Simplify this complex algebraic fraction

Question

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?

Answer 1

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

You will get the books answer:

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

Answer 2

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

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

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

Answer 3

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:

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

Answer 4

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

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