# How do you simplify (x/9 - 1/x) / (1+ 3/x) ?

Jul 13, 2016

$\frac{\frac{x}{9} - \frac{1}{x}}{\frac{1}{1} + \frac{3}{x}} = \frac{x - 3}{9}$

#### Explanation:

For this particular problem, we'd have to take note of some algebraic properties such as

$\frac{a}{b} - \frac{c}{d} = \frac{a d - b c}{b d}$

In this case, we follow this property to start simplifying:

$\frac{\frac{x}{9} - \frac{1}{x}}{\frac{1}{1} + \frac{3}{x}} = \frac{\frac{{x}^{2} - 9}{9 x}}{\frac{x + 3}{x}}$

We can simplify further by multiplying top and bottom (numerator and denominator) by the reciprocal of $\frac{x + 3}{x}$, namely $\frac{x}{x + 3}$.

(x/9 - 1/x)/(1/1+3/x) = (((x^2-9)/(9x)) * (x/(x+3)))/((cancel((x+3)/(x))) * (cancel(x/(x+3)))) = (cancel(x)(x^2-9))/(9cancel(x)(x+3)

We've now simplified our expression, although we're not done yet.
Note that we can still factor the top (numerator), that is, ${x}^{2} - 9$.
In order to factor this expression, we proceed as follows:

${x}^{2} - 9 = 0$

${x}^{2} = 9$

Solving for $x$ gives us

x = ± sqrt(9)

x = ± 3

Since the solutions are $x = 3$ and $x = - 3$, we now have our factors and can rewrite the numerator in the following way:

$\frac{\cancel{x} \left({x}^{2} - 9\right)}{9 \cancel{x} \left(x + 3\right)} = \frac{\left(x - 3\right) \cancel{\left(x + 3\right)}}{\cancel{\left(x + 3\right)} \cdot 9} = \frac{x - 3}{9}$