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[tex]\qquad\qquad\huge\underline{{\sf Answer}}☂[/tex]

Let's solve ~

[tex]\qquad \sf  \dashrightarrow \: \dfrac{6x + 1}{3} + 1 = \dfrac{x - 3}{6} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{6x + 1}{3} - \dfrac{x - 3}{6} =- 1[/tex]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{12x + 2 - x + 3}{6} =- 1[/tex]

[tex]\qquad \sf  \dashrightarrow \: \dfrac{11x + 5}{6} =- 1[/tex]

[tex]\qquad \sf  \dashrightarrow \: {11x + 5}{} =- 6[/tex]

[tex]\qquad \sf  \dashrightarrow \: 11x = -6 - 5[/tex]

[tex]\qquad \sf  \dashrightarrow \: x = -1 [/tex]

davidmaharjan62

Step-by-step explanation:

[tex] \frac{6x + 1}{3} + 1 = \frac{x - 3}{6} \\ \frac{6x + 1 + 3}{3} = \frac{x - 3}{6} \\ 6( 6x + 4) = 3(x - 3) \\ 36x + 24 = 3x - 9 \\ 36x - 3x = - 9 - 24 \\ 33x = - 33 \\ x = - 1[/tex]

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