Question

(x+2)/6-{(11-x)/3-1/4}=(3x-4)/9

Answer 1

Answer: By solving the equation (x+2)/6-{(11-x)/3-1/4}=(3x-4)/9 , the value of x  is 16.84

Step-by-step explanation:

(x+2)/6-{(11-x)/3-1/4}=(3x-4)/9

first we solve the bracket

⇒ (x+2)/6 - [4(11 - x) -3 ] /12 = (3x - 4)/9

⇒ (x+2)/6- [44 - 4x -3 ] / 12 = (3x - 4)/9

⇒ (x+2)/6 - [ 41- 4x]/12 = (3x - 4)/9

Taking LCM  of 6 and 12 is 12

⇒ [2(x+2) - 41 + 4x ] / 12 =  (3x - 4)/9

⇒ [2x +2 -41 +4x ]/12 = (3x - 4)/9

⇒ [ 6x -39 ]/12 = (3x-4) / 9

By cross multiplication , we get

9 ( 6x-39 ) = 12 ( 3x-4)

54x - 351 = 36x - 48

54x - 36x = -48 + 351

18x = 303

x = 16.84

To know more about multiplication from the given link

https://brainly.in/question/51354985

To know more about LCM from the given link

https://brainly.in/question/54179983

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Answer 2

EXPLANATION.

$\sf \displaystyle \frac{(x + 2)}{6} - \bigg[\frac{(11 - x)}{3} - \frac{1}{4} \bigg] = \frac{(3x - 4)}{9}$

We can write expression as,

$\sf \displaystyle \frac{(x + 2)}{6} - \bigg[\frac{4(11 - x)- 3}{12} \bigg] = \frac{(3x - 4)}{9}$

$\sf \displaystyle \frac{(x + 2)}{6} - \bigg[\frac{44 - 4x - 3}{12} \bigg] = \frac{(3x - 4)}{9}$

$\sf \displaystyle \frac{(x + 2)}{6} - \bigg[\frac{41 - 4x}{12} \bigg] = \frac{(3x - 4)}{9}$

$\sf \displaystyle \frac{2(x + 2) - (41 - 4x)}{12} = \frac{(3x - 4)}{9}$

$\sf \displaystyle \frac{2x + 4 - 41 + 4x}{12} = \frac{(3x - 4)}{9}$

$\sf \displaystyle \frac{6x - 37 }{12} = \frac{(3x - 4)}{9}$

$\sf \displaystyle \frac{(6x - 37) }{4} = \frac{(3x - 4)}{3}$

$\sf \displaystyle 3(6x - 37) = 4(3x - 4)$

$\sf \displaystyle 18x - 111 = 12x - 16$

$\sf \displaystyle 18x - 12x = - 16 + 111$

$\sf \displaystyle 6x = 95$

$\sf \displaystyle \boxed{x = \frac{95}{6} }$

∴ The value of x is equal to 95/6.