Answer :
To solve this problem, we must set up the appropriate system of equations using the distance formula [tex]\( d = rt \)[/tex], where [tex]\( d \)[/tex] is distance, [tex]\( r \)[/tex] is rate (speed), and [tex]\( t \)[/tex] is time.
Given:
- The motorboat travels 9 miles downstream in 30 minutes (0.5 hours).
- The return trip upstream takes 90 minutes (1.5 hours).
Let [tex]\( x \)[/tex] be the speed of the boat in still water in miles per hour, and [tex]\( y \)[/tex] be the speed of the current in miles per hour.
When traveling downstream, the effective speed of the boat is the speed of the boat plus the speed of the current: [tex]\( x + y \)[/tex].
Using the distance-rate-time formula [tex]\( d = rt \)[/tex] for the downstream trip:
[tex]\[ 9 = (x + y) \cdot 0.5 \][/tex]
When traveling upstream, the effective speed of the boat is the speed of the boat minus the speed of the current: [tex]\( x - y \)[/tex].
Using the distance-rate-time formula for the upstream trip:
[tex]\[ 9 = (x - y) \cdot 1.5 \][/tex]
Thus, we have the following system of equations:
[tex]\[ \begin{aligned} 9 & = 0.5 (x + y) \\ 9 & = 1.5 (x - y) \end{aligned} \][/tex]
Therefore, the system of equations that can be used to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is:
[tex]\[ \begin{aligned} 9 & = 0.5 (x - y) \\ 9 & = 1.5 (x + y) \end{aligned} \][/tex]
Given:
- The motorboat travels 9 miles downstream in 30 minutes (0.5 hours).
- The return trip upstream takes 90 minutes (1.5 hours).
Let [tex]\( x \)[/tex] be the speed of the boat in still water in miles per hour, and [tex]\( y \)[/tex] be the speed of the current in miles per hour.
When traveling downstream, the effective speed of the boat is the speed of the boat plus the speed of the current: [tex]\( x + y \)[/tex].
Using the distance-rate-time formula [tex]\( d = rt \)[/tex] for the downstream trip:
[tex]\[ 9 = (x + y) \cdot 0.5 \][/tex]
When traveling upstream, the effective speed of the boat is the speed of the boat minus the speed of the current: [tex]\( x - y \)[/tex].
Using the distance-rate-time formula for the upstream trip:
[tex]\[ 9 = (x - y) \cdot 1.5 \][/tex]
Thus, we have the following system of equations:
[tex]\[ \begin{aligned} 9 & = 0.5 (x + y) \\ 9 & = 1.5 (x - y) \end{aligned} \][/tex]
Therefore, the system of equations that can be used to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is:
[tex]\[ \begin{aligned} 9 & = 0.5 (x - y) \\ 9 & = 1.5 (x + y) \end{aligned} \][/tex]
