Question

solve :- (x+2)/6 - [(11-x)/3 - 1/4] =
(3x-4)/12

Answer 1

first open brackets

=>(x+2)/6-11-x/3+1/4=3x-4/12
=>now we have to take lcm of left hand side
=>Lcm=12
=>[2(x+2)-4(11-x)+3(1)]/12
=>now,
=>(2x+4-44-4x+4)/12
=>(-2x-37)/12
=>now whole equation
=>(-2x-37)/12=(3x-4)/12
=>we will cancel out 12 which is on both side
=>-2x-37=3x-4
=>x=41
Answer
Mark this as brainliest thanks for patience!

Answer 2

Step-by-step explanation:

$\mathsf{Given : \dfrac{x + 2}{6} - \bigg(\dfrac{11 - x}{3} - \dfrac{1}{4}\bigg) = \dfrac{3x - 4}{12}}$

$\mathsf{\implies \dfrac{x + 2}{6} - \bigg(\dfrac{4(11 - x) - 3}{12}\bigg) = \dfrac{3x - 4}{12}}$

$\mathsf{\implies \dfrac{x + 2}{6} - \bigg(\dfrac{44 - 4x - 3}{12}\bigg) = \dfrac{3x - 4}{12}}$

$\mathsf{\implies \dfrac{x + 2}{6} - \bigg(\dfrac{41 - 4x}{12}\bigg) = \dfrac{3x - 4}{12}}$

$\mathsf{\implies \bigg(\dfrac{x + 2}{6} - \dfrac{(41 - 4x)}{12}\bigg) = \dfrac{3x - 4}{12}}$

$\mathsf{\implies \dfrac{2(x + 2) - (41 - 4x)}{12} = \dfrac{3x - 4}{12}}$

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

$\mathsf{\implies \dfrac{-37 + 6x}{12} = \dfrac{3x - 4}{12}}$

$\mathsf{\implies {-37 + 6x} = {3x - 4}}$

$\mathsf{\implies 6x - 3x = 37 - 4}$

$\mathsf{\implies 3x = 33}$

$\mathsf{\implies x = \dfrac{33}{3}}$

$\mathsf{\implies x = 11}$