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Simultaneous Equations

Date: 07/01/2002 at 14:06:29
From: Vickie
Subject: Word Problem

The problem is: a man has 20 coins. All of the coins are dimes and 
quarters.  If the dimes were quarters and the quarters were dimes he 
would have 90 cents more than he has now.  How many dimes and 
quarters does he have now?

My solution was to start with two equations: 

  #1      D +    Q = 20

  #2   .10D + .25Q = (I don't know what to put here)


Date: 07/03/2002 at 10:13:33
From: Doctor Ian
Subject: Re: Word Problem

Hi Vickie,

You've made a good start.  Can I make a suggestion?  When working 
with money, it's often a good idea to express everything in terms 
of pennies, since that lets you work with integers. 

Your first equation lets you express the number of dimes in terms 
of the number of quarters, and vice versa.  That is, all of the 
following equations say the same thing:

  d + q = 20

      d = 20 - q

      q = 20 - d

Setting up the second equation is a little trickier.  If the guy 
HAS d dimes and q quarters, then the amount of money he has 
(expressed in pennies) is 

  10*d + 25*q

Now, if the guy HAD q dimes and d quarters, the amount of money 
he would have would be 

  25*d + 10*q

Does that make sense?  What the second equation needs to express 
is that the second option would add up to 90 pennies more than 
the first option.  In other words, you'd have to ADD 90 cents to 
what he actually HAS, to get what he WOULD have:

  (10*d + 25*q) + 90 = (25*d + 10*q)

Here's where the first equation comes in handy!  Since you can 
express q in terms of d, you can substitute (20-q) everywhere you 
see a d to get

  (10*(20-q) + 25*q) + 90 = (25*(20-q) + 10*q)

Can you take it from here? 
  
- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Linear Equations
Middle School Algebra
Middle School Equations

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