Question

$(1 - x) \div 4 + (3x - 1) \div 6 = (x \div 2) - (7 \div 6)$
​please answer.. its urgent

Answer 1

Answer:

Step-by-step explanation:

(1-x)÷4 + (3x-1) ÷6 = (x÷2) - (7÷6)

(6-6x + 12x-4 ) ÷24 = (3x-7) ÷6        <LCM>

(2+6X) ÷24 = (3X-7)÷6

(2+6x)÷ (3x-7) = 4

2+6x = 12x-28

2+28 = 12x-6x

30÷6 = x

⇒ x = 5

regards

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

As, it is given that,

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

So, we can solve this question by taking the L.C.M of all the terms.

$\sf{→ \frac{(1 - x)}{4} + \frac{(3x - 1)}{6} = \frac{x}{2} - \frac{7}{6}} \\ \\ \bf{Taking \: L.C.M} \\ \\ \sf{→\frac{3(1 - x) + 2(3x - 1)}{12} = \frac{3x - 7}{6}} \\ \\ \sf{→\frac{3 - 3x + 6x - 2}{12} = \frac{3x - 7}{6}} \\ \\ \sf{→ \frac{3x + 1}{\cancel{12}} = \frac{3x - 7}{\cancel{6}}} \\ \\ \sf{→\frac{3x + 1}{2} = 3x - 7} \\ \\ \sf{→3x + 1 = 2(3x - 7)} \\ \\ \sf{→3x + 1 = 6x - 14} \\ \\ \sf{→6x - 3x = 1 + 14} \\ \\ \sf{→3x = 15} \\ \\ \sf{→x = \frac{\cancel{15}}{\cancel{3}}} \\ \\ \sf{→x = 5} \\ \\ \Large{\implies{\boxed{\boxed{\sf{x = 5}}}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\put(1,1.2){\line(1,0){6.5}}\end{picture}$

$\setlength{\unitlength}{0.19 cm}}\begin{picture}(12,4)\thicklines\put(12,40){ \#\:STAY\:HOME }\put(11,35){ \#\:STAY\: SAFETY }\put(20,20){\circle{10}}\put(19.5,21){\line(-1,0){1.5}}\put(20.5,21){\line(1,0){1.5}}\put(18.8,20.3){\circle*{0.7}}\put(21.1,20.3){\circle*{0.7}}\put(20,20.4){\line(0,-1){1.4}}\put(20.7,18.4){\line(-1,0){1.57}}\put(18.6,16.56){\line(0,-1){2}}\put(21.4,16.56){\line(0,-1){2}}\put(19,14.55){\line(-1,0){5}}\put(21,14.55){\line(1,0){5}}\put(14,14.55){\line(0,-1){16}}\put(26,14.55){\line(0,-1){16}}\put(14,14.55){\line(-4,3){5}}\put(26,14.55){\line(4,3){5}}\put(14,11){\line(-4,3){7.5}}\put(26,11){\line(4,3){7.5}}\put(9,18.3){\line(0,1){8.3}}\put(31,18.3){\line(0,1){8.3}}\put(33.5,16.6){\line(0,1){10}}\put(6.5,16.6){\line(0,1){10}}\put(3.5,26.5){\line(1,0){33}}\put(3.5,26.5){\line(0,1){20}}\put(3.5,46.5){\line(1,0){33}}\put(36.5,26.5){\line(0,1){20}}\put(14,-0.6){\line(1,0){12}}\put(14,-1.5){\line(1,0){12}}\put(14.4,-1.6){\line(0,-1){18.5}}\put(25.6,-1.6){\line(0,-1){18.5}}\put(20,-1.6){\line(0,-1){3}}\put(20,-4.6){\line(-1,-6){2.6}}\put(20,-4.6){\line(1,-6){2.6}}\put(14.4,-20){\line(-5,-6){3}}\put(25.6,-20){\line(5,-6){3}}\put(17.3,-20){\line(0,-1){3.7}}\put(17.3,-20){\line(-1,0){3}}\put(17.3,-23.6){\line(-1,0){6}}\put(22.6,-20){\line(0,-1){3.7}}\put(22.6,-20){\line(1,0){3}}\put(22.6,-23.6){\line(1,0){6}}\end{picture}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\put(1,1.2){\line(1,0){6.5}}\end{picture}$