# How do you simplify and find the restrictions for ((3x)/y -x/1) / ((2y)/1 - y/x)?

May 17, 2017

$= \frac{{x}^{2} \left(3 - y\right)}{{y}^{2} \left(2 x - 1\right)}$

$x \ne 0 \mathmr{and} y \ne 0$

#### Explanation:

$\frac{\frac{3 x}{y} - \frac{x}{1}}{\frac{2 y}{1} - \frac{y}{x}} \text{ } x \ne 0 \mathmr{and} y \ne 0$

Before we start with any simplifying we can see that the denominators at the top and bottom have an $x$ and a $y$
Therefore, they may not be $0$
We might find further restrictions later.

Write the complex fraction as a division of two fractions:

$\left(\frac{3 x}{y} - \frac{x}{1}\right) \div \left(\frac{2 y}{1} - \frac{y}{x}\right)$ find LCD and subtract as usual,.

$= \frac{\left(3 x - x y\right)}{y} \div \frac{\left(2 x y - y\right)}{x} \text{ } \leftarrow$ factorise the numerators:

$= \frac{x \left(3 - y\right)}{y} \div \frac{y \left(2 x - 1\right)}{x} \text{ } \leftarrow$ multiply by the reciprocal

$= \frac{x \left(3 - y\right)}{y} \times \frac{x}{y \left(2 x - 1\right)}$

$= \frac{{x}^{2} \left(3 - y\right)}{{y}^{2} \left(2 x - 1\right)}$

The original restrictions are confirmed.