Associated Topics || Dr. Math Home || Search Dr. Math

Renting Bike Safety Equipment

```
Date: 10/09/2001 at 17:34:45
From: Jessica
Subject: Solving equations with variables on both sides

Also, can you help me on this word problem:

Susppose you live near a park that has a bike trail you like to ride.
The Park Department rents a bike with safety equipment for \$5 a day.
If you provide your own safety equipment, the bike rental is \$3 a day.
You could buy equipment at a sports store for \$28. How many times
must you use the trail to justify buying your own safety equipment?

I have no idea how to even start these problems.
Jessica
```

```
Date: 10/10/2001 at 14:09:38
From: Doctor Ian
Subject: Re: Solving equations with variables on both sides

Hi Jessica,

I think this entry from the Dr. Math archives will help you see how to
deal with equations that have the same variable appearing on both
sides:

Basic Tips on Solving for X
http://mathforum.org/dr.math/problems/megan.11.16.00.html

As for the story problem, it's better to come up with a 'dumb'
solution that you understand than try to come up with a tricky one
that you have to sort of hope turns out to be right.

If you buy equipment, it costs you \$28 + \$3 = \$31 to go riding the
first time, and then \$3 more for each time after that.

Times riding  Total cost
------------  ----------
1            31
2            34
3            37
4            40
5            43
6            46

On the other hand, if you rent equipment, it costs you \$5 the first
time, and \$5 each time after that:

Times riding  Total cost (buy)  Total cost (rent)
------------  ----------------  -----------------
1            31                 5
2            34                10
3            37                15
4            40                20
5            43                25
6            46                30

If you keep adding rows to this table, you'll eventually get to a
point where the cost of renting is higher than the cost of buying.
That's the number you're looking for.

Once you've found an answer that you know must be correct, _then_ you
can start thinking about easier ways to find it. For example, each
time you rent instead of buying, it costs you an extra \$2. So a
different way to ask the question is: If you spend an extra \$2 each
time you rent safety equipment, how many times will you have to rent
it before you've spent \$28?

A third way is to just write down the equations for the two
situations,

cost of buying = 28 + n*3

cost of renting = 0 + n*5

set the two costs equal to each other,

28 + n*3 = 0 + n*5

and solve this equation to find the value of n that represents the
'break even' point. (So probably this story is supposed to get you to
set up the kind of equation that you asked about in the first part of

Note that each of the two original equations,

cost of buying = 28 + n*3

cost of renting = 0 + n*5

is the equation of a line. If you plotted the lines (with the number
of times on the trail along the x-axis, and the total cost along the
y-axis), you'd find that they intersect at a point. The value of n at
this point is the number of times that you have to go riding to reach
the 'break even' point.

All of these approaches should lead you to the same answer - and in
fact, finding a single answer in (at least) two completely different
ways is one of the best techniques to use for making sure that you've
worked a problem correctly.

more, or if you have any other questions.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Word Problems

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search