Solve the following

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


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:

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


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



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 .




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