Question

Evaluate each of the following when x = 2, y = −1.
$\left(\frac{3}{5}{x}^{2}y\right)\times \left(-\frac{15}{4}x{y}^2\right)\times \left(\frac{7}{9}{x}^2{y}^2\right)$

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:

$$\begin{gathered} \left(\frac{3}{5}{x}^{2}y\right)\times \left(-\frac{15}{4}x{y}^2\right)\times \left(\frac{7}{9}{x}^2{y}^2\right) \\ =\left\{\frac{3}{5}\times \left(-\frac{15}{4}\right)\times \frac{7}{9}\right\}\times \left(x^2\times x\times x^2\right)\times \left(y\times y^2\times y^2\right) \\ =\left\{\frac{3}{5}\times \left(-\frac{15}{4}\right)\times \frac{7}{9}\right\}\times \left(x^{2+1+2}\right)\times \left(y^{1+2+2}\right) \\ =-\frac{7}{4}{x}^5{y}^5 \end{gathered}$$

$\therefore$ $\left(\frac{3}{5}{x}^{2}y\right)\times \left(-\frac{15}{4}x{y}^2\right)\times \left(\frac{7}{9}{x}^2{y}^2\right)=-\frac{7}{4}{x}^5{y}^5$.

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

$$\begin{gathered} -\frac{7}{4}{x}^5{y}^5 \\ =-\frac{7}{4}{\left(2\right)}^5{\left(-1\right)}^5 \\ =\left(-\frac{7}{4}\right)\times 32\times \left(-1\right) \\ =56 \end{gathered}$$

Thus, the answer is 56.