Question

plz answer this question​

Answer 1

Answer:

Above is your answer.......

Answer 2

Answer :-

Value of x = -1.

Solution :-

$\implies\sf{ \dfrac{6x + 1}{3} + 1 = \dfrac{x - 3}{6} }$

$\implies\sf{ \dfrac{6x + 1}{3} + \dfrac{1}{1} = \dfrac{x - 3}{6} }$

$\implies\sf{ \dfrac{(6x + 1) + 3}{3}= \dfrac{x - 3}{6} }$

$\implies\sf{ \dfrac{6x + 4}{3}= \dfrac{x - 3}{6} }$

$\implies\sf{ \dfrac{6x}{3} + \dfrac{4}{3} = \dfrac{x}{6} - \dfrac{3}{6} }$

$\implies\sf{2x+ \dfrac{4}{3} = \dfrac{x}{6} - \dfrac{1}{2} }$

$\implies\sf{2x - \dfrac{x}{6} = - \dfrac{1}{2} - \dfrac{4}{3} }$

$\implies\sf{ \dfrac{12x - x}{6} = \dfrac{ - 3 - 8}{6} }$

$\implies\sf{ \dfrac{11x}{6} = \dfrac{ - 11}{6} }$

$\implies\sf{11x = - 11x}$

$\implies\boxed{\sf\red{x = - 1}}$

Therefore, Value of x = -1.

Verification :-

$\implies\sf{ \dfrac{6x + 1}{3} + 1 = \dfrac{x - 3}{6} }$

$\implies\sf{ \dfrac{6( - 1)+ 1}{3} + 1 = \dfrac{ - 1 - 3}{6} }$

$\implies\sf{ \dfrac{ - 5}{3} + 1 = \dfrac{ - 4}{6} }$

$\implies\sf{ \dfrac{ - 5 + 3}{3} = \dfrac{ - 2}{3} }$

$\implies\sf{ \dfrac{ - 2}{3}= \dfrac{ - 2}{3} }$

LHS = RHS.

Hence, Verified.