# How do you simplify (2-1/y)/(4-1/ y^2)?

It is $\frac{2 - \frac{1}{y}}{4 - \frac{1}{y} ^ 2} = \frac{2 - \frac{1}{y}}{\left(2 - \frac{1}{y}\right) \cdot \left(2 + \frac{1}{y}\right)} = \frac{1}{2 + \frac{1}{y}}$

Sep 22, 2015

Multiply both numerator and denominator by ${y}^{2}$ and use the difference of square identity ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$ to find:

$\frac{2 - \frac{1}{y}}{4 - \frac{1}{{y}^{2}}} = \frac{y}{2 y + 1}$

with exclusions $y \ne 0$ and $y \ne \frac{1}{2}$

#### Explanation:

$f \left(y\right) = \frac{2 - \frac{1}{y}}{4 - \frac{1}{{y}^{2}}} = \frac{\left(2 y - 1\right) y}{4 {y}^{2} - 1} = \frac{\left(2 y - 1\right) y}{{\left(2 y\right)}^{2} - {1}^{2}}$

$= \frac{\left(2 y - 1\right) y}{\left(2 y - 1\right) \left(2 y + 1\right)} = \frac{y}{2 y + 1}$

with exclusions $y \ne 0$ and $y \ne \frac{1}{2}$

Notice that if $y = 0$ or $y = \frac{1}{2}$ then $f \left(y\right)$ is undefined, but $\frac{y}{2 y + 1}$ is defined. Hence the exclusions.