rough: \( (-2,1) \), parallel to \( y=3 x-1 \)

To find the equation of a line that is **parallel** to \( y = 3x - 1 \) and **passes through the point** \( (-2, 1) \), follow these steps: 1. **Determine the slope of the given line:** The equation \( y = 3x - 1 \) is in the slope-intercept form \( y = mx + b \), where \( m \) is the slope. So, the slope \( m \) of the given line is **3**. 2. **Use the slope-point form for the new line:** Since parallel lines have the **same slope**, the new line will also have a slope of **3**. The point-slope form of a line is: \[ y - y_1 = m(x - x_1) \] where \( (x_1, y_1) \) is a point on the line, and \( m \) is the slope. Plugging in the slope and the given point \( (-2, 1) \): \[ y - 1 = 3(x - (-2)) \\ y - 1 = 3(x + 2) \] 3. **Simplify the equation to slope-intercept form:** Distribute the slope on the right side: \[ y - 1 = 3x + 6 \] Add **1** to both sides to solve for \( y \): \[ y = 3x + 7 \] **Final Equation:** \[ y = 3x + 7 \]