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:

\[\left( \frac{3}{5} x^2 y \right) \times \left( - \frac{15}{4}x y^2 \right) \times \left( \frac{7}{9} x^2 y^2 \right)\]

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

\[ = \left\{ \frac{3}{5} \times \left( - \frac{15}{4} \right) \times \frac{7}{9} \right\} \times \left( x^{2 + 1 + 2} \right) \times \left( y^{1 + 2 + 2} \right)\]

\[ = - \frac{7}{4} x^5 y^5\]

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

Substituting x = 2 and y = \[-\] 1 in the result, we get:

\[- \frac{7}{4} x^5 y^5 \]

\[ = - \frac{7}{4} \left( 2 \right)^5 \left( - 1 \right)^5 \]

\[ = \left( - \frac{7}{4} \right) \times 32 \times \left( - 1 \right)\]

\[ = 56\]

Thus, the answer is 56.