Question

plz solve the following equation

Answer 1

$\underline \bold {Solution:-}$

$(i) \: \frac{ \frac{3}{7}z - \frac{1}{2} }{z - \frac{1}{4} } - \frac{3}{14} = \frac{1}{7} \\ \\ \frac{ \frac{6z - 7}{14} }{ \frac{4z - 1}{4} } = \frac{1}{7} + \frac{3}{14} \\ \\ \frac{ \frac{6z - 7}{14} }{ \frac{4z - 1}{4} } = \frac{2 + 3}{14} \\ \\ \frac{4(6z - 7)}{14(4z - 1)} = \frac{5}{14} \\ \\ on multiplying \: both \: sides \: by \: 14 \\ \\ \frac{4(6z - 7)}{14(4z - 1)} \times 14 = \frac{5}{14} \times 14 \\ \\on \: cross \: multiplication \\ \\ 24z - 28 = 20z - 5 \\ \\ 24z - 20z = 28 - 5 \\ \\ 4z = 23 \\ \\ z = \frac{23}{4}$

Answer 2

$(v) \: \frac{ \frac{3}{7}z - \frac{1}{2} }{z - \frac{1}{4} } - \frac{3}{14} = \frac{1}{7} \\ \\ \frac{ \frac{3z \times 2 - 1 \times 7}{7 \times 2} }{ \frac{z \times 4 - 1 \times 1}{4} } = \frac{1}{7} + \frac{3}{14} \\ \\ \frac{ \frac{6z - 7}{14} }{ \frac{4z - 1}{4} } = \frac{1 \times 2 + 3 \times 1}{14} \\ \\ \frac{6z - 7}{14} \times \frac{4}{4z - 1} = \frac{2 + 3}{14} \\ \\ \frac{2(6z - 7)}{7(4z - 1)} = \frac{5}{14} \\ \\ \frac{12z - 14}{28z - 7} = \frac{5}{14} \\ \\ \bf On \: Cross \: Multiplying, \: we \: get \\ \\ 14(12z - 14) = 5(28z - 7) \\ \\ 168z - 196 = 140z - 35 \\ \\ 168z - 140z = 196 - 35 \\ \\ 28z = 161 \\ \\ z = \frac{161}{28} \\ \\ z = \frac{23}{4}$

$(vii) \: \: \frac{(x + 1)(4x - 3) - 4 {x}^{2} + 5 }{4x + 1} = 2 \\ \\ \frac{4 {x}^{2} - 3x + 4x - 3 - 4 {x}^{2} + 5}{4x + 1} = 2 \\ \\ \frac{4 {x}^{2} - 4 {x}^{2} + x + 5 }{4x + 1} = 2 \\ \\ \frac{x + 5}{4x + 1} = \frac{2}{1} \\ \\ \bf On \: Cross \: multiplying \: we \: get \\ \\ x + 5 = 2(4x + 1) \\ \\ x + 5 = 8x + 2 \\ \\ 8x - x = 5 - 2 \\ \\ 7x = 3 \\ \\ x = \frac{3}{7}$

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