Simplify this complex algebraic fraction I'm stumped on this problem, I need to know how this answer was arrived at but my text book doesn't show this. 
$$\frac{\frac{1}{x+y}}{\frac{x}{y}}$$
The text book says the answer is this:
$$\frac{y}{x(x+y)}$$
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?
 A: Hint: Multiply by $\dfrac{\dfrac{y}{x}}{\dfrac{y}{x}}$, that is, the reciprocal.
You will get the books answer:
$$\frac{y}{x(x+y)}$$
A: Besides to @Amzoti's post, you may do one of the following ones:
$$\large \dfrac{\dfrac ab}{\dfrac cd} = \dfrac{ad}{bc}$$
$$\large \dfrac ab \div \dfrac cd = \dfrac ab \times \dfrac dc$$
A: 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:
$$\dfrac{\dfrac{1}{x+y}}{\dfrac{x}{y}}=\frac{1}{x+y}\cdot \frac{y}{x}=\frac{y}{x(x+y)}.$$
A: 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
$$y\over x(x+y)$$
