Solve the following absolute value inequality. Write the solution set using interval notation and graph it. \[ \left|\frac{x-4}{3}\right|>1 \]

To solve the absolute value inequality: \[ \left|\frac{x-4}{3}\right| > 1 \] we can break it down into two separate inequalities: 1. **Case 1:** \(\frac{x-4}{3} > 1\) 2. **Case 2:** \(\frac{x-4}{3} < -1\) Let's solve each case individually. ### Case 1: \(\frac{x-4}{3} > 1\) Multiply both sides by 3 to isolate \(x\): \[ x - 4 > 3 \] Add 4 to both sides: \[ x > 7 \] ### Case 2: \(\frac{x-4}{3} < -1\) Multiply both sides by 3: \[ x - 4 < -3 \] Add 4 to both sides: \[ x < 1 \] ### Combining Both Cases The solution set consists of all real numbers less than 1 or greater than 7. In interval notation, this is: \[ (-\infty, 1) \cup (7, \infty) \] ### Graphing the Solution Here's how the solution looks on a number line: \[ \begin{array}{c|cccccccccc} x & \cdots & -2 & -1 & 0 & \mathbf{1} & 2 & \cdots & 6 & \mathbf{7} & 8 & \cdots \\ \hline \text{Solution} & \bullet & \longleftarrow & & & & & & \longrightarrow & \bullet & \\ \end{array} \] - **Open circles** at \(x = 1\) and \(x = 7\) indicate that these points are not included in the solution. - **Shading** extends to the left of 1 and to the right of 7, representing all values \(x < 1\) and \(x > 7\). ### Final Answer \[ \boxed{\,(-\infty,\ 1)\ \cup\ (7,\ \infty)\,} \] **Graph:** \[ \begin{array}{c|cccccccccc} & \cdots & 1 & \cdots & 7 & \cdots \\ \hline \text{Number Line} & \text{<----} & \circ & \text{----} & \circ & \text{---->} \end{array} \]