How do you multiply and simplify \frac { y ^ { 2} - 4} { y ^ { 2} } \cdot \frac { y ^ { 2} - 2y } { y ^ { 2} + 7y - 18}?

Dec 17, 2017

$\setminus \frac{\left(y - 2\right) \left(y + 2\right)}{y \left(y + 9\right)}$
OR
$= \setminus \frac{{y}^{2} - 4}{{y}^{2} + 9 y}$

Explanation:

$\setminus \frac{{y}^{2} - 4}{{y}^{2}} \setminus \cdot \setminus \frac{{y}^{2} - 2 y}{{y}^{2} + 7 y - 18}$

We know $\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)$

so $\left({y}^{2} - 4\right) = \left(y - 2\right) \left(y + 2\right)$

Also if we factorise the denominator term, $\left({y}^{2} + 7 y - 18\right)$,
we get $\left(y + 9\right) \left(y - 2\right)$

So give expression can be written as:

$= \setminus \frac{\left(y - 2\right) \left(y + 2\right)}{{y}^{2}} \setminus \cdot \setminus \frac{y \left(y - 2\right)}{\left(y + 9\right) \left(y - 2\right)}$

$= \setminus \frac{\left(y - 2\right) \left(y + 2\right)}{{y}^{\cancel{2}}} \setminus \cdot \setminus \frac{{\cancel{y}}^{1} \cancel{\left(y - 2\right)}}{\left(y + 9\right) \cancel{\left(y - 2\right)}}$

$= \setminus \frac{\left(y - 2\right) \left(y + 2\right)}{y} \setminus \cdot \setminus \frac{1}{\left(y + 9\right)}$

$= \setminus \frac{\left(y - 2\right) \left(y + 2\right)}{y \left(y + 9\right)}$
OR
$= \setminus \frac{{y}^{2} - 4}{{y}^{2} + 9 y}$

$\frac{\left(y - 2\right) \left(y + 2\right)}{y} ^ 2 \cdot \frac{y \left(y - 2\right)}{\left(y - 2\right) \left(y + 9\right)}$

$\frac{\left(y - 2\right) \left(y + 2\right)}{y \left(y + 9\right)}$