Given the line with equation [tex]y=\dfrac{2}{3}x-5.[/tex]
Parallel lines have the same slopes, so the parallel line has equation
[tex]y=\dfrac{2}{3}x+a.[/tex]
This line passes through the point (-6,-1), then its coordinates satisfy the equation:
[tex]-1=\dfrac{2}{3}\cdot (-6)+a,\\ \\a=-1+4,\\ \\a=3.[/tex]
Therefore, the equation of needed line is
[tex]y=\dfrac{2}{3}x+3.[/tex]
