Answer :
EXPLANATION
Given that the distance between Ayvile and Beeville is 119 miles.
The distance between Beeville and Charlwoods is 90 miles.
If Russell drives 10 miles faster from Beeville to Charlwoods than from Ayville to Beeville we can apply the following relationships:
(Beeville to Charlwoods)_speed = 10*(Ayville to Beeville)_speed
We know that the speed is the division between the distance and the elapsed time:
[tex]\frac{dis\tan ce_{BeevilletoCharlwoods}}{time_{BeevilletoCharlwoods}_{}}=10\cdot\frac{dis\tan ce_{AyvilletoBeeville}}{time_{AyvilletoBeeville}}[/tex]Distance_Beeville to Charlwoods/time_Beeville to Charlwoods = 10* Distance_Ayville to Beeville/time_Ayville to Beeville
As the total elapsed time is equal we can apply the following equation:
time_Beeville to Charlwoods + time_Ayville to Beeville = 5.4
Let's call x to the time elapsed from Beeville to Charlwoods and y to the time elapsed from Ayville to Beeville.
Replacing terms:
x + y = 5.4
Now, we can replace terms on the speed equation and rewrite one time in function of the other as follows:
[tex]\frac{90}{x}=10\cdot\frac{119}{5.4-x}[/tex]Multiplying 10 by 119:
[tex]\frac{90}{x}=\frac{1190}{5.4-x}[/tex]Reversing the fractions:
[tex]\frac{x}{90}=\frac{5.4-x}{1190}[/tex]Multiplying both sides by 90 and by 1190:
[tex]1190x=90(5.4-x)[/tex]Applying the distributive property:
[tex]1190x=486-90x[/tex]Adding +90x to both sides:
[tex]1190x+90x=486[/tex]Adding like terms:
[tex]1280x=486[/tex]Dividing both sides by 1280:
[tex]x=\frac{486}{1280}=0.379\text{ hours}[/tex]Then, the value of y would be:
y = 5.4 - 0.379 = 5.02 hours
Now, we can compute the needed speeds as follows:
[tex]Speed\text{ from }Ayville\text{ to Be}eville=\frac{119}{5.02}=23.7\text{ mph}[/tex][tex]\text{Speed from }Beeville\text{ to Charlwoods=}\frac{90}{0.379}=237\text{ mph}[/tex]We can check this answers by dividing the speed from Beeville to Charlwoods from the speed from Ayville to Beeville = 237mph/23.7 mph = 10
As the scale factor is 10, the solutions are well computed.
