Clueless on n/(n+3) + (n+9)/(n^2+8n+15), etc.

999x2/3

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Jul 9, 2007
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3
__n__......___n+9___
..n+3........n^2 + 8n + 15

4..........__2a__
...............a+1


sorry its the only way i could post that up, in between the 2 fractions on problem number 1 there is a +

and a minus on the second problem... most of my work consists of problems like these but i never learned this in school... im pretty sure i have an answer to #2 but if someone could please run me through the process or simplifying these, it would help so much and i would be able to do the rest of my assignment...
 

pka

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Jan 29, 2005
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\(\displaystyle \L\begin{array}{rcl}
\frac{n}{{n + 3}} + \frac{{n + 9}}{{n^2 + 8n + 15}} &=& \frac{n}{{n + 3}} + \frac{{n + 9}}{{(n + 3)(n + 5)}} \\
&=& \frac{{n(n + 5)}}{{\left( {n + 3} \right)\left( {n + 5} \right)}} + \frac{{n + 9}}{{(n + 3)(n + 5)}} \\
&=& \frac{{n^2 + 6n + 9}}{{\left( {n + 3} \right)\left( {n + 5} \right)}} \\
&=& \frac{{\left( {n + 3} \right)^2 }}{{\left( {n + 3} \right)\left( {n + 5} \right)}} \\
&=& \frac{{\left( {n + 3} \right)}}{{\left( {n + 5} \right)}},\quad x \not= - 3\quad \\
\end{array}\)

You really should learn to use LaTeX!
 

999x2/3

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Jul 9, 2007
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sorry but i cant look at a problem and understand it completely,

i got lost right about here, where did that come from and why did u throw it next to it


i appreciate ure help but i really need someone to quickly do a simple step by step of what they did, and why they did it... sorry but i cant just learn it by seeing a crapload of numbers bunched together it confuses me
 

galactus

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Sep 28, 2005
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Like any fraction, you have to make the denominators the same to simplify or solve. pka muliplied the top and bottom of the left side by n+5 so the denominators are the same.

Suppose you had

\(\displaystyle \frac{1}{2}+\frac{1}{6}\)

You could write this as

\(\displaystyle \L\\\frac{1}{2\cdot{1}}+\frac{1}{3\cdot{2}}\)

The denominators must be the same to add, so make them both 6 by mulitplying the top and bottom of the left side by 3. That makes the denominator 6

\(\displaystyle \L\\\frac{3}{3}\cdot\frac{1}{2}+\frac{1}{6}\)

\(\displaystyle \L\\\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\)

See?. Same principle with your problem.
 

999x2/3

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Jul 9, 2007
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aright i understand now, thank you

What about the second one. 4 minus (2a over a+1) is it the same principle? i know you have to turn the 4 into a fraction by making it 4/1 but after that im having trouble getting a common denomonator between a number and a variable
 

Denis

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Feb 17, 2004
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1,465
RULE:
x - a/b
= (bx - a) / b

SO:
4 - 2a / (a +1)
= [4(a +1) - 2a] / (a + 1)
 

soroban

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Jan 28, 2005
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5,588
Hello, 2999/3!

\(\displaystyle \L4\,-\,\frac{2a}{a\,+\,1}\)

Same rule as before . . . get a common denominator.

We have: \(\displaystyle \L\:\frac{4}{1}\,-\,\frac{2a}{a\,+\,1}\)

. . The common denominator is \(\displaystyle (a\,+\,1)\)


Multiply the first fraction by \(\displaystyle \frac{a+1}{a+1}\)

. . \(\displaystyle \L\frac{4}{1}\,\cdot\,\frac{a\,+\,1}{a\,+\,1}\:-\:\frac{2a}{a\,+\,1} \;=\;\frac{4a\,+\,4}{a\,+\,1}\:-\:\frac{2a}{a\,+\,1}\)


Now they have the same denominator; we can combine the numerators.

. . \(\displaystyle \L\frac{4a\,+\,4\,-\,2a}{a\,+\,1} \;=\;\frac{2a\,+\,4}{a\,+\,1}\;\;\text{or}\;\;\frac{2(a\,+\,2)}{a\,+\,1}\)

 
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