<PRE><B>Question 330450</B>
&lt;br&gt;Lucy has acollection of 300 coins of 5 cents and ten cents coins. If the total value of her collection is 45.00, how many of each kind are there? &lt;br&gt;

This problem as written has no &quot;real&quot; answer.&lt;br&gt;

If you had 300 10 cent coins with no 5 cent coins, that would only add up to 30 dollars.  But the question says that the total value of the 300 coins is 45.00.&lt;br&gt;

This is an impossible real world situation.  Please check the question and submit again.  :)&lt;br&gt;

But if we go through the motions, this is what we are looking at:&lt;br&gt;

Let x=number of 5 cent coins
Let y=number of 10 cent coins&lt;br&gt;

Thus&lt;br&gt;

x+y=300 given
.05*x+.10*y=45.00 given&lt;br&gt;

We need to solve this system of equations!&lt;br&gt;

Multiply the bottom equation by 20.  Then we have:&lt;br&gt;

x+y=300
x+2y=900&lt;br&gt;

Then subtract the bottom equation from the top.&lt;br&gt;

-y=-600 or y=600&lt;br&gt;

Then x+y=300 or x-600=300 and x=-300&lt;br&gt;

The only answer to this system of equations would be x=-300, y=600&lt;br&gt;

Unfortunately, in real world terms, it is not possible to have a negative amount of coins!</PRE>