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I'm stumped on this problem, I need to know how this answer was arrived at but my text book doesn't show this.


The text book says the answer is this:


I think the problem is that I lack the insight to find the correct LCD, but I tried this, and it's obviously wrong. $$y(x+y)$$

How can the correct LCD for this be obtained and how is this problem solved assuming the text book's answer is even correct?

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up vote 4 down vote accepted

Hint: Multiply by $\dfrac{\dfrac{y}{x}}{\dfrac{y}{x}}$, that is, the reciprocal.

You will get the books answer:


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The point of this section is to find LCD and simplify, this is what the other questions in the set are doing. I really would like to know the how from that context if possible. – James Dec 21 '13 at 6:14
Why are you trying to complicate the problem? There are two approaches that are posted that are the preferred ways to approach this. Use LCD when LCD is called for, but it is not here. – Amzoti Dec 21 '13 at 6:21
So I can understand it fully, and pass the test. you're solution technically gives the answer of $$\frac{\frac{y}{x(x+y)}}{1}$$ which isn't how it's answered in the text book. – James Dec 21 '13 at 6:24
What is anything divided by 1, itself? Also, technically, this is what multiplying by reciprocal means, it is just not shown that way. You can just say multiply by $y/x$. – Amzoti Dec 21 '13 at 6:24
+1 for each of my friends! – amWhy Dec 21 '13 at 13:16

Besides to @Amzoti's post, you may do one of the following ones:

$$\large \dfrac{\dfrac ab}{\dfrac cd} = \dfrac{ad}{bc}$$

$$\large \dfrac ab \div \dfrac cd = \dfrac ab \times \dfrac dc$$

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Nice hints too my friend +1 – Amzoti Dec 21 '13 at 6:21
@B.S. I suppose that the c' are just c. If not, we are entering a new era ! Cheers. – Claude Leibovici Dec 21 '13 at 8:22
@ClaudeLeibovici: I made the pictures in MS Office and it is just a ,. :D – Babak S. Dec 21 '13 at 11:07
No time at all! My pleasure! ;-) – amWhy Dec 21 '13 at 13:21

I hate to spoil your efforts here by giving a straight solution, but I feel this is important, so I will add some small perspective. Dividing a real number $p$ by another real number $q \neq 0$ is equivalent to multiplying $p$ by the multiplicative inverse of $q$. Specifically (and as a generalization), $$\frac{p}{q}=p \cdot \frac{1}{q}.$$

You have probably already internalized this thought. Now consider your expression in these terms:

$$\dfrac{\dfrac{1}{x+y}}{\dfrac{x}{y}}=\frac{1}{x+y}\cdot \frac{y}{x}=\frac{y}{x(x+y)}.$$

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To approach this problem using an LCD, we are considering

$$\frac {1\over x+y}{x\over y}=\frac 1{\frac xy(x+y)}$$

which means we are looking for the least common denominator of $\frac xy$ and $x+y$. In the $\frac xy$ portion, the $\frac 1y$ part is at best reducing the size of the LCD, while $x+y$ is not guaranteed to have any common factor with $x,y$, therefore the $\frac 1y$ portion does not add a factor to the LCD. The $x$ part, however, is a multiplier, and therefore does add to the LCD consideration as $x(x+y)$. The only thing remaining is to determine if the $\frac 1y$ part takes away a factor, which can be accomplished as $\frac {x(x+y)}y$. This is our denominator, and then the fraction looks like

$$\frac 1{x(x+y)\over y}$$

But this is familiar, we can rewrite it as

$$y\over x(x+y)$$

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