Question:
Multiply $-\frac{3}{2} x^{2} y^{3}$ by $(2 x-y)$ and verify the answer for $x=1$ and $y=2$.
Solution:
To find the product, we will use distributive law as follows:
$-\frac{3}{2} x^{2} y^{3} \times(2 x-y)$
$=\left(-\frac{3}{2} x^{2} y^{3} \times 2 x\right)-\left(-\frac{3}{2} x^{2} y^{3} \times y\right)$
$=\left(-3 x^{2+1} y^{3}\right)-\left(-\frac{3}{2} x^{2} y^{3+1}\right)$
$=-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$
Substituting x = 1 and y = 2 in the result, we get:
$-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$
$=-3(1)^{3}(2)^{3}+\frac{3}{2}(1)^{2}(2)^{4}$
$=-3 \times 1 \times 8+\frac{3}{2} \times 1 \times 16$
$=-24+24$
$=0$
Thus, the product is $-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$, and its value for $x=1$ and $y=2$ is 0 .