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Answer :

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.

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