Question

Simplify the complex fraction

Answer 1

\[\large \dfrac{\left(\dfrac{4}{x+3}\right)}{\left(\dfrac{1}{x}+3\right)}\]
Let's start by working on the bottom first,\[\large \left(\dfrac{1}{x}+3\right) \qquad \rightarrow \qquad \left(\dfrac{1}{x}+\dfrac{3}{1}\right)\]We need a common denominator,
It looks like that common denominator will be \(x\) yes?
So we need to multiply the 3 by \(\dfrac{x}{x}\).

Answer 2

\[\large \left(\dfrac{1}{x}+\dfrac{3}{1}\cdot\color{royalblue}{\dfrac{x}{x}}\right) \qquad = \qquad \left(\dfrac{1}{x}+\dfrac{3x}{x}\right) \qquad = \qquad \left(\dfrac{1+3x}{x}\right)\]

Answer 3

So now our problem looks like this,\[\large \dfrac{\left(\dfrac{4}{x+3}\right)}{\left(\dfrac{1+3x}{x}\right)}\]

Remember when you `divide` fractions, you can rewrite it as `multiplication` if you simply `flip` the bottom fraction.

So this will change like so,\[\large \dfrac{\left(\dfrac{4}{x+3}\right)}{\left(\dfrac{1+3x}{x}\right)} \qquad \rightarrow \qquad \left(\dfrac{4}{x+3}\right)\left(\dfrac{x}{1+3x}\right)\]

Answer 4

Multiplying these two terms gives us,\[\large \dfrac{4x}{(x+3)(1+3x)}\]

Some further simplification can be done on the bottom.
I'll leave you to do that part ^^

Answer 5

@zepdrix wonderful explanation !