# simplifying this expression...

• March 10th 2008, 06:19 AM
jvignacio
simplifying this expression...
hey guys i only got upto the second stage of simplifying this question.
any help.. thank u.

$

(\frac {2} {y} - \frac {3y-2} {y-1}) / (\frac {y} {y-1})

$

Note: "/" = these two fractions over this fraction.
• March 10th 2008, 07:27 AM
colby2152
Quote:

Originally Posted by jvignacio
hey guys i only got upto the second stage of simplifying this question.
any help.. thank u.

$

(\frac {2} {y} - \frac {3y-2} {y-1}) / (\frac {y} {y-1})

$

Note: "/" = these two fractions over this fraction.

$(\frac {2} {y} - \frac {3y-2} {y-1}) / (\frac {y} {y-1})$

$(\frac{2(y-1)}{y^2-y} - \frac{(3y-2)y}{y^2-y} )/ (\frac {y}{y-1})$

$\frac{2(y-1)-(3y-2)y}{y^2-y} / (\frac {y}{y-1})$

$\frac{[2(y-1)-(3y-2)y](y-1)}{y^3-y^2}$

$\frac{2(y-1)-(3y-2)y}{y^2}$

$\frac{2y-2-3y^2+2y}{y^2}$

$\frac{-3y^2+4y-2}{y^2}$
• March 10th 2008, 11:43 AM
Soroban
Hello, jvignacio!

Quote:

Simplify: . $\frac{\dfrac {2} {y} - \dfrac {3y-2} {y-1}} {\dfrac {y} {y-1}}$

Multiply top and bottom by $y(y-1)$

. . $\frac{y(y-1)}{y(y-1)} \cdot\frac{\dfrac{2}{y} - \dfrac{3y-2}{y-1}}{\dfrac{y}{y-1}} \;\;=\;\;\frac{2(y-1) - y(3y-2)}{y^2}$

. . $= \;\;\frac{2y-2 - 3y^2 + 2y}{y^2} \;\;=\;\frac{-3y^2 + 4y - 2}{y^2}$