1. Algebra Problem Solving

Peter leaves his home at 10 AM and bicycles towards Sarah's home. At 11AM, Sarah leaved her home by car to deliver an urgent message to Peter. She and Peter meet at 11:30AM. If Peter and Sarah live 34 miles apart, and Sarah drives 20 mph faster than Peter bicycles, find Peter's bicycle speed.

Thank you for any help in advance.

2. When Peter meets Sarah, peter has traveled 1.5 hours and covered x miles.Sarah travels 0.5 hour and covers (34 - x) miles.

Can you proceed now?

3. I honestly still do not know what to do. I also have a question with the miles Sarah covered.....why did you use (34-x) what does this stand for? Would I now set up an equation with these variables? If I did set up an equation how would I do it using 34-x this is tricky to me. Thank you so much for your help. It is greatly appreciated!

4. Originally Posted by matgrl
I honestly still do not know what to do. I also have a question with the miles Sarah covered.....why did you use (34-x) what does this stand for? Would I now set up an equation with these variables? If I did set up an equation how would I do it using 34-x this is tricky to me. Thank you so much for your help. It is greatly appreciated!
The distance between them is initially 34 miles. If Peter travels x miles then clearly Sarah has travelled 34 - x miles.

Using speed = distance/time:

Peter: v = x/1.5

Sarah: v + 20 = (34 - x)/0.5 => v + 20 = 68 - 2x => v = 48 - 2x.

Equate those two equations and solve for x and hence get v.

5. Peters speed

Hello,
I did this problem differently and it may help a student understand it better.

Let x = Peters speed and x+20 = Sarahs speed
Peter travels 1.5 hrs Sarah travels 1/2 hr
Peters distance =1.5 x Sarahs distance = 1/2 ( x +20 )
Add the two distances and equate to 34 miles
2x + 10 = 34
x=12 mph

bjh

6. Thank you very much...this is a huge help!