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

999x2/3

New member
Joined
Jul 9, 2007
Messages
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...
 
\(\displaystyle \L\begin{array}{rcl}
\frac{n}{{n + 3}} + \frac{{n + 9}}{{n^2 + 8n + 15}} &=& \frac{n}{{n + 3}} + \frac{{n + 9}}{{(n + 3)(n + 5)}} \\
&=& \frac{{n(n + 5)}}{{\left( {n + 3} \right)\left( {n + 5} \right)}} + \frac{{n + 9}}{{(n + 3)(n + 5)}} \\
&=& \frac{{n^2 + 6n + 9}}{{\left( {n + 3} \right)\left( {n + 5} \right)}} \\
&=& \frac{{\left( {n + 3} \right)^2 }}{{\left( {n + 3} \right)\left( {n + 5} \right)}} \\
&=& \frac{{\left( {n + 3} \right)}}{{\left( {n + 5} \right)}},\quad x \not= - 3\quad \\
\end{array}\)

You really should learn to use LaTeX!
 
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
math1.gif


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
 
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.
 
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
 
RULE:
x - a/b
= (bx - a) / b

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

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

 
Top