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It costs $\$ 0.05$ to send a text message and $\$ 0.10$ to send a picture on your cell phone. You spend $\$ 4.00$ and send $5$ more text messages than pictures.

How many text messages $x$ and pictures $y$ did you send?

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do you have an idea? Did you ever solve such problems before? – sigmatau Jul 16 '13 at 16:13

Hint: Try to translate the question from English into Mathematics. (I'll do everything in cents)


I got this from "You spend $\$4.00$" and "$\$0.05$ to send a text message and $\$0.10$ to send a picture". Also, we have


This is from "$5$ more text messages than pictures".

So, with this information, can you solve it?

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We give a solution of the traditional kind, before the "algebra" approach became standard. Much more sophisticated methods of the same general nature were known in China at least $1800$ years ago, and probably much earlier.

Similar methods were taught elsewhere for many centuries, and disappeared from Western European curricula about two hundred years ago.

The solution below is almost certainly not the one you should use.

Let us guess that we sent $5$ text messages and no pictures. That would have cost $25$ cents.

So we are off by $375$ cents. Poor guess!

For every additional text message we send, we must send a picture, to keep text messages $5$ ahead. And each such combo will cost us $15$ cents. The number of pictures is the number of combos, $\frac{375}{15}$. Thus $25$ pictures were sent, and $30$ text messages.

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$$\begin{align} 0.05x+0.1y&=4.00 \\ x-y&=5 \end{align}$$

This can be solved with substitution or elimination.

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