Solve the equation (1/3)(x - 3) - [(x + 1)/3 - (3 + x)/3] = 1/3 - (2 - x)/3 + x/3 + 1.

Step-by-step solution

1. 1

   Simplify the bracket first. Both fractions inside it already have denominator $3$, so combine the numerators: $\frac{(x+1)-(3+x)}{3}=\frac{x+1-3-x}{3}=-\frac{2}{3}$. The $x$ terms cancel, leaving a pure constant.

2. 2

   Finish the left side. Subtracting $-\frac23$ means adding it: $\frac{x-3}{3}+\frac{2}{3}=\frac{x-3+2}{3}=\frac{x-1}{3}$.

3. 3

   Combine the three thirds on the right. $\frac{1-(2-x)+x}{3}=\frac{1-2+x+x}{3}=\frac{2x-1}{3}$. Watch the sign: subtracting $(2-x)$ contributes $-2+x$, not $-2-x$.

4. 4

   Absorb the trailing $+1$. Write $1=\frac33$, so the right side becomes $\frac{2x-1+3}{3}=\frac{2x+2}{3}$.

5. 5

   Clear the common denominator and solve. The equation $\frac{x-1}{3}=\frac{2x+2}{3}$ multiplies out to $x-1=2x+2$. Collecting terms gives $-x=3$, so $x=-3$.

6. 6

   Check the answer. At $x=-3$ the left side is $\frac{1}{3}(-6)-\left(\frac{-2}{3}-\frac{0}{3}\right)=-2+\frac23=-\frac43$, and the right side is $\frac13-\frac{5}{3}-1+1=-\frac43$. Both sides equal $-\frac43$.