Question

Multiply $\frac{2}{3}{x}^3{y}^3$ by $(3x-15y)$ and verify the result for $x=2,y=-1$.

Answer 1

$\frac{2}{3}{x}^3{y}^3\times (3x-15y)$

$=\frac{2}{3}{x}^3{y}^3\cdot 3x-\frac{2}{3}{x}^3{y}^3\cdot 15y$

$=2x^4{y}^3-10x^3{y}^4$

Now, for $x=2,y=-1$, we have

L.H.S.-

$\frac{2}{3}{x}^3{y}^3\times (3x-15y)$

$=\frac{2}{3}{\left(2\right)}^3{(-1)}^3\times (3\cdot 2-15\cdot (-1))$

$=\frac{2}{3}\cdot 8\cdot (-1)\times (6+15)$

$=(-\frac{16}{3})\times 21$

$=-112$

R.H.S.-

$2x^4{y}^3-10x^3{y}^4$

$=2{\left(2\right)}^4{(-1)}^3-10{\left(2\right)}^3{(-1)}^4$

$=2\cdot 16\cdot (-1)-10\cdot 8\cdot 1$

$=-32-80$

$=-112$

$\because$ L.H.S. $=$ R.H.S.

Hence verified.