## anonymous one year ago PLEASE HELP WILL MEDAL AND FAN NEED ASAP FOR FINAL GRADE !! Sophie drove 225 miles south through the smoky mountains .Snow during her return trip made her average speed on her return trip 30 mph slower, The total driving time was 8 hours . Find Sophie's average speed on each part of the trip .

1. kropot72

Let x be her average speed on the outgoing trip. Then her average speed on the return trip = x - 30. Time for outgoing trip = 225/x. Time for return trip = 225/(x - 30). The total driving time was 8 hours. Therefore we can write the following equation: $\large \frac{225}{x}+\frac{225}{(x-30)}=8 ..................(1)$ Now you need to solve equation (1) to find the value of x.

2. anonymous

calculating it would add to x=45/4 ?

3. kropot72

I used x(x - 30) as a common denominator to add the fractions on the left hand side of equation (1), giving: 225x - 6750 + 225x = 8x(x - 30) ............(2) When equation (2) is expanded and rearranged we get the quadratic: $\large 8x^{2}-690x+6750=0\ .............(3)$ Now you need to solve the quadratic in (3) to find the value of x.

4. anonymous

I got x=75

5. kropot72

Yes, 75 mph is correct for the outgoing trip. Now you need to find the average speed for the return trip.

6. anonymous

how would I go about finding the average speed . I am terrible at this

7. kropot72

If you look at my first post to this thread, I wrote: "Then her average speed on the return trip = x - 30."

8. anonymous

I see ! ok and where would I plug that in

9. kropot72

You already posted that x = 75. So just plug that value for x into: average speed on the return trip = x - 30

10. anonymous

so return trip is just x=30 no equation involved ?

11. anonymous

or average speed for return trip =30

12. kropot72

Average speed on the return trip = x - 30 ...........(4) We found the value for x, her average speed on the outgoing trip, was 75 mph. Just plug that value for x into equation (4) to find the average speed on the return trip.