Write the equation of a line that is parallel to the line \(y = -x + 2\) and passes through the point (-2, 3).

To find the equation of a line that is parallel to the line given by \(y = -x + 2\), we first need to identify the slope of the given line. The equation is in slope-intercept form \(y = mx + b\), where \(m\) is the slope. From the equation \(y = -x + 2\), we can see that the slope \(m\) is \(-1\). Since parallel lines have the same slope, the slope of the line we want to find will also be \(-1\). Next, we will use the point-slope form of the equation of a line, which is given by: \[ y - y_1 = m(x - x_1) \] where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. We have the point \((-2, 3)\) and the slope \(-1\). Substituting these values into the point-slope form: \[ y - 3 = -1(x - (-2)) \] This simplifies to: \[ y - 3 = -1(x + 2) \] Distributing the \(-1\): \[ y - 3 = -x - 2 \] Now, we add \(3\) to both sides to solve for \(y\): \[ y = -x - 2 + 3 \] This simplifies to: \[ y = -x + 1 \] Thus, the equation of the line that is parallel to \(y = -x + 2\) and passes through the point \((-2, 3)\) is: \[ \boxed{y = -x + 1} \]