Pages: 1 2

 

2008-05-08 01:17:37
Funky_Finny
Banned
+456|6604|Carnoustie, Scotland
5(2x-3)
______

4x(squared)-9

So that's 5 brackets 2 ex minus 3 over 4 ex squared minus nine.

I don't need the answer only, how you did it too please.

karma for good answers nd thanksbye.
2008-05-08 03:26:16
_raab
Member
+28|6705|Western Aust.
i'm assuming that equation is = 0 and you are solving for x? Re-arrange the equation so its do-able I guess. Remember, 1/x = x^(-1). Eg 1/5 = .2, 5^(-1) = .2

Expand the top, 5(2x-3) = 10x-15.

And yeah, im stuck/bored here. Anyone else?
2008-05-08 04:40:18
TimmmmaaaaH
Damn, I... had something for this
+725|6911|Brisbane, Australia

_raab wrote:

5(2x-3) = 10x-15.
Thats what she said.
https://bf3s.com/sigs/5e6a35c97adb20771c7b713312c0307c23a7a36a.png
2008-05-08 05:11:31
Mint Sauce
Frighteningly average
+780|6757|eng
Expand out the brackets. Put it = 0 and bring the bottom equation up. Isolate the x's on the same side, and there you go, a value for x.
#rekt
2008-05-08 05:13:15
d.Luxe
Banned
+64|6322

Funky_Finny wrote:

5(2x-3)
______

4x(squared)-9

So that's 5 brackets 2 ex minus 3 over 4 ex squared minus nine.

I don't need the answer only, how you did it too please.

karma for good answers nd thanksbye.
How old are you? We've done that at school already...
2008-05-08 05:15:54
DesertFox-
The very model of a modern major general
+796|7156|United States of America

Funky_Finny wrote:

5(2x-3)
______

4x(squared)-9

So that's 5 brackets 2 ex minus 3 over 4 ex squared minus nine.

I don't need the answer only, how you did it too please.

karma for good answers nd thanksbye.
Just to make sure you have:

5(2x-3)
4x2-9               correct?

Well if you factor that out, you get:

   5(2x-3)   
(2x-3)(2x+3)

Cancel, the (2x-3) and get:
    5   
(2x+3)

I'm not sure what you wanted us to do, but I hope that helped.
2008-05-08 05:30:06
Tetrino
International OMGWTFBBQ
+200|7202|Uhh... erm...
Going from:

__5__
(2x+3)

You use x^-1 = 1/x and bring the (2x+3) up to make

5(2x+3)^-1

Assuming that 5(2x+3)^-1 = 0, you can now divide both sides by 5, and then multiply both sides by (2x+3) twice to get

2x+3 = 0

Then, just transfer the coefficients over to get x = 6.
2008-05-08 05:37:20
AWSMFOX
Banned
+405|6934|A W S M F O X
This will be clearly helpful when being mugged by hobos.
2008-05-08 06:20:27
Vub
The Power of Two
+188|6966|Sydney, Australia

Tetrino wrote:

Going from:

__5__
(2x+3)

You use x^-1 = 1/x and bring the (2x+3) up to make

5(2x+3)^-1

Assuming that 5(2x+3)^-1 = 0, you can now divide both sides by 5, and then multiply both sides by (2x+3) twice to get

2x+3 = 0

Then, just transfer the coefficients over to get x = 6.
Incorrect.

(2x+3)^-1 = 0 is true only when x = infinity. And how the hell did you get 6 haha?
2008-05-08 06:55:58
Tetrino
International OMGWTFBBQ
+200|7202|Uhh... erm...

Vub wrote:

Tetrino wrote:

Going from:

__5__
(2x+3)

You use x^-1 = 1/x and bring the (2x+3) up to make

5(2x+3)^-1

Assuming that 5(2x+3)^-1 = 0, you can now divide both sides by 5, and then multiply both sides by (2x+3) twice to get

2x+3 = 0

Then, just transfer the coefficients over to get x = 6.
Incorrect.

(2x+3)^-1 = 0 is true only when x = infinity. And how the hell did you get 6 haha?
Crap, I forgot about that. In that case, integrate it somewhere along those lines, I guess. Can't remember the damn math for it.
2008-05-08 07:12:41
Ender2309
has joined the GOP
+470|7042|USA
no integration. this looks like algebra 2 which means they ignore a lot of the rules and do the work purely for the theory/experience.

i do believe what he's looking for is bringing up the bottom half like others have said and then just doing the basic algebra.

Last edited by Ender2309 (2008-05-08 07:14:54)

2008-05-08 07:13:52
bob9f6
Member
+11|6460|A sea above the map of Tassie
x=-1.5 and x=0

Just using logic, not the correct/maths way;

5/(2x+3)=0   , so

5/what = 0    ,

   5/0 = 0      ,and

=> 2x+3=0   ,

2x=-3    ,  so

x=-3/2   or -1.5 


Plug back into original equation to check.
2008-05-08 07:18:42
DrunkFace
Germans did 911
+427|7152|Disaster Free Zone
Yeah, but 5/0 doesn't equal anything. Haven't you learnt yet that you can't divide by zero.

Anyway, until Funky actually explains what he wants done to the fraction (you know the actual question) there is nothing that can be done.

Its like saying: 4x/3 I don't need the answer only, how you did it too please.
2008-05-08 07:18:54
Ender2309
has joined the GOP
+470|7042|USA

bob9f6 wrote:

x=-1.5 and x=0

Just using logic, not the correct/maths way;

5/(2x+3)=0   , so

5/what = 0    ,

   5/0 = 0      ,and

=> 2x+3=0   ,

2x=-3    ,  so

x=-3/2   or -1.5 


Plug back into original equation to check.
no. although that is the correct answer, you did the work wrong.
5/0!=0. it equals no solution, or solution dne. you have to bring up (2x+3).
2008-05-08 07:35:31
bob9f6
Member
+11|6460|A sea above the map of Tassie
"no. although that is the correct answer"  ???

Look its been a long time since year 8 but yeah i spose in this case you would say, x=0 is undefined,

or even better set initial limits

x>0 or x<0,

but however

x=-1.5   

edit: oh yeah you dont have to bring (2x+3) up. That is one method but there is many ways to build a table and you still get a table.

Last edited by bob9f6 (2008-05-08 07:37:26)

2008-05-08 11:03:56
stkhoplite
Banned
+564|6951|Sheffield-England
911
2008-05-08 11:09:51
kylef
Gone
+1,352|6964|N. Ireland
5(2x-3)
______

4x(squared)-9
10x - 15
_______                         = 0
4x2-9

###

10x - 15 = -4x2 + 9
4x2 + 10x - 15 = 9
4x2 + 10x - 15 - 9 = 0
4x2 + 10x - 24 = 0

Then, use the quadratic equation... -b +/- sqroot. b2 - 4ac all over 2a

X = -2.22
X = -17.78

Last edited by kylef (2008-05-08 11:21:09)

2008-05-08 11:12:51
mkxiii
online bf2s mek evasion
+509|6707|Uk

kylef wrote:

5(2x-3)
______

4x(squared)-9
10x - 15
_______                         = 0
4x2-9

###

10x - 15 = -4x + 9
16x = 24
x = 1.5
the squared for the x disappeared and 10x--4x = 14x
2008-05-08 11:13:22
Hope_is_lost117
Psy squad
+49|6468|Belgium

bob9f6 wrote:

x>0 or x<0,
Awsm new smiley's
2008-05-08 11:15:17
AWSMFOX
Banned
+405|6934|A W S M F O X

Funky_Finny wrote:

5(2x-3)
______

4x(squared)-9
x=34 {-87} 4* > 8 =

x.4 -9 + x = ∞+1
2008-05-08 11:15:44
kylef
Gone
+1,352|6964|N. Ireland
Whoops, forgot about that squared! Thanks, mkx. I'll edit my post shortly.
2008-05-08 11:16:49
S.Lythberg
Mastermind
+429|6918|Chicago, IL

Funky_Finny wrote:

5(2x-3)
______

4x(squared)-9

So that's 5 brackets 2 ex minus 3 over 4 ex squared minus nine.

I don't need the answer only, how you did it too please.

karma for good answers nd thanksbye.
4x2-9 = (2x+3)(2x-3)

so

5(2x-3)
---------
(2x+3)(2x-3)

2x-3 cancels

5
--
2x+3

is the simplest form
as for the answer, the equation is undefined at x=3/2 and x=-3/2

the equation is never equal to zero people, stop feeding him lies.

Last edited by S.Lythberg (2008-05-08 11:21:57)

2008-05-08 11:20:24
kylef
Gone
+1,352|6964|N. Ireland
Updated my post...
2008-05-08 11:48:51
Sgt.Davi
Touches Himself At Night.
+300|7114|England
The equasion does not equal zero, its an expression which he has to cancel down.

Factorize the top, difference of squares on the bottom, cancel out the two similar numbers in brackets and there.
2008-05-08 13:01:49
[-DER-]Omega
membeR
+188|7298|Lithuania

DesertFox- wrote:

Funky_Finny wrote:

5(2x-3)
______

4x(squared)-9

So that's 5 brackets 2 ex minus 3 over 4 ex squared minus nine.

I don't need the answer only, how you did it too please.

karma for good answers nd thanksbye.
Just to make sure you have:

5(2x-3)
4x2-9               correct?

Well if you factor that out, you get:

   5(2x-3)   
(2x-3)(2x+3)

Cancel, the (2x-3) and get:
    5   
(2x+3)

I'm not sure what you wanted us to do, but I hope that helped.
Uhm, desertfox has it right people.
https://bf3s.com/sigs/fe717ed1eb823c939460a42f15bced7dd0057c51.png

 

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