Rational Expression Confusion.

**Cabletown** · September 11th 2009, 01:46 PM

Hello, I'm just doing a quick review of old material for my pre-calc class and I believe I'm in for a rude awakening. It's been at least 2 years since I've done algebra and I've pretty much forgotten most of it. If somebody could be so kind as to walk me through the steps necessary to solve this I'd be very great-full, thank you.

Instructions are to simplify:

b - a^-1 / a - b^-1

**artvandalay11** · September 11th 2009, 01:59 PM

$a^{-1}=\frac{1}{a}$ and likewise $b^{-1}=\frac{1}{b}$

Therefore

$\frac{b-a^{-1}}{a-b^{-1}}=\frac{b-\frac{1}{a}}{a-\frac{1}{b}}$

Now you can always multiply anything by the number 1, and since $\frac{ab}{ab}=1$ , let's multiply by that to get rid of the $\frac{1}{a}$ in the numerator and the $\frac{1}{b}$ in the denominator
$<br /> <br /> \frac{b-\frac{1}{a}}{a-\frac{1}{b}}\cdot \frac{ab}{ab}=<br />$

$=\frac{(b-\frac{1}{a})ab}{(a-\frac{1}{b})ab}$

$=\frac{ab^2-b}{a^2b-a}$ and we can factor both the top and the bottom

$=\frac{b(ab-1)}{a(ab-1)}$ and now we can cancel

$=\frac{b}{a}$

**Cabletown** · September 11th 2009, 02:23 PM

Thank you for the quick reply, however, I'm a bit confused when you multiply both by ab. What exactly happened to the 1/a and 1/b in the problem. Thank you!

I understand the part below.

**artvandalay11** · September 11th 2009, 02:27 PM

$\frac{1}{x} \cdot x=\frac{x}{x}=1$

$\frac{1}{x}\cdot xy=\frac{1}{x}\cdot x\cdot y=1\cdot y=y$

$\frac{1}{a}\cdot ab=b$ because the a's cancel (assuming a is not 0)

$(b-\frac{1}{a})\cdot ab=b\cdot ab -\frac{1}{a} \cdot ab$

$=b\cdot b\cdot a -\frac{ab}{a}=b^2a-b=ab^2-b$

and a similar approach is done for the denominator

**Cabletown** · September 11th 2009, 02:34 PM

Ok, that cleared it up for me, thank you very much for the assistance!

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