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Moon Rabbits
02-28-2008, 02:44 AM
Simplify.

(7y^2 / y^2 - 9) * (4y + 2 / 14y^2)

Tavrobel
02-28-2008, 02:55 AM
Begin by distributing terms to both the numerator and the denominator. Since this is multiplication, there shouldn't be anything wrong with that.

(7y<sup>2</sup>)(4y) + (7y<sup>2</sup>)(2) / (4y<sup>2</sup>)(y<sup>2</sup>) - (4y<sup>2</sup>)(9)

Combine like terms, if applicable. Then simplify.

(28y<sup>3</sup>+14y<sup>2</sup>) / (4y<sup>4</sup>-36y<sup>2</sup>)

Since we have a Y squared in every term, remove them from the problem.

(28y+14) / (4y<sup>2</sup>-36)

Factor 14 from the top (optional step).

14*(2y+1) / (4y<sup>2</sup>-36)

Factor out 2 from both the top and the bottom.

7*(2y+1) / (2y<sup>2</sup>-18)

Optionally:
Factor out a 2 from the bottom. That leaves you with a difference of squares.

7(2y+1) / 2(y<sup>2</sup>-9)
7(2y+1) / 2(y+3)(y-3)

If you need to do something else with the problem, now would be the time to do so.

Math sucks. Your only hope is suicide.

rubah
02-28-2008, 03:29 AM
It'd be a lot easier if you divided seven y squared by fourteen y squared at the onset, leaving you with a 2 in the denominator :]] Then you just have (2y-1)/(y-3)(y+3) don't you?

Tavrobel
02-28-2008, 04:02 AM
Ohh :skull::skull::skull::skull:. I misread it as 4y^2. Waah.

crono_logical
02-28-2008, 08:32 AM
It'd be a lot easier if you divided seven y squared by fourteen y squared at the onset, leaving you with a 2 in the denominator :]] Then you just have (2y-1)/(y-3)(y+3) don't you?Where'd the 2y-1 come from, rubah? :p

o_O
02-28-2008, 09:13 AM
Yeah, Allie. :p

Here's the working for anyone that wants it:
<pre>(7y<sup>2</sup> / y<sup>2</sup> - 9) * (4y + 2 / 14y<sup>2</sup>)
= (7y<sup>2</sup> / 14y<sup>2</sup>) * (4y + 2 / y<sup>2</sup> - 9) Rearranging the terms into a more desirable order, by commutativity.
= (1 / 2) * (4y + 2 / y<sup>2</sup> - 9) Reducing left multiplicand.
= (4y + 2 / 2y<sup>2</sup> - 18) Multiplying multiplicands.
= (2y + 1 / y<sup>2</sup> - 9) Reducing again.
= (2y + 1) / (y + 3)(y - 3) Factorising.
</pre>

blackmage_nuke
02-28-2008, 09:32 AM
If you take the 2 out of 4y+2 to get 2(2y+1) then use that 2 to multiply with the 7y^2 to get 14y^2 then that cancels out with the 14y^2 on the bottom and your left with what o_O said

edit: and put your brackets correctly. i guess most people assume you meant
(7y^2 / (y^2 - 9)) * ((4y + 2) / 14y^2)

But sometimes that wont be so obvious

rubah
02-28-2008, 03:48 PM
It'd be a lot easier if you divided seven y squared by fourteen y squared at the onset, leaving you with a 2 in the denominator :]] Then you just have (2y-1)/(y-3)(y+3) don't you?Where'd the 2y-1 come from, rubah? :p

gosh, didn't you see the +2 right afterwards? I must have forgotten it 8)

Flying Mullet
02-28-2008, 03:57 PM
Similar approach to what's above.

(7y^2 / y^2 - 9) * (4y + 2 / 14y^2)

= (7y^2 / y^2 - 9) * (2y + 1 / 7y^2) - factor out a 2 from the second fraction

= (1/ y^2 - 9) * (2y + 1 / 1) = cancel the 7y^2

= (2y + 1) / (y^2 - 9) = multiply

= (2y + 1) / (y +/- 3)^2 = factor the denominator if need be

crono_logical
02-28-2008, 07:59 PM
Though once you get good at it, you can do these in your head :p

Tavrobel
02-28-2008, 11:11 PM
But only if you copy down the problem correctly.

Flying Mullet
02-28-2008, 11:21 PM
Details, details...

o_O
02-28-2008, 11:49 PM
If you're good enough at maths, you can solve things even if you have the wrong problem to begin with. :p

rubah
02-29-2008, 02:43 AM
*bad enough

geeknick
11-15-2011, 07:44 AM
I was wondering, in this question to simplifying rational expressions (http://rational-expressions-solutions-online.blogspot.com/2011/11/online-solver-solves-rational.html) ??

Find (-6x^2 - 4xy + 8x)/(2x)

do they want me to find out what the variables are, or do they want me
to do something else with it? That is the exact way they have the
question written down.

rubah
11-15-2011, 04:53 PM
They want you to simplify it, get rid of the 2x. if it doesn't come out nicely, then you just say that it's simplified already.

that y term is kinda weird, though