Math 10 Pure Fractions! Add/Multiply/Factor

**SCT** · September 30th 2008, 06:05 PM

alright, my math is easy but theres always a few hard ones. just wondering if anyone can do these, show restrictions and show as much work as possible, or explain the steps.

4 | 5
--------- | - -----------
y^2+5y+6 | y^2-y-12

4 | 3
--------- | + -------------
w^2+5w+4 | w^2+2w+8

6b^3 |+ 9g^5
------- | ---------
8g^5 | 12b^3

1 + | 4 |- 6
-------|------ | -------
2x^2 | 3x | x^3

Thanks for any help

**Soroban** · September 30th 2008, 06:54 PM

Hello, SCT!

I must assume you know about finding the Lowest Common Denominator
. . and converting the fractions to the LCD.

$\frac{4}{y^2+5y+6} - \frac{5}{y^2-y-12}$

We have: . $\frac{4}{(y+2)(y+3)} - \frac{5}{(y+3)(y-4)} \qquad LCD \:=\:(y+2)(y+3)(y-4)$

Convert: . $\frac{4}{(y+2)(y+3)}\cdot{\color{blue}\frac{y-4}{y-4}} \;- \;\frac{5}{(y+3)(y-4)} \cdot{\color{blue}\frac{y+2}{y+2}}$

. . . . $=\;\frac{4(y-4)}{(y+2)(y+3)(y+4)} - \frac{5(y+2)}{(y+3)(y-4)(y+2)}$

. . . . $= \;\frac{4y - 16 - 5y - 10}{(y+2)(y+3)(y-4)}$

. . . . $= \;\frac{-y-26}{(y+2)(y+3)(y-4)}$

$\frac{6b^3}{8g^5} + \frac{9g^5}{12b^3}$

Reduce the fractions: . $\frac{3b^3}{4g^5} + \frac{3g^5}{4b^3} \qquad LCD \:=\:4b^3g^5$

Convert: . $\frac{3b^3}{4g^5}\cdot{\color{blue}\frac{b^3}{b^3} } \;+\; \frac{3g^5}{4b^3}\cdot{\color{blue}\frac{g^5}{g^5} } \;= \;\frac{3b^6}{4b^3g^5} + \frac{3g^{10}}{4b^3g^5} \;=\;\frac{3b^6 + 3g^{10}}{4b^3g^5}$

$\frac{1}{2x^2} + \frac{4}{3x} - \frac{6}{x^3}$

The LCD is: . $6x^3$

Convert: . $\frac{1}{2x^2}\cdot{\color{blue}\frac{3x}{3x}} \;+\;\frac{4}{3x}\cdot{\color{blue}\frac{2x^2}{2x^ 2}} \;-\;\frac{6}{x^3}\cdot{\color{blue}\frac{6}{6}}$

. . . . $= \;\frac{3x}{6x^3} + \frac{8x^2}{6x^3} - \frac{36}{6x^3} \;=\;\frac{8x^2 + 3x - 36}{6x^3}$

**SCT** · September 30th 2008, 07:12 PM

thanks!
any idea about

m - 3
------ -----
2m-4 30-6

and

4 3
------ + --------
w^2+5w+4 w^2+2w-8

just not getting the 2nd cause of the lcd and such and first one i keep getting different answers

**icemanfan** · September 30th 2008, 07:17 PM

Did you factor the denominators?

$\frac{4}{w^2 + 5w + 4} + \frac{3}{w^2 + 2w - 8} =$

$\frac{4}{(w+4)(w+1)} + \frac{3}{(w+4)(w-2)}$

The LCD is $(w+4)(w+1)(w-2)$ .

**SCT** · September 30th 2008, 07:23 PM

Ah, thank you.
Figured out that question. (factored +3 +2 into the 5w+4 for some reason. 5, 3+2 instinct!)
thank you.
anyone know about the other one?

**icemanfan** · September 30th 2008, 07:36 PM

Is the problem to simplify

$\frac{m}{2m-4} - \frac{3}{30-6}$ ?

**SCT** · September 30th 2008, 07:38 PM

no, i need to find the LCD and solve.
m doesnt need to be evaluated

**icemanfan** · September 30th 2008, 07:41 PM

$\frac{m}{2m-4} - \frac{3}{30-6} = \frac{m}{2(m-2)} - \frac{1}{8}$ .

The LCD is $8(m-2)$ .

Can you do the rest?

**SCT** · September 30th 2008, 07:43 PM

im confused.. the way our teacher taught us (maybe im incorrect)
but is to go
m
---
2(m-2)
and
3
-----
3(m-2)

making the LCD 6(m-2)
or cancelling the (m-2)'s out and making it 6?