To find the equation of a line that passes through the point \((-2, -5)\) and is parallel to the line given by the equation \(y = x + 3\), we first note that parallel lines have the same slope.
The slope of the line \(y = x + 3\) is 1 (since it can be written in the slope-intercept form \(y = mx + b\), where \(m\) is the slope).
Since the new line is parallel, it also has a slope \(m = 1\).
Now, we can 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. Plugging in the point \((-2, -5)\) and the slope \(1\):
\[ y - (-5) = 1(x - (-2)) \]
Simplifying this:
\[ y + 5 = 1(x + 2) \] \[ y + 5 = x + 2 \] \[ y = x + 2 - 5 \] \[ y = x - 3 \]
Thus, the equation of the line that passes through the point \((-2, -5)\) and is parallel to the line \(y = x + 3\) is:
\[ y = x - 3 \]