# Evaluate each of the following when x = 2, y = −1.

Question:

Evaluate each of the following when x = 2, y = −1.

$(2 x y) \times\left(\frac{x^{2} y}{4}\right) \times\left(x^{2}\right) \times\left(y^{2}\right)$

Solution:

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:

$(2 x y) \times\left(\frac{x^{2} y}{4}\right) \times\left(x^{2}\right) \times\left(y^{2}\right)$

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

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

$=\frac{1}{2} x^{5} y^{4}$

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

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

$\frac{1}{2} x^{5} y^{4}$

$=\frac{1}{2}(2)^{5}(-1)^{4}$

$=\frac{1}{2} \times 32 \times 1$

$=16$

Thus, the answer is 16 .