Answer:
The jogging rate of Matthew is 5 mph.
Step-by-step explanation:
Let the jogging rate of Matthew be x mph.
It is given that his jogging rate was 25 mph slower than the rate when he was riding. So, the riding rate is (x+25) mph.
The distance between Matthew and his friend's house is 12 miles.
[tex]Speed=\frac{Distance}{Time}[/tex]
[tex]Time=\frac{Distance}{Speed}[/tex]
The time taken by Matthew in jogging is [tex]\frac{12}{x}[/tex] and the time taken by Matthew in riding is [tex]\frac{12}{x+25}[/tex].
It took him 2 hours longer to jog there than ride back.
[tex]\frac{12}{x}=\frac{12}{x+25}+2[/tex]
[tex]\frac{12}{x}-\frac{12}{x+25}=2[/tex]
[tex]\frac{12(x+25)-12x}{x(x+25)}=2[/tex]
[tex]12x+300-12x=2x(x+25)[/tex]
[tex]300=2x^2+50x[/tex]
[tex]0=2x^2+50x-300[/tex]
[tex]0=x^2+25x-150[/tex]
[tex]0=x^2+30x-5x-150[/tex]
[tex]0=(x+30)(x-5)[/tex]
Equate each factor equal to 0.
[tex]x=5,-30[/tex]
The speed cannot be negative, therefore the jogging rate of Matthew is 5 mph.
