Question

solve each of the following equations.
$\frac{1}{6} (1 - x) + \frac{2}{3} x - \frac{1}{4} (1 - 7x) = \frac{13}{6}$
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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

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$\red{\underline \bold{To \: Find:}}$

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$\frac{1}{6} (1 - x) + \frac{2}{3} x - \frac{1}{4} (1 - 7x) = \frac{13}{6}$

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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⇛ $\sf \large{ \frac { 1 } { 6 } - \frac { x } { 6 } + \frac { 2x } { 3 } - \frac { 1 } { 4 } + \frac { 7x } { 4 } = \frac { 13 } { 6 }}$

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$\underline{\bold{\texttt{Taking LCM = 12 }}}$

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⇛ $\sf \large{ \frac { 2} { 12} - \frac { 2x } { 12} + \frac { 8x } { 12} - \frac { 3 } { 12} + \frac { 21x } { 12 } = \frac { 13 } { 6 }}$

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⇛ $\sf \large{ \frac { 2 - 2x + 8x - 3 + 21x } { 12 } = \frac { 13 } { 6 }}$

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⇛ $\sf \large{ \frac { -1 + 27x } { 12 } = \frac { 13 } { 6 }}$

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$\underline{\bold{\texttt{Cross multiplying }}}$

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⇛ $\tt \large{ -6 + 162x = 156 }$

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⇛ $\sf \large{ 162x = 162 }$

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⇛ $\tt \large{ \boxed{x = 1} }$

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Hence we got that the value of 'x' is 1