Question

Solve:-
$\frac{x+2}{6} -[\frac{11-x}{3} -\frac{1}{4} ]=\frac{3x-4}{12}$


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Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt Solve:-$

$\sf \bf \dfrac{x+2}{6} -\bigg[\dfrac{11-x}{3} -\dfrac{1}{4}\bigg ]=\dfrac{3x-4}{12}$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf\implies \dfrac{x+2}{6} -\bigg[\dfrac{11-x}{3} -\dfrac{1}{4} \bigg]=\dfrac{3x-4}{12}$

$\tt\underline {By\: taking \:LCM}$

$\sf \bf\implies \dfrac{x+2}{6} -\bigg[\dfrac{44-4x-3}{12} \bigg]=\dfrac{3x-4}{12}$

$\sf \bf\implies \dfrac{x+2}{6} -\bigg[\dfrac{41-4x}{12}\bigg] =\dfrac{3x-4}{12}$

$\tt\underline {By\: taking \:LCM}$

$\sf \bf\implies \dfrac{2x+4-41+4x}{12}=\dfrac{3x-4}{12}$

$\sf \bf\implies \dfrac{6x-37}{\cancel {12}} =\dfrac{3x-4}{\cancel {12}}$

$\sf \bf \implies 6x-37=3x-4$

$\sf \bf \implies 6x-3x=37-4$

$\sf \bf \implies 3x=33$

$\sf \bf \implies x=\dfrac{33}{3}$

$\implies {\boxed {\color{blue} {\sf \bf x=11}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\tt {\underbrace {\overbrace {Identities}}}$

$\sf \bf {(a+b)}^{2}={a}^{2}+2ab+{b}^{2}$

$\sf \bf {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}$

$\sf \bf (a+b) (a-b) ={a}^{2}-{b}^{2}$

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Answer 2

★Given:-

$\displaystyle \mapsto \sf \frac{x+2}{6} - \bigg(\frac{11-x}{3} - \frac{1}{4} \bigg)=\frac{3x-4}{12}$

★To find:-

• Value of x.

★Solution:-

$\displaystyle :\implies \sf \frac{x+2}{6} - \bigg(\frac{11-x}{3} - \frac{1}{4} \bigg)=\frac{3x-4}{12}$

$\displaystyle :\implies \sf \frac{x+2}{6} -\bigg(\frac{44-4x-3}{12} \bigg) = \frac{3x-4}{12} \\\\\\:\implies \sf \frac{x+2}{6} -\bigg(\frac{41-4x}{12} \bigg)=\frac{3x-4}{12} \\\\\\:\implies \sf \frac{6x-37}{\cancel{12}} = \frac{3x-4}{\cancel{12}} \\\\\\:\implies \sf 6x-37=3x-4 \\\\\\:\implies \sf 6x - 3x = 37 - 4\\\\\\:\implies \sf 3x = 33\\\\\\:\implies \underline{\boxed{\sf x = 11.}}$

Hence,

The value of x is 11.

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