Question

[(6x +1) ÷ 3] +1 = (x-3) ÷ 6 solve and check​

Answer 1

$\huge\mathbb{SOLUTION}$

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

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

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

$\implies\sf \frac{12x+2-x+3}{6}=-1$

$\implies\sf \frac{11x+5}{6}=-1$

$\implies\sf 11x+5=-6$

$\implies\sf 11x=-11$

$\implies\sf x=\cancel\frac{-11}{11}=-1$

Checking:-

Substitute the value of x

LHS

$\implies\sf \frac{6x+1}{3}+1$

$\implies\sf \frac{6×(-1)+1}{3}+1$

$\implies\sf \frac{-5}{3}+1$

$\implies\sf \frac{-5+3}{3}$

$\implies\sf \frac{-2}{3}$

RHS

$\implies\sf \frac{(x-3)}{6}$

$\implies\sf \frac{(-1-3)}{6}$

$\implies\sf \cancel\frac{-4}{6}$

$\implies\sf \frac{-2}{3}$

Hence, LHS = RHS checked

Answer 2

$\bold\orange{\underline{\underline{:SOLUTION:}}}$

$⟶(6x + 1) \div3 ) + 1 = (x - 3) \div 6 \\ \\ ⟶ \frac{6x + 1}{3} + 1 = \frac{x - 3}{6} \\ \\ ⟶ \frac{6x + 1}{3} - \frac{x - 3}{6} = - 1 \\ \\ ⟶ \frac{12x + 2 - x + 3}{6} = - 1 \\ \\ ⟶ \frac{11x + 5}{6} = - 1 \\ \\ ⟶11x + 5 = - 6 \\ \\⟶11x = - 6 - 5 \\ \\ ⟶11x = - 11 \\ \\ ⟶x = - 1$

☆$\bold\purple{\boxed{<em>VERI</em><em>FICATION</em>}}$☆

• here putting the value of X=-1

$⟶ \frac{6x + 1}{3} + 1 = \frac{x - 3}{6} \\ \\ ⟶ \frac{6 \times ( - 1) + 1}{3} + 1 = \frac{ - 1 - 3}{6} \\ \\ ⟶ \frac{ - 6 + 1}{3} + 1 = \frac{ - 4}{6} \\ \\ ⟶ \frac{ - 5}{3} + 1 = \frac{ - 4}{6} \\ \\ ⟶ \frac{ - 5 + 3}{3} = \frac{ - 4}{6} \\ \\ ⟶ \frac{ - 2}{3} = \frac{ - 2}{3}$

☆HENCE, VERIFIED☆