Complex Fraction Help

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October 23rd 2007, 04:22 PM
fluffy_penguin

Complex Fraction Help

Can someone help me in detail showing their steps to get the answer to this problem:

http://img152.imageshack.us/img152/3...ractiongg2.jpg
October 23rd 2007, 04:45 PM
Krizalid

Here's a Hint:

$\frac{{y - \dfrac{{5y - 7}}<br /> {7}}}<br /> {{\dfrac{6}<br /> {{14}} + \dfrac{3}<br /> {{2y}}}} = \frac{{14\left( {y - \dfrac{{5y - 7}}<br /> {7}} \right)}}<br /> {{14\left( {\dfrac{6}<br /> {{14}} + \dfrac{3}<br /> {{2y}}} \right)}}.$

Can you see that it is easier now?
October 23rd 2007, 04:53 PM
fluffy_penguin

Not really, no. > >

I see where you got the 14 from since it's the lcd but I've no clue what to do next.
October 23rd 2007, 07:21 PM
Soroban

Hello, fluffy_penguin!

Quote:

Use any method to simplify the complex fraction.

$\frac{y - \frac{5y-7}{7}}{\frac{6}{14} - \frac{3}{2y}}$

Multiply top and bottom by the LCD, $14y$

$\frac{14y\left(y - \frac{5y-7}{7}\right)}{14y\left(\frac{6}{14} - \frac{3}{2y}\right)} \;=\;\frac{14y\cdot y \:- \:14y\cdot\frac{5y-7}{7}}{14y\cdot\frac{6}{14} \:- \:14y\cdot\frac{3}{2y}} \;= \;\frac{14y - 2y(5y-7)}{6y - 7\cdot3}$

. . $=\;\frac{14y^2 - 10y^2 + 14y}{6y-21} \;= \;\frac{4y^2+14y}{6y-21} \;=\;\frac{2y(2y + 7)}{3(2y-7)}$

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