The given equation of a line is,
[tex]5x-y=-3[/tex]The above equation can be rewritten as,
[tex]y=5x+3\text{ -----(1)}[/tex]The general equation of a line in slope-intercept form is given by,
[tex]y=mx+c\text{ ------(2)}[/tex]Here, m is the slope of the line and c is the y intercept.
Comparing equations (1) and (2), we get slope m=5.
We have to find the equation of a line parallel to the given line 5x-y=-3 and passing through point (x1, y1)=(-3,2).
The slopes of two parallel lines are always equal. Hence, the slope of a line parallel to 5x-y=-3 is m=5.
Now, the point-slope form of a line with slope m=5 and passing through point (x1, y1)=(-3,2) can be written as,
[tex]\begin{gathered} m=\frac{y1-y}{x1-x} \\ 5=\frac{2-y}{-3-x} \end{gathered}[/tex]Rearrange the above equation in slope intercept form.
[tex]\begin{gathered} 5(-3-x)=2-y \\ 5\times(-3)-5x=2-y \\ -15-5x=2-y \\ y=5x+2+15 \\ y=5x+17 \end{gathered}[/tex]Therefore, the equation of a line parallel to the given line 5x-y=-3 and passing through point (-3,2) is y=5x+17.