<PRE><B>Question 370991</B>
&lt;br&gt;&lt;font face=&quot;Tahoma&quot;&gt;Let one number be x.  The other number is then (105-x). &lt;br&gt;

Then we have:&lt;br&gt;

{{{x*(1/2)=(105-x)*(1/3)}}} given&lt;br&gt;

{{{6*x*(1/2)=6*(105-x)*(1/3)}}} multiply by 6 on each side to eliminate fractions&lt;br&gt;

{{{3x=2*(105-x)}}}&lt;br&gt;

{{{3x=210-2x}}}&lt;br&gt;

{{{5x=210}}}&lt;br&gt;

{{{x=42}}}&lt;br&gt;

So if one number is 42, the other number would be 63.&lt;br&gt;

However, we are not quite finished, because we can also set it up this way:&lt;br&gt;

{{{x*(1/3)=(105-x)*(1/2)}}} given&lt;br&gt;

{{{6*x*(1/3)=6*(105-x)*(1/2)}}} multiply by 6 on each side to eliminate fractions&lt;br&gt;

{{{2x=3*(105-x)}}}&lt;br&gt;

{{{2x=315-3x}}}&lt;br&gt;

{{{5x=315}}}&lt;br&gt;

{{{x=63}}}&lt;br&gt;

And of course, if one number is 63, the other number must be 42.&lt;br&gt;

This proves that no matter which number you multiply by the fractions of 1/2 and 1/3, you will get the same answer.&lt;br&gt;

In other words, the two numbers will always be 63 and 42.&lt;br&gt;

I hope this helps!&lt;br&gt;</PRE>