Question

Evaluate each of the following when x = 2, y = −1.
$\left(2xy\right)\times \left(\frac{x^{2}y}{4}\right)\times \left(x^2\right)\times \left(y^2\right)$

Answer 1

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ $a^m\times a^n=a^{m+n}$.

We have:

$$\begin{gathered} \left(2xy\right)\times \left(\frac{x^{2}y}{4}\right)\times \left(x^2\right)\times \left(y^2\right) \\ =\left(2\times \frac{1}{4}\right)\times \left(x\times x^2\times x^2\right)\times \left(y\times y\times y^2\right) \\ =\left(2\times \frac{1}{4}\right)\times \left(x^{1+2+2}\right)\times \left(y^{1+1+2}\right) \\ =\frac{1}{2}{x}^5{y}^4 \end{gathered}$$

$\therefore$ $\left(2xy\right)\times \left(\frac{x^{2}y}{4}\right)\times \left(x^2\right)\times \left(y^2\right)=\frac{1}{2}{x}^5{y}^4$

Substituting x = 2 and y = $-$1 in the result, we get:

$$\begin{gathered} \frac{1}{2}{x}^5{y}^4 \\ =\frac{1}{2}{\left(2\right)}^5{\left(-1\right)}^4 \\ =\frac{1}{2}\times 32\times 1 \\ =16 \end{gathered}$$

Thus, the answer is $16$.