hey guys i only got upto the second stage of simplifying this question.
any help.. thank u.
$\displaystyle
(\frac {2} {y} - \frac {3y-2} {y-1}) / (\frac {y} {y-1})
$
Note: "/" = these two fractions over this fraction.
$\displaystyle (\frac {2} {y} - \frac {3y-2} {y-1}) / (\frac {y} {y-1}) $
$\displaystyle (\frac{2(y-1)}{y^2-y} - \frac{(3y-2)y}{y^2-y} )/ (\frac {y}{y-1})$
$\displaystyle \frac{2(y-1)-(3y-2)y}{y^2-y} / (\frac {y}{y-1})$
$\displaystyle \frac{[2(y-1)-(3y-2)y](y-1)}{y^3-y^2}$
$\displaystyle \frac{2(y-1)-(3y-2)y}{y^2}$
$\displaystyle \frac{2y-2-3y^2+2y}{y^2}$
$\displaystyle \frac{-3y^2+4y-2}{y^2}$
Hello, jvignacio!
Simplify: .$\displaystyle \frac{\dfrac {2} {y} - \dfrac {3y-2} {y-1}} {\dfrac {y} {y-1}} $
Multiply top and bottom by $\displaystyle y(y-1)$
. . $\displaystyle \frac{y(y-1)}{y(y-1)} \cdot\frac{\dfrac{2}{y} - \dfrac{3y-2}{y-1}}{\dfrac{y}{y-1}} \;\;=\;\;\frac{2(y-1) - y(3y-2)}{y^2}$
. . $\displaystyle = \;\;\frac{2y-2 - 3y^2 + 2y}{y^2} \;\;=\;\frac{-3y^2 + 4y - 2}{y^2}$