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

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Solution

## To find the product, we will use distributive law as follows: $-\frac{3}{2}{x}^{2}{y}^{3}×\left(2x-y\right)\phantom{\rule{0ex}{0ex}}=\left(-\frac{3}{2}{x}^{2}{y}^{3}×2x\right)-\left(-\frac{3}{2}{x}^{2}{y}^{3}×y\right)\phantom{\rule{0ex}{0ex}}=\left(-3{x}^{2+1}{y}^{3}\right)-\left(-\frac{3}{2}{x}^{2}{y}^{3+1}\right)\phantom{\rule{0ex}{0ex}}=-3{x}^{3}{y}^{3}+\frac{3}{2}{x}^{2}{y}^{4}$ Substituting x = 1 and y = 2 in the result, we get: $-3{x}^{3}{y}^{3}+\frac{3}{2}{x}^{2}{y}^{4}\phantom{\rule{0ex}{0ex}}=-3{\left(1\right)}^{3}{\left(2\right)}^{3}+\frac{3}{2}{\left(1\right)}^{2}{\left(2\right)}^{4}\phantom{\rule{0ex}{0ex}}=-3×1×8+\frac{3}{2}×1×16\phantom{\rule{0ex}{0ex}}=-24+24\phantom{\rule{0ex}{0ex}}=0$ 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.  Suggest Corrections  0      Similar questions
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