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

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To do:

We have to multiply $-\frac{3}{2}x^2y^3$ by $(2x-y)$ and verify the answer for $x = 1$ and $y = 2$.

Solution:

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

$=\frac{-3}{2} \times 2 \times x^{2+1} y^{3}-\frac{3}{2} \times(-y) x^{2} y^{3}$

$=-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{3+1}$

$=-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$

If $x=1, y=2$, then

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

$=\frac{-3}{2}(1)^{2}(2)^{3}(2 \times 1-2)$

$=\frac{-3}{2} \times 1 \times 8 \times 0$

$=0$

RHS $=-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$

$=-3(1)^{3}(2)^{3}+\frac{3}{2}(1)^{2}+(2)^{4}$

$=-3 \times 1 \times 8+\frac{3}{2} \times 1 \times 16$

$=-24+24$

$=0$

Therefore,

LHS $=$ RHS

Updated on 10-Oct-2022 13:19:29