# Rational Expressions and Radical Expressions

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• July 9th 2009, 07:33 PM
illeatyourxface
I just want to check my answers with someone because im not sure i am correct...

1. 12u cubed/7v to the fifth * 14v squared/3u =
8 u squared/ v cubed

2. 9x * 4/x =
36?

3. x squared -3x +2/x squared -4x +4 divided by x squared -2x +1/ 3x squared -12 =

3(x+2)/ (x-1)

4. 15 divided by 5x/x+1 =
3(x+1) / x

5. 6x/x-2 divided by 3x =
2/x-2
• July 9th 2009, 07:43 PM
yeongil
You really should learn to type in LaTex. I can barely read your post.

Quote:

1. 12u cubed/7v to the fifth * 14v squared/3u =
8 u squared/ v cubed
You mean this?
$\frac{12u^3}{7v^5} \cdot \frac{14v^2}{3u}$
If so, then
$\frac{12u^3}{7v^5} \cdot \frac{14v^2}{3u} = \frac{4u^2}{v^3} \cdot \frac{2}{1} = \frac{8u^2}{v^3}$
It's correct.

Quote:

2. 9x * 4/x =
36?
You mean this?
$9x \cdot \frac{4}{x}$
If so, and assuming that x isn't 0, then 36 is right.

Quote:

3. x squared -3x +2/x squared -4x +4 divided by x squared -2x +1/ 3x squared -12 =

3(x+2)/ (x-1)
I'm not even touching this one. (Headbang)

Quote:

4. 15 divided by 5x/x+1 =
3(x+1) / x
You mean this?
$15 \div \frac{5x}{x + 1}$
$= 15 \cdot \frac{x + 1}{5x} = \frac{3(x + 1)}{x}$

Quote:

5. 6x/x-2 divided by 3x =
2/x-2
You mean this?
$\frac{6x}{x - 2} \div 3x$
$= \frac{6x}{x - 2} \cdot \frac{1}{3x} = \frac{2}{x - 2}$

01
• July 9th 2009, 07:48 PM
illeatyourxface
Thanks again,
how would you solve problem involving addition???

ex) x/ (x squared-9) + 3/(x squared-9)
• July 9th 2009, 07:49 PM
pickslides
Quote:

Originally Posted by illeatyourxface
I just want to check my answers with someone because im not sure i am correct...

1. 12u cubed/7v to the fifth * 14v squared/3u =
8 u squared/ v cubed

$\frac{12u^3}{7v^5}\times\frac{14v^2}{3u}$

$=\frac{12u^3\times 14v^2}{7v^5\times 3u}$

$=\frac{12u^3\times 14v^2}{7v^5\times 3u}$

$=\frac{8u^3\times v^2}{v^5\times u}$

$=8u^2\times v^{-3}$

$=\frac{8u^2}{ v^{-3}}$
• July 9th 2009, 07:51 PM
yeongil
Quote:

Originally Posted by illeatyourxface
Thanks again,
how would you solve problem involving addition???

ex) x/ (x squared-9) + 3/(x squared-9)

If the denominators are the same, just add the numerators up, like this:
$\frac{x}{x^2 - 9} + \frac{3}{x^2 - 9} = \frac{x + 3}{x^2 - 9}$

But we're not done, because the denominator can be factored:
$\frac{x + 3}{x^2 - 9} = \frac{x + 3}{(x + 3)(x - 3)}$

Cancel out the x + 3, and the answer is
$\frac{1}{x - 3}$

01
• July 9th 2009, 07:53 PM
illeatyourxface
Quote:

Originally Posted by pickslides
$\frac{12u^3}{7v^5}\times\frac{14v^2}{3u}$

$=\frac{12u^3\times 14v^2}{7v^5\times 3u}$

$=\frac{12u^3\times 14v^2}{7v^5\times 3u}$

$=\frac{8u^3\times v^2}{v^5\times u}$

$=8u^2\times v^{-3}$

$=\frac{8u^2}{ v^{-3}}$

I thought negative exponents were used only if you moved it from the top or bottom...??
• July 9th 2009, 07:55 PM
Stroodle
Quote:

Originally Posted by illeatyourxface

3. x squared -3x +2/x squared -4x +4 divided by x squared -2x +1/ 3x squared -12 =

Do you mean:

$\frac{x^2-3x+2}{x^2-4x+4}\div\frac{x^2-2x+1}{3x^2-12}$

$=\frac{(x-2)(x-1)}{(x-2)^2}\div\frac{(x-1)^2}{3(x+2)(x-2)}$

$=\frac{3(x-2)^2(x+2)(x-1)}{(x-2)^2(x-1)^2}$

$=\frac{3(x+2)}{x-1}$
• July 9th 2009, 07:57 PM
illeatyourxface
okayyy how would you add or subtract if the demoninator was different?

ex)

4/x cubed y + 7/x y squared
• July 9th 2009, 07:58 PM
illeatyourxface
and im really sorry if this is confusing i dont know how to type in LAtex, im sorrry
• July 9th 2009, 08:06 PM
Stroodle
$\frac{12u^3}{7v^5}\times\frac{14v^2}{3u}$

$=\frac{168u^3v^2}{21v^5u}$

$=\frac{8u^2}{v^3}$

edit* Oops I didn't notice that Yeongil already answered this one correctly...
• July 9th 2009, 08:13 PM
pickslides
Quote:

Originally Posted by illeatyourxface
I thought negative exponents were used only if you moved it from the top or bottom...??

yep my bad, forgot to remove the "-" sign
• July 9th 2009, 08:34 PM
illeatyourxface
to solve it.... would it be like

4/ x cubed y + 7/ x y squared =

(4)(y) + (7) (x squared) / (x cubed y)(x y squared) ????
• July 9th 2009, 08:44 PM
Stroodle
Do you mean:

$\frac{4}{x^3y}+\frac{7}{xy^2}$

$=\frac{4y}{x^3y^2}+\frac{7x^2}{x^3y^2}$

$=\frac{4y+7x^2}{x^3y^2}$
• July 9th 2009, 09:11 PM
illeatyourxface
okay soo if you are subtracting...

11/3x - 5/6x =

33x -5 / 6x ?
• July 9th 2009, 09:16 PM
Stroodle
$\frac{11}{3x}-\frac{5}{6x}$

To make the denominator the same for both terms you multiply the top and bottom of the first term by 2

$=\frac{22}{6x}-\frac{5}{6x}$

$=\frac{22-5}{6x}$

$=\frac{17}{6x}$
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