Answer:
[tex]y=2x\\y=3(x-\frac{1}{2})[/tex]
Paul will catch Fred after [tex]1\ hours[/tex] at a distance [tex]3\ miles[/tex].
Step-by-step explanation:
Distance is represented by [tex]y[/tex] and the time is represented by [tex]x[/tex]. Here after [tex]x[/tex] hours(when Fred started) the distance covered by Fred and Paul is same that is [tex]y[/tex].
For Fred:
time=[tex]x[/tex]
distance=[tex]y[/tex]
Speed[tex]=2\ mph[/tex]
[tex]Distance=speed\times time\\y=2\times x \\y=2x[/tex]
Paul:
Since Fred started 30 minutes earlier, so Paul's travel time will be 30 minutes less than that of Fred.
Time[tex]=x-\frac{1}{2}[/tex]
Speed[tex]=3\ mph[/tex]
Distance[tex]=y\\[/tex]
[tex]y=3\times (x-\frac{1}{2})\\y=3(x-\frac{1}{2})[/tex]
Hence system of equation
[tex]y=2x\\y=3(x-\frac{1}{2})[/tex]
Solution of this system of equations gives the time, when Paul will catch Fred and distance covered by each of them.
[tex]y=2x\\y=3(x-\frac{1}{2})\\\\2x=3x-\frac{3}{2}\\x=\frac{3}{2}\\\\y=2x\\y=2\times \frac{3}{2}\\y=3[/tex]
Time taken by Paul[tex]=x-\frac{1}{2}=\frac{3}{2}-\frac{1}{2}=1[/tex]
Hence Paul will catch Fred after [tex]1\ hours[/tex] at a distance [tex]3\ miles[/tex].
