# Simplify this equation

• Aug 21st 2008, 04:44 AM
jerry
Simplify this expression
Simplify: $\displaystyle (a^{-1}+b^{-2})^{-1}$

I know that's simple. I can express that as:

$\displaystyle \frac{1}{\frac{1}{a}+\frac{1}{b^{2}}}$

The problem is that, to me, that doesn't look any simpler.

I can't think of how else to express this. Is there another more appropriate answer?
• Aug 21st 2008, 05:04 AM
Moo
Hi,
Quote:

Originally Posted by jerry
Simplify: $\displaystyle (a^{-1}+b^{-2})^{-1}$

I know that's simple. I can express that as:

$\displaystyle \frac{1}{\frac{1}{a}+\frac{1}{b^{2}}}$

The problem is that, to me, that doesn't look any simpler.

I can't think of how else to express this. Is there another more appropriate answer?

You can just transform 1/a + 1/bē into a single fraction (common denominator), it depends on what you mean by "simple" :p

By the way, this is not an equation. An equation contains the = sign ! This is an expression ^^
• Aug 21st 2008, 06:35 AM
Soroban
Hello, jerry!

Quote:

Simplify: .$\displaystyle (a^{-1}+b^{-2})^{-1}$

I can express that as: .$\displaystyle \frac{1}{\dfrac{1}{a}+\dfrac{1}{b^{2}}}$

The problem is that, to me, that doesn't look any simpler.

You have a complex fraction . . . one with more than two "levels".
. . We can and must simplify it further.

Multiply top and bottom by the LCD, $\displaystyle ab^2$

. . $\displaystyle \frac{{\color{blue}ab^2}\cdot(1)}{{\color{blue}ab^ 2}\cdot\left(\dfrac{1}{a}+\dfrac{1}{b^2}\right)} \;=\;\frac{ab^2}{b^2+a} \quad\hdots\quad see?$

• Aug 21st 2008, 03:56 PM
jerry
OK. Thank you both for that!