# How do you evaluate 2x ^ { 2} y if  x = 2\frac { 1} { 2} ,y = - 3\frac { 3} { 5}?

##### 2 Answers
Oct 4, 2017

$2 {x}^{2} y = - 45$

#### Explanation:

You can evaluate $2 {x}^{2} y$ by subbing in the given points, $x = 2 \frac{1}{2}$ and $y = - 3 \frac{3}{5}$.

First, it's best to turn these fractions into improper fractions:
$x = \frac{5}{2}$
$y = - \frac{18}{5}$

Subbing these into the expression we get:
$2 {x}^{2} y$
$= 2 {\left(\frac{5}{2}\right)}^{2} \left(- \frac{18}{5}\right)$
$= 2 \left(\frac{25}{4}\right) \left(- \frac{18}{5}\right)$

These fractions can be further simplified before you multiply them together:

$= 1 \left(\frac{25}{2}\right) \left(- \frac{18}{5}\right)$
$= \frac{5}{2} \left(- \frac{18}{1}\right)$
$= \frac{5}{1} \left(- 9\right)$
$= - 45$

So in short, $2 {x}^{2} y = - 45$

Oct 4, 2017

$- 45$

#### Explanation:

$\text{change the mixed numbers into "color(blue)"improper fractions}$

$\Rightarrow 2 \frac{1}{2} = \frac{5}{2} \text{ and } 3 \frac{3}{5} = \frac{18}{5}$

$\Rightarrow 2 {x}^{2} y$

$= 2 \times {\left(\frac{5}{2}\right)}^{2} \times - \frac{18}{5}$

$= 2 \times \frac{25}{4} \times - \frac{18}{5}$

$\textcolor{b l u e}{\text{cancel common factors }}$ on numerators/denominators.

$= {\cancel{2}}^{1} \times {\cancel{25}}^{5} / {\cancel{4}}^{2} \times - \frac{18}{\cancel{5}} ^ 1$

$= 1 \times \frac{5}{\cancel{2}} ^ 1 \times - {\cancel{18}}^{9} / 1$

$= 1 \times 5 \times - 9 = - 45$