Solution:
Given that;
A line is parallel to y=x-3 and contains the point that (3,-2).
To find the equation of the line, the slope-intercept form of a line is
[tex]\begin{gathered} y=mx+c \\ Where\text{ } \\ m\text{ is the slope} \\ c\text{ is the y-intercept} \end{gathered}[/tex]The slope of the given equation is
[tex]\begin{gathered} y=x-3 \\ m=1 \end{gathered}[/tex]Since the line is parallel to the given equation, the slope of the line is 1
Where
[tex](x,y)=(3,-2)[/tex]Substitute the coordinates into the slope-intercept form of a line above
[tex]\begin{gathered} y=mx+c \\ -2=1(3)+c \\ -2=3+c \\ Collect\text{ like terms} \\ c=-2-3=-5 \\ c=-5 \end{gathered}[/tex]The equation of the line becomes
[tex]\begin{gathered} y=mx+c \\ y=1(x)+(-5) \\ y=x-5 \end{gathered}[/tex]Hence, the answer is
[tex]y=x-5[/tex]