Question

$8x^{2} y^{4} \times | - \frac{3x}{4y^{3} } | =$
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Answer 1

$8x^2y^4\left|-\frac{3x}{4y^3}\right|\\\\=8x^2y^4\left|\frac{3x}{4y^3}\right|\\\\=8x^2y^4\frac{\left|3x\right|}{\left|4y^3\right|}\\\\=8x^2y^4\frac{3\left|x\right|}{4\left|y^3\right|}\\=\frac{3\left|x\right|\times \:8x^2y^4}{4\left|y^3\right|}\\\\=\frac{24x^2y^4\left|x\right|}{4\left|y^3\right|}\\\\\\\\\\\\\\=\frac{6x^2y^4\left|x\right|}{\left|y^3\right|}$