To answer this question, we will set and solve a system of linear equations.
Let x be the length of route X, and y be the length of route Y. In one week Cyd drove route X five times and route Y three times, therefore, we can set the following equation:
[tex]5x+3y=284.[/tex]
In another week, she drove route x twice and route y seven times, therefore:
[tex]2x+7y=363.[/tex]
Solving the second equation for x, we get:
[tex]\begin{gathered} 2x=363-7y, \\ x=\frac{363-7y}{2}\text{.} \end{gathered}[/tex]
Substituting the above result, in the first equation, we get:
[tex]5(\frac{363-7y}{2})+3y=284.[/tex]
Solving the above equation for y, we get:
[tex]\begin{gathered} 5(\frac{363}{2})-5(\frac{7}{2})y+3y=284, \\ \frac{1815}{2}-\frac{35}{2}y+3y=284, \\ 1815-35y+6y=568, \\ -29y=-1247, \\ y=43. \end{gathered}[/tex]
Now, we substitute y=43 in x=(363-7y)/2 and get:
[tex]x=\frac{363-7\cdot43}{2}=31.[/tex]
Answer: x is the length of route X, and y is the length of route Y.
[tex]\begin{gathered} x=31, \\ y=43. \end{gathered}[/tex]
