1. ## Math 10 Pure Fractions! Add/Multiply/Factor

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

2. Hello, SCT!

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

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

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

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

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

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

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

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

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

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

$\displaystyle \frac{1}{2x^2} + \frac{4}{3x} - \frac{6}{x^3}$
The LCD is: .$\displaystyle 6x^3$

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

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

3. thanks!

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

4. Did you factor the denominators?

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

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

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

5. 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?

6. Is the problem to simplify

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

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

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

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

Can you do the rest?

9. 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?

10. You should always reduce the fractions to lowest terms before figuring out the LCD. I'm still not sure if we're looking at the same problem, though.

11. oops. typed it wrong.

m over 2m-4 subtract 3 over 3m-6

can you get the LCD and then what to multiply the numerators by?

12. $\displaystyle \frac{m}{2m-4} - \frac{3}{3m-6} =$

$\displaystyle \frac{m}{2(m-2)} - \frac{3}{3(m-2)} =$

$\displaystyle \frac{m}{2(m-2)} - \frac{1}{m-2}$

The LCD is $\displaystyle 2(m-2)$.

13. would that make the answer
m-2
-----
2(m-2) therefore .:. one half?