - 2x) + (1 + 2x)}{(4x + 1) + (x - 3)} &= \frac{1}{2} \\[6pt] \frac{1 - 2x + 1 + 2x}{4x + 1 + x - 3} &= \frac{1}{2} \\[6pt] \frac{2}{5x - 2} &= \frac{1}{2} \\[6pt] 2 \times 2 &= 1 \times (5x - 2) \\[6pt] 4 &= 5x - 2 \\[6pt] 4 + 2 &= 5x \\[6pt] 6 &= 5x \\[6pt] \color{green} x &= \color{green} \frac{6}{5} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{(1 - 2x) + (1 + 2x)}{(4x + 1) + (x - 3)} \\[6pt] &= \frac{2}{5x - 2} \\[6pt] &= \frac{2}{5(\frac{6}{5}) - 2} \\[6pt] &= \frac{2}{6 - 2} \\[6pt] &= \frac{2}{4} \\[6pt] \textbf{L.H.S } &= \frac{1}{2} \\[6pt] \textbf{R.H.S } &= \frac{1}{2} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{6}{5}\)

Solution

\[ \begin{aligned} \frac{x^2 - (x + 2)(x + 3)}{7x + 1} &= \frac{2}{3} \\[6pt] \frac{x^2 - (x^2 + 5x + 6)}{7x + 1} &= \frac{2}{3} \\[6pt] \frac{x^2 - x^2 - 5x - 6}{7x + 1} &= \frac{2}{3} \\[6pt] \frac{-5x - 6}{7x + 1} &= \frac{2}{3} \\[6pt] 3 \times (-5x - 6) &= 2 \times (7x + 1) \\[6pt] -15x - 18 &= 14x + 2 \\[6pt] -15x - 14x &= 2 + 18 \\[6pt] -29x &= 20 \\[6pt] \color{green} x &= \color{green} \frac{-20}{29} \\ \\ &\color{magenta} \textbf{Check } \\[6pt] \textbf{L.H.S } &= \frac{-5x - 6}{7x + 1} \\[6pt] &= \frac{-5(\frac{-20}{29}) - 6}{7(\frac{-20}{29}) + 1} \\[6pt] &= \frac{\frac{100}{29} - 6}{\frac{-140}{29} + 1} \\[6pt] &= \frac{\frac{100 - 174}{29}}{\frac{-140 + 29}{29}} \\[6pt] &= \frac{\frac{-74}{29}}{\frac{-111}{29}} \\[6pt] &= \frac{74}{29} \times \frac{29}{111} \\[6pt] &= \frac{74}{111} \\[6pt] \textbf{L.H.S } &= \frac{2}{3} \\[6pt] \textbf{R.H.S } &= \frac{2}{3} \\ \\ L.H.S &= R.H.S \\ \text{Hence } & \text{proved} \end{aligned} \]

Answer \( \color{red} x = \dfrac{-20}{29}\)

Solution

\[ \begin{aligned} \frac{3-x^2}{8+x^2} &= \frac{-3}{4} \\[6pt] \textbf{Put } x^2 = y & \ (\text{in above equaiton}) \\[6pt] \frac{3-y}{8+y} &= \frac{-3}{4} \\[6pt] 4 (3-y) &= -3 (8+y) \\[6pt] 12 - 4y &= -24 - 3y \\[6pt] - 4y + 3y &= -24 - 12 \\[6pt] - y &= -36 \\[6pt] y &= 36 \\[6pt] x^2 &= 36 \ \ \ [\therefore {\color{magenta} y = x^2} ] \\[6pt] x &= \sqrt{36} \\[6pt] \color{green} x &= \color{green} 6 \end{aligned} \]

Answer \( \color{red} x = 6 \)

Solution

\[ \begin{aligned} \frac{y^2+6}{8y^2+3} &= \frac{1}{5} \\[6pt] \textbf{Put } y^2 = a & \ (\text{in above equaiton}) \\[6pt] \frac{a+6}{8a+3} &= \frac{1}{5} \\[6pt] 5 (a+6) &= 1 (8a+3) \\[6pt] 5a + 30 &= 8a + 3 \\[6pt] 30 - 3 &= 8a - 5a \\[6pt] 27 &= 3a \\[6pt] a &= \frac{27}{3} \\[6pt] a &= 9 \\[6pt] y^2 &= 9 \ \ \ [{ \therefore \color{magenta} a = y^2}] \\[6pt] y &= \sqrt{9} \\[6pt] \color{green} y &= \color{green} 3 \end{aligned} \]

Answer \( \color{red} y = 3 \)

Solution

\[ \begin{aligned} \frac{x^2-9}{5+x^2} &= \frac{5}{9} \\[6pt] \textbf{Put } x^2 = b & \ (\text{in above equaiton}) \\[6pt] \frac{b-9}{5+b} &= \frac{5}{9} \\[6pt] 9 (b-9) &= 5 (5+b) \\[6pt] 9b - 81 &= 25 + 5b \\[6pt] 9b - 5b &= 25 + 81 \\[6pt] 4b &= 106 \\[6pt] b &= \frac{106}{4} \\[6pt] b &= \frac{53}{2} \\[6pt] x^2 &= \frac{53}{2} \ \ \ [\therefore {\color{magenta} b = x^2}] \\[6pt] \color{green} x &= \color{green} \sqrt{\frac{53}{2}} \end{aligned} \]

Answer \( \color{red} x = \sqrt{\dfrac{53}{2}} \)

Solution

\[ \begin{aligned} \frac{y^2+4}{3y^2+7} &= \frac{1}{2} \\[6pt] \textbf{Put } y^2 = m & \ (\text{in above equaiton}) \\[6pt] \frac{m+4}{3m+7} &= \frac{1}{2} \\[6pt] 2 (m+4) &= 1 (3m+7) \\[6pt] 2m + 8 &= 3m + 7 \\[6pt] 8 - 7 &= 3m - 2m \\[6pt] 1 &= m \\[6pt] m &= 1 \\[6pt] y^2 &= 1 \ \ \ [\therefore {\color{magenta} m = y^2}] \\[6pt] y &= \sqrt{1} \\[6pt] \color{green} y &= \color{green} 1 \end{aligned} \]

Answer \( \color{red} y = 1 \)