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

Question:

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

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

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:

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

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

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

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

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

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

$-\frac{7}{4} x^{5} y^{5}$

$=-\frac{7}{4}(2)^{5}(-1)^{5}$

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

$=56$

Thus, the answer is 56 .

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