"Canceling out" in division doesn't always work the same way does it?I've been working on Nested Fractions at the Khan Academy. Recently I was doing a routine problem and came to the correct conclusion but I realized I didn't understand why I wouldn't keep dividing. Here's what I begin with. I'm supposed to reduce it to an equivalent simplified expression. I wish I could link to the problem but I had moved on after posting here. Here's the start: ( 1 + x y ) x y Part way through my problem now looks like: y ⋅ ( y + x ) y ( x ) I decide to "cancel out" the y's out side the parentheses. I end up with... y + x x I understood this the be the right answer. But for a moment I was very tempted to cancel the remaining x's and be left with a y. Even though I do this canceling operation all the time I realized then that there is something about it I don't understand. So here I am. Can you explain why it is that I was able to cancel the y earlier but not the x near the end?Much appreciated.

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&quot;Canceling out&quot; in division doesn&#039;t always work the same way does it?I&#039;ve been working on Nested Fractions at the Khan Academy. Recently I was doing a routine problem and came to the correct conclusion but I realized I didn&#039;t understand why I wouldn&#039;t keep dividing. Here&#039;s what I begin with. I&#039;m supposed to reduce it to an equivalent simplified expression. I wish I could link to the problem but I had moved on after posting here. Here&#039;s the start:                    (        1        +                  x          y                )                    x        y            Part way through my problem now looks like:                    y        ⋅        (        y        +        x        )                    y        (        x        )            I decide to &quot;cancel out&quot; the y&#039;s out side the parentheses. I end up with...                    y        +        x            x      I understood this the be the right answer. But for a moment I was very tempted to cancel the remaining x&#039;s and be left with a y. Even though I do this canceling operation all the time I realized then that there is something about it I don&#039;t understand. So here I am. Can you explain why it is that I was able to cancel the y earlier but not the x near the end?Much appreciated.

Donavan Scott · Accepted Answer

In order to &quot;cancel&quot; something on the top with something on the bottom, the thing must be multiplied to the top, and multiplied to the bottom. Here, this x is added to something on top, so it cannot be cancelled immediately. Sometimes, you can factor(this is the reverse of distributing, where you kind of turn addition into multiplication) something in order to get it to &quot;cancel&quot;. One way to think about this, is rather than &quot;cancelling&quot;, try reducing to 1, so in this case            y      (      y      +      x      )              y      (      x      )        =      y    y    ∗            y      +      x        x  and       y    y    =  1 if   y  ≠  0, so       y    y    ∗            y      +      x        x    =  1  ∗            y      +      x        x    =            y      +      x        x  This is what is actually going on when you &quot;cancel&quot;.Also, in this case, you cannot factor the top in any productive way, so you can&#039;t reduce anything else to 1. Thus, this is as simple of an answer as you&#039;re going to get.

sviraju6d · Accepted Answer

Short answer: you cannot cancel out things that are added, only things that are multiplied.Long answer: When you are canceling out factors in division, you are really doing two things: first you are factoring them out, then you are using the fact that       a    a   is equal to   1 if   a  ≠  0. The result is that, if   A,   B and   C are some (nonzero) expressions, then            A      ⋅      B              A      ⋅      C        =      B    C  which is a direct consequence of the fact that            a      b              c      d        =      a    c    ⋅      b    d  for arbitrary values of   a  ,  b  ,  c  ,  d since it means that            A      ⋅      B              A      ⋅      C        =      A    A        B    C    =  1  ⋅      B    C    =      B    C    .In your case, you have the expression            y      (      y      +      x      )              y      ⋅      x      You can use the rule I cited by seting   A  =  y,   B  =  (  y  +  x  ) and   C  =  xYou cannot use the rule for the x-es, because the top (numerator) of the fraction is not of the form   x  ⋅  A , where A is some expression.

"Canceling out" in division doesn't always work the same way does it? I've been working on Nested F

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