Recall that two equations in standard form represent parallel lines if they are as follows:
[tex]\begin{gathered} Ax+By=C_1, \\ Ax+By=C_2, \end{gathered}[/tex]
Where A>0, and all the coefficients are integers.
Therefore the equation of a parallel line to the given line is as follows:
[tex]3x+5y=k[/tex]
Since the parallel line passes through (15,4) then:
[tex]3*15+5*4=k.[/tex]
Simplifying the above result we get:
[tex]\begin{gathered} 45+20=k, \\ k=65. \end{gathered}[/tex]
Therefore:
[tex]3x+5y=65.[/tex]
Solving the above equation for y we get:
[tex]\begin{gathered} 3x+5y-3x=65-3x, \\ 5y=-3x+65, \\ \frac{5y}{5}=-\frac{3x}{5}+\frac{65}{5}, \\ y=-\frac{3}{5}x+13. \end{gathered}[/tex]
Answer: Last option.
