   SEARCH HOME Math Central Quandaries & Queries  Subject: Simultaneous equations Name: Mick There were 17 envelopes bought, some were brown, some were white. The brown envelopes cost one cent more per envelope than the white ones. The total cost was 80 cents. How much of each type of envelope was bought? --Many thanks! Hi Mick.

Start by assigning variables to the unknowns:

Let B = the number of brown envelopes,
let W = the number of white envelopes,
let w = the cost of one white envelope and
let b = the cost of one brown envelope.

Then translate the question into equation terms:

There were 17 envelopes bought, some were brown, some were white.
becomes B + W = 17

The brown envelopes cost one cent more per envelope than the white ones.
becomes b = w + 1

The total cost was 80 cents.
becomes Bb + Ww = 80

Now we can solve for B using the first equations:

B = 17 - W

And then substitute for b and B in the total cost equation:

(17-W)(w+1) + Ww = 80
which reduces to
W = 17w - 63

which still leaves two variables unresolved. However, we know that W is a whole positive number (we are interpreting the word "some" in the question to mean "at least one") and w is a whole positive number as well.

We also know that W is at most 16 (there must be at least one brown envelope to qualify as "some") and at least 1. So let's write this as an inequality:

1 ≤ W ≤ 16

substituting for W, we have

1 ≤ 17w - 63 ≤ 16
which reduces:
64 ≤ 17w ≤ 79
3.76 ≤ w ≤ 4.64

however, we know w is an integer number of cents, so

w = 4

Finally, we can use this to find W and B:

W = 17w - 63 = 17(4) - 63 = 5
B = 17 - W = 17 - 5 = 12

Always remember to check at the end:

b = w + 1 = 4 + 1 = 5

Bb + Ww = (12)(5) + (5)(4) = 60 + 20 = 80 correct.

Hope this helps,
Stephen La Rocque.     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.