Algebra II: 30 mi w/ wind in 3/5 hr; 5/6 hr against wind

[Tay]

Can you help me with this problem?

If Togo flew his ultralight airplane 30 miles with the wind & it took 3/5 hours. He flew back against the wind and it took 5/6 hours. Solve for x & y.

 

[galactus]

Since d=rt, you can use d/r=t

Let x=his rate in no wind and y=wind rate

His rate with the wind is x+y

Against the wind, x-y

\(\displaystyle \L\\\frac{30}{x+y}=\frac{3}{5}\)
\(\displaystyle \L\\\frac{30}{x-y}=\frac{5}{6}\)

Solve for x and y.

 

[stapel]

If Togo flew his ultralight airplane 30 miles with the wind & it took 3/5 hours. He flew back against the wind and it took 5/6 hours. Solve for x & y.

According to the above statement, your assignment must have defined "x" and "y". The previous reply, providing explanations and the set-up, may not have used "x" and "y" in the way your text intended. (You did not supply this information.) Make sure you understand the reasoning, and have used the variables in accordance with the assignment's expectations! :shock:

Eliz.