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

Question

asked Feb 4, 2023 80.0k views

1 Answer

Steve Tarver · Answer 1 · 2023-02-10T12:03:58+0000

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

(x^2 - 3x)/(x^2 + 13x + 36) / (x^2+7x)/(x^2+16x+63) = (x(x-3))/((x+4)(x+9)) / (x(x+7))/((x+7)(x+9))

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

(x^2 - 3x)/(x^2 + 13x + 36) / (x^2+7x)/(x^2+16x+63) = (x(x-3))/((x+4)(x+9)) / (x)/(x+9)

Now, use the property

$\frac ab / \frac cd = \frac ab * \frac dc$

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

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

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

$(x^2 - 3x)/(x^2 + 13x + 36) / (x^2+7x)/(x^2+16x+63) = \boxed{(x-3)/(x+4)}$