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

#### 999x2/3

##### New member
__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

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

You really should learn to use LaTeX!

#### 999x2/3

##### New member
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

##### Super Moderator
Staff member
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.

$$\displaystyle \frac{1}{2}+\frac{1}{6}$$

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

##### New member
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

##### Senior Member
RULE:
x - a/b
= (bx - a) / b

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

#### soroban

##### Elite Member
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.

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