\( \llbracket \frac { | x - 1 | - 1 } { 3 } ] = 2 \)

To solve the equation \[ \left\llbracket \frac{|x - 1| - 1}{3} \right\rrbracket = 2, \] we first need to understand what the notation \(\llbracket y \rrbracket\) means. This notation typically represents the greatest integer less than or equal to \(y\), also known as the floor function, denoted as \(\lfloor y \rfloor\). Therefore, we can rewrite the equation as: \[ \left\lfloor \frac{|x - 1| - 1}{3} \right\rfloor = 2. \] This means that \[ 2 \leq \frac{|x - 1| - 1}{3} < 3. \] We can solve this compound inequality step by step. 1. **Multiply the entire inequality by 3** (since 3 is positive, the direction of the inequalities remains the same): \[ 6 \leq |x - 1| - 1 < 9. \] 2. **Add 1 to all parts of the inequality**: \[ 7 \leq |x - 1| < 10. \] Now we can break this down into two separate inequalities: ### First Inequality: \( |x - 1| \geq 7 \) This implies two cases: - Case 1: \( x - 1 \geq 7 \) \[ x \geq 8. \] - Case 2: \( x - 1 \leq -7 \) \[ x \leq -6. \] ### Second Inequality: \( |x - 1| < 10 \) This also implies two cases: - Case 1: \( x - 1 < 10 \) \[ x < 11. \] - Case 2: \( x - 1 > -10 \) \[ x > -9. \] ### Combining the Results Now we combine the results from both inequalities: 1. From \( |x - 1| \geq 7 \): - \( x \geq 8 \) or \( x \leq -6 \). 2. From \( |x - 1| < 10 \): - \( x < 11 \) and \( x > -9 \). Now we analyze the intervals: - For \( x \geq 8 \) and \( x < 11 \): \[ 8 \leq x < 11. \] - For \( x \leq -6 \) and \( x > -9 \): \[ -9 < x \leq -6. \] ### Final Solution Thus, the solution set for the original equation is: \[ x \in [8, 11) \cup (-9, -6]. \]