We know that two lines are parallel if they have the same slope.
Then, we can write the given equations of the lines in their slope-intercept form.
[tex]\begin{gathered} y=mx+b \\ \text{ Where} \\ \text{ m is the slope} \\ b\text{ is the y-intercept} \end{gathered}[/tex]Then, we solve for y each equation:
• First equation
[tex]\begin{gathered} 6y-x=14 \\ \text{ Add x from both sides} \\ 6y-x+x=14+x \\ 6y=14+x \\ \text{ Divide by 14 from both sides} \\ \frac{6y}{6}=\frac{14+x}{6} \\ y=\frac{14}{6}+\frac{1}{6}x \\ \text{ Simplify} \\ y=\frac{7\cdot2}{3\cdot2}+\frac{1}{6}x \\ y=\frac{7}{6}+\frac{1}{6}x \\ \text{ Reorder} \\ y=\frac{1}{6}x+\frac{7}{6} \end{gathered}[/tex]• Second equation
[tex]\begin{gathered} -4y+2x=1 \\ \text{ Subtract 2x from both sides} \\ -4y+2x-2x=1-2x \\ -4y=1-2x \\ \text{ Divide by -4 from both sides} \\ \frac{-4y}{-4}=\frac{1-2x}{-4} \\ y=-\frac{1}{4}+\frac{2x}{4} \\ y=-\frac{1}{4}+\frac{1}{2}x \\ \text{ Reorder} \\ y=\frac{1}{2}x-\frac{1}{4} \end{gathered}[/tex]• Third equation
[tex]\begin{gathered} 2y-\frac{1}{3}x=12 \\ \text{ Add }\frac{1}{3}x\text{ from both sides} \\ 2y-\frac{1}{3}x+\frac{1}{3}x=12+\frac{1}{3}x \\ 2y=12+\frac{1}{3}x \\ \text{ Divide by 2 from both sides} \\ \frac{2y}{2}=\frac{12+\frac{1}{3}x}{2} \\ y=\frac{12}{2}+\frac{\frac{1}{3}}{2}x \\ y=6+\frac{\frac{1}{3}}{\frac{2}{1}}x \\ y=6+\frac{1\cdot1}{3\cdot2}x \\ y=6+\frac{1}{6}x \\ \text{ Reorder} \\ y=\frac{1}{6}x+6 \end{gathered}[/tex]• Fourth equation
[tex]\begin{gathered} 5x+3y=1 \\ \text{ Subtract 5x from both sides} \\ 5x+3y-5x=1-5x \\ 3y=1-5x \\ \text{ Divide by 3 from both sides} \\ \frac{3y}{3}=\frac{1-5x}{3} \\ y=\frac{1}{3}-\frac{5}{3}x \\ \text{ Reorder} \\ y=-\frac{5}{3}x+\frac{1}{3} \end{gathered}[/tex]Then, the lines have the following slopes:
• First
[tex]m=\frac{1}{6}[/tex]• Second
[tex]m=\frac{1}{2}[/tex]• Third
[tex]m=\frac{1}{6}[/tex]• Fourth
[tex]m=-\frac{5}{3}[/tex]As we can see, only the first and third lines have the same slope.
Therefore, only the first and third lines are parallel.
