Question
Evaluate following when $x = 2, y = −1.$
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$
✓
Answer
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.$a^m \times a^n=a^{m+n}$.
We have:
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}×\text{x}^2×\text{x}^2\big)×\big(\text{y}×\text{y}×\text{y}^2)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}^{1+2+2}\big)×\big(\text{y}^{1+1+2}\big)$
$=\frac{1}{2}\text{x}^5\text{y}^4$
$\therefore(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)=\frac{1}{2}\text{x}^5\text{y}^4$
Substituting $x = 2$ and $y = -1$ in the result,we get:
$\frac{1}{2}\text{x}^5\text{y}^4$
$=\frac{1}{2}(2)^5(−1)^4$
$=\frac{1}{2}×32×1$
$= 16$
Thus, the answer is $16.$
We have:
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}×\text{x}^2×\text{x}^2\big)×\big(\text{y}×\text{y}×\text{y}^2)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}^{1+2+2}\big)×\big(\text{y}^{1+1+2}\big)$
$=\frac{1}{2}\text{x}^5\text{y}^4$
$\therefore(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)=\frac{1}{2}\text{x}^5\text{y}^4$
Substituting $x = 2$ and $y = -1$ in the result,we get:
$\frac{1}{2}\text{x}^5\text{y}^4$
$=\frac{1}{2}(2)^5(−1)^4$
$=\frac{1}{2}×32×1$
$= 16$
Thus, the answer is $16.$
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