# Thread: Find simple form of algebraic expression

1. ## Find simple form of algebraic expression

Question is in the link. Just zoom it in for a clearer view. I was trying to multiply the numerator and the denominator each separate term by the denominators of the other two terms, but it was becoming too complex.
help is much appreciated!

http://img19.imageshack.us/img19/5452/69306157.jpg

Thanks,

Auger

2. $\frac{ x - 1 }{\sqrt{x} - 1 } - \frac{ x^2 - 2x -3 }{ x + 1 } - \sqrt{x} =$

$= \frac{(\sqrt{x} - 1)(\sqrt{x} + 1 )}{\sqrt{x} - 1} - \frac{(x -3)(x + 1 )}{ x + 1 } - \sqrt{x} =$

$= \sqrt{x} + 1 - ( x - 3 ) - \sqrt{x} = 4 - x$

3. Originally Posted by Auger
Question is in the link. Just zoom it in for a clearer view. I was trying to multiply the numerator and the denominator each separate term by the denominators of the other two terms, but it was becoming too complex.
help is much appreciated!

http://img19.imageshack.us/img19/5452/69306157.jpg

Thanks,

Auger

$\frac{x-1}{\sqrt{x}-1}-\frac{x^2-2x-3}{x+1}-\sqrt{x}$

Multiply the numerator and denominator of first term by the conjugate of the denominator.
$\frac{(x-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{(x-1)(\sqrt{x}+1)}{x-1}=\sqrt{x}+1$

For the next term, we have $\frac{x^2-2x-3}{x+1}=\frac{(x+1)(x-3)}{x+1}=x-3$

4. Originally Posted by josipive
$\frac{ x - 1 }{\sqrt{x} - 1 } - \frac{ x^2 - 2x -3 }{ x + 1 } - \sqrt{x} =$

$= \frac{(\sqrt{x} - 1)(\sqrt{x} + 1 )}{\sqrt{x} - 1} - \frac{(x -3)(x + 1 )}{ x + 1 } - \sqrt{x} =$

$= \sqrt{x} + 1 - ( x - 3 ) - \sqrt{x} = 4 - x$
Beat me by six minutes. lol