2.12: Algebra Tips and Tricks IV (Tips for dealing with fractions)
A quick fraction hint
A couple of ideas while working with fractions. Note that if you distribute a number times a fraction, you multiply on top:
\(x \left( \frac{1}{2} + y \right) = \frac{x}{2} + xy\)
The reason is when we multiply fractions, we multiply straight across, and we can always think of \(x\) as \(\frac{x}{1}\). Hence \(x \cdot \frac{1}{2} = \frac{x}{1} \cdot \frac{1}{2} = \frac{x}{2}\).
Complex Fractions
If you have “fractions within fractions”, this calls out to be simplified. One way to do it is to multiply top and bottom of the outer fraction by the same number so that the inner fractions go away. For example,
\(\cfrac{\cfrac{1}{3} - \cfrac{\sqrt{3}}{3} }{ \cfrac{1}{9} }\)
If we multiply top and bottom by \(9\) in the following example, that gets rid of the \(3\) and the \(9\) denominators:
\[\begin{align*} \left(\cfrac{\cfrac{1}{3} - \cfrac{\sqrt{3}}{3} }{ \cfrac{1}{9} } \right) \cdot \cfrac{9}{9} & = \cfrac{9 \cdot \cfrac{1}{3} - 9 \cdot \cfrac{\sqrt{3}}{3} }{ 9 \cdot \cfrac{1}{9} } \\ & = \frac{3 - 3 \sqrt{3}}{1} \\ & = 3 - 3 \sqrt{3} \end{align*}\]
You can also multiply by the reciprocal instead of dividing. Like this:
\[\begin{align*} \left(\cfrac{\cfrac{1}{3} - \cfrac{\sqrt{3}}{3} }{ \cfrac{1}{9} } \right) & = \left(\cfrac{1}{3} - \cfrac{\sqrt{3}}{3}\right) \cdot \frac{9}{1} \\ & = 9 \cdot \cfrac{1}{3} - 9 \cdot \cfrac{\sqrt{3}}{3} \\ & = 3 - 3 \sqrt{3} \end{align*}\]
The same thing happens with variables. Consider this problem:
\(\cfrac{\cfrac{1}{x} + \cfrac{1}{y}}{x}\)
We can simplify this by multiplying by \(xy\) to get rid of the \(x\) and \(y\) denominators on top.
\[\begin{align*} \cfrac{\cfrac{1}{x} + \cfrac{1}{y}}{x} & = \cfrac{\cfrac{1}{x} + \cfrac{1}{y}}{x} \cdot \frac{xy}{xy} \\ & = \cfrac{\cfrac{xy}{x} + \cfrac{xy}{y}}{xxy} \\ & = \cfrac{y + x}{x^2y} \end{align*}\]
Here’s one more example.
To simplify this one, we need to clear all the denominators by multiply by \(x\), \((x+1)\) and \((x-1)\). Not easy, but we can do it.
\[\begin{align*} \cfrac{\cfrac{1}{x+1} - \cfrac{1}{x-1}}{\cfrac{1}{x}} \cdot \frac{x(x-1)(x+1)}{x(x-1)(x+1)} & = \cfrac{\cfrac{x(x+1)(x-1)}{x+1} - \cfrac{x(x+1)(x-1)}{x-1}}{\cfrac{x(x+1)(x-1)}{x}}\\ & = \cfrac{x(x-1) - x(x+1)}{(x-1)(x+1)}\\ & = \cfrac{x^2 - x - (x^2 + x)}{x^2 - 1} \\ & = \cfrac{x^2 - x - x^2 - x}{x^2 - 1} \\ & = \cfrac{-2x}{x^2 - 1} \end{align*}\]