Simplify

[rsheedy1]

The question is (2+3/x)/(2/x-3/x^2) * (3/x^2-2/x)/(3/x^2+2/x^3).
I factored out -1 from the numerator of the second fraction, so I could cross out the denominator of the first and the numerator of the second.

I now have -1(2+3/x) / (3/x^2+2/x^3). Is this all I can do, or do I put, for the numerator, -1(2+3/x)?

If I can do that, and then multiply the by the denominator's reciprical, I end up crossing out the 5's and one of the x. For a final answer I got -x(x+1)...

Did I do this right?

 

[rsheedy1]

Okay. I understand how to get to that point. Now multiply by the reciprical and get x^2(2x-3) / 3x+2...

Is that all I can do?

 

[soroban]

Hello, rsheedy1!

\(\displaystyle \text{Simplify: }\:\dfrac{2+\frac{3}{x}}{\frac{2}{x}-\frac{3}{x^2}} \cdot\dfrac{\frac{3}{x^2}-\frac{2}{x}}{\frac{3}{x^2}+\frac{2}{x^3}}\)

Multiply the first fraction by \(\displaystyle \frac{x^2}{x^2}\); multiply the second fraction by \(\displaystyle \frac{x^3}{x^3}\)

. . \(\displaystyle \dfrac{x^2\left(2 + \frac{3}{x}\right)}{x^2\left(\frac{2}{x} - \frac{3}{x^2}\right)} \cdot \dfrac{x^3\left(\frac{3}{x^2} - \frac{2}{x}\right)}{x^3\left(\frac{3}{x^2} + \frac{2}{x^3}\right)} \;=\;\dfrac{2x^2+3x}{2x-3}\cdot\dfrac{3x-2x^2}{3x+2} \)

. . \(\displaystyle =\;\dfrac{x(2x+3)}{2x-3}\cdot\dfrac{-x(2x -3)}{3x+2} \;=\; -\dfrac{x^2(2x+3)}{3x+2} \)