Question

Help please! I already tried doing this but I didn't get the answer right, and I don't know where I went wrong

Answer 1

I think factorizing everything you can first will make the simplification ... well, simpler.

$\dfrac{x^2 - 3x}{x^2 + 13x + 36} \div \dfrac{x^2+7x}{x^2+16x+63} = \dfrac{x(x-3)}{(x+4)(x+9)} \div \dfrac{x(x+7)}{(x+7)(x+9)}$

The factors of $x+7$ in the second rational expression cancel:

$\dfrac{x^2 - 3x}{x^2 + 13x + 36} \div \dfrac{x^2+7x}{x^2+16x+63} = \dfrac{x(x-3)}{(x+4)(x+9)} \div \dfrac{x}{x+9}$

Now, use the property

$\dfrac ab \div \dfrac cd = \dfrac ab \times \dfrac dc$

(this is the property of multiplication having to do with multiplicative inverse, or "inverting the divisor" as the question calls it) to write

$\dfrac{x^2 - 3x}{x^2 + 13x + 36} \div \dfrac{x^2+7x}{x^2+16x+63} = \dfrac{x(x-3)}{(x+4)(x+9)} \times \dfrac{x+9}x$

and we see some more cancellation, namely of the factors of $x$ and $x+9$.

$\dfrac{x^2 - 3x}{x^2 + 13x + 36} \div \dfrac{x^2+7x}{x^2+16x+63} = \boxed{\dfrac{x-3}{x+4}}$