# How do you simplify this algebraically?

• Sep 16th 2010, 09:14 PM
yess
How do you simplify this algebraically?
Algebraically simplify: -x^(-1) +1-(x-1)(x^-2) to ((1/x)-1)^2
• Sep 16th 2010, 09:36 PM
pickslides
You have $\displaystyle \frac{-1}{x}+1-\frac{x-1}{x^2}$

I would suggest starting by making a common denominator of $x^2$
• Sep 17th 2010, 01:28 AM
Educated
Remember that multiplying by negative powers is the same as dividing by that positive power.

So you would have the equation of:

$\dfrac{-1}{x}+1-\dfrac{x-1}{x^2}$

Make each of them have a common denominator by multiplying by corresponding values of x:

$= \dfrac{-x}{x^2}+\dfrac{x^2}{x^2}-\dfrac{x-1}{x^2}$

$=\dfrac{-x+x^2-x+1}{x^2}$

$=\dfrac{x^2-2x+1}{x^2}$

$=\dfrac{(1-x)^2}{x^2}$

Note, x^2-2x+1 can be factorised to (x-1)^2 or (1-x)^2. I used (1-x)^2 because it helps in getting the final answer you gave.

$=\left(\dfrac{1-x}{x}\right)^2$

$=(\frac{1}{x}-\frac{x}{x})^2$

$=(\frac{1}{x}-1)^2$
• Sep 17th 2010, 05:22 AM
Soroban
Hello, yess!

Quote:

$\text{Simplify }\,-x^{-1} +1-(x-1)x^{-2}\:\text{ to }\:\left(\dfrac{1}{x}-1\right)^2$

We have: . $-\dfrac{1}{x} + 1 - \left(\dfrac{x-1}{x^2}\right) \;\;=\;\;-\dfrac{1}{x} + 1 - \left(\dfrac{x}{x^2} - \dfrac{1}{x^2}\right)$

. . . . . . $=\;\;-\dfrac{1}{x} + 1 - \dfrac{1}{x} + \dfrac{1}{x^2} \;\;=\;\;\dfrac{1}{x^2} - \dfrac{2}{x} + 1$

Factor: . $\left(\dfrac{1}{x} - 1\right)^2$
• Sep 17th 2010, 05:46 AM
Quote:

Originally Posted by yess
Algebraically simplify: -x^(-1) +1-(x-1)(x^-2) to ((1/x)-1)^2

$-x^{-1}+1-(x-1)x^{-2}$

$=-\frac{1}{x}-(-1)-(x-1)x^{-2}$

$=-\left(\frac{1}{x}-1\right)+(1-x)x^{-2}$

$=-\left(\frac{1}{x}-1\right)+\frac{1}{x}(1-x)x^{-2}x$

$=-\left(\frac{1}{x}-1\right)+\left(\frac{1}{x}-1\right)x^{-1}$

$=\left(\frac{1}{x}-1\right)\left(x^{-1}-1\right)=\left(\frac{1}{x}-1\right)\left(\frac{1}{x}-1\right)$