Question

PLEASE HELP ME!!!

Answer 1

The answer is:

$\frac{3y(2x-1)(x-3}{x(x-6)(x+1)}$

Answer 2

Answer: $\frac{3y(2x-1)}{x(x-2)}$

Step-by-step explanation:

Given problem to simplify is $\frac{\frac{15xy^2}{x^2-5x+6} }{\frac{5x^2y}{2x^2-7x+3} } \\\\=\frac{15xy^2}{x^2-5x+6}\cdot\ \frac{2x^2-7x+3} {5x^2y}............\text{(by fractional division property,the denominator invert and its reciprocal multiply to numerator)}\\\\=\frac{15xy^2}{(x-3)(x-2)}\cdot\frac{(x-3)(2x-1)} {5x^2y}.........\ \text{(factorizing quadratic equation)}$

$=\frac{3y(2x-1)}{x(x-2)}........\text{(cancelling like terms with same degree and coefficients from numerator and denominator)}$

so the simplified form of the given problem is $\frac{3y(2x-1)}{x(x-2)}$