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Danteshek

Posts: 2170
Joined: Wed Dec 10, 2008 4:40 pm

Solve for X?

(6,666,667 + X) / (30,666,667 + X) = .25

Renzo

Posts: 4254
Joined: Tue Dec 02, 2008 3:23 am

Man. You do suck at math.

camelcrema

Posts: 281
Joined: Fri Oct 01, 2010 2:43 pm

1,333,333

UnitarySpace

Posts: 197
Joined: Tue Mar 31, 2009 2:18 am

1,333,333 bro.

buckilaw

Posts: 839
Joined: Fri May 07, 2010 1:27 am

--ImageRemoved--

Times 424,628

mbusch22

Posts: 255
Joined: Fri Jan 29, 2010 1:08 pm

camelcrema wrote:
1,333,333

+1. had to try it to see if i had any traces of math skills left.

ptblazer

Posts: 376
Joined: Sun Oct 31, 2010 11:27 pm

Danteshek wrote:Solve for X?

(6,666,667 + X) / (30,666,667 + X) = .25

Just in case.

multiply BOTH SIDES of the equation by (30,666,667 + X).

So. (6,666,667 + X) = .25*(30,666,667 + X)

factor the .25 and get X on one side of the equation

999,999.75 = .75X

divide by .75

Renzo

Posts: 4254
Joined: Tue Dec 02, 2008 3:23 am

(6,666,667 + X) / (30,666,667 + X) = .25

(6,666,667 + X) /1 = .25(30,666,667 + X)

6,666,667 + X = .25(30,666,667) + .25X

6,666,667 +0.75X = .25(30,666,667)

0.75X = .25(30,666,667) - 6,666,667 = 999 999.75

X= 999 999.75/ 0.75

x= 1 333 333

Danteshek

Posts: 2170
Joined: Wed Dec 10, 2008 4:40 pm

Yeah, I figured it out. Hahaha

No wonder I went to law school. Got 97V and ~40M on GMAT (660).

glewz

Posts: 781
Joined: Tue Jun 08, 2010 4:32 pm

Danteshek wrote:Solve for X?

(6,666,667 + X) / (30,666,667 + X) = .25

Other answers are all correct, but only because X does not equal to -30,666,667. (denominator of the fraction can't be zero) Always check for those conditions.

Danteshek

Posts: 2170
Joined: Wed Dec 10, 2008 4:40 pm

Btw, this was part of an anti dilution problem in my Business Planning class

Renzo

Posts: 4254
Joined: Tue Dec 02, 2008 3:23 am

glewz wrote:
Danteshek wrote:Solve for X?

(6,666,667 + X) / (30,666,667 + X) = .25

Other answers are all correct, but only because X does not equal to -30,666,667. (denominator of the fraction can't be zero) Always check for those conditions.

Nerd.

ptblazer

Posts: 376
Joined: Sun Oct 31, 2010 11:27 pm

glewz wrote:
Danteshek wrote:Solve for X?

(6,666,667 + X) / (30,666,667 + X) = .25

Other answers are all correct, but only because X does not equal to -30,666,667. (denominator of the fraction can't be zero) Always check for those conditions.

I'm not sure what you are saying here, maybe I'm just not understanding what you mean. The answer is correct because thats what X needs to be to make the left side equal the right side. I agree with the rule that the denominator cannot equal zero, but what if the question was written like this.
(6,666,667 + X) = .25(30,666,667 + X)

The denominator wouldn't be zero, because there isn't one. If you used -30,666,667 for X you'd get 30,666,667 = 6,666,667, which is obviously not right. X has only one possible answer.

random5483

Posts: 684
Joined: Sat Mar 20, 2010 8:17 pm

ptblazer wrote:
glewz wrote:
Danteshek wrote:Solve for X?

(6,666,667 + X) / (30,666,667 + X) = .25

Other answers are all correct, but only because X does not equal to -30,666,667. (denominator of the fraction can't be zero) Always check for those conditions.

I'm not sure what you are saying here, maybe I'm just not understanding what you mean. The answer is correct because thats what X needs to be to make the left side equal the right side. I agree with the rule that the denominator cannot equal zero, but what if the question was written like this.
(6,666,667 + X) = .25(30,666,667 + X)

The denominator wouldn't be zero, because there isn't one. If you used -30,666,667 for X you'd get 30,666,667 = 6,666,667, which is obviously not right. X has only one possible answer.

If you use -30,666,667 for X you would get -24,000,000 / 0 = .25. A zero in the denominator would give you an undefined answer. I think glewz was talking about checking to see that X does not make the denominator zero when posting problems. However, his comment is not relevant since the denominator is not zero and the "condition" he mentioned does not apply.

stratocophic

Posts: 2204
Joined: Tue Dec 22, 2009 6:24 pm

random5483 wrote:
ptblazer wrote:
glewz wrote:
Danteshek wrote:Solve for X?

(6,666,667 + X) / (30,666,667 + X) = .25

Other answers are all correct, but only because X does not equal to -30,666,667. (denominator of the fraction can't be zero) Always check for those conditions.

I'm not sure what you are saying here, maybe I'm just not understanding what you mean. The answer is correct because thats what X needs to be to make the left side equal the right side. I agree with the rule that the denominator cannot equal zero, but what if the question was written like this.
(6,666,667 + X) = .25(30,666,667 + X)

The denominator wouldn't be zero, because there isn't one. If you used -30,666,667 for X you'd get 30,666,667 = 6,666,667, which is obviously not right. X has only one possible answer.

If you use -30,666,667 for X you would get -24,000,000 / 0 = .25. A zero in the denominator would give you an undefined answer. I think glewz was talking about checking to see that X does not make the denominator zero when posting problems. However, his comment is not relevant since the denominator is not zero and the "condition" he mentioned does not apply.
Yeah, pretty sure you can manipulate any first order algebraic equation into having an invalid point if you set it up like this, but only because it isn't simplified. You have to simplify all the way before you can start throwing around "can't divide by zeros."

glewz

Posts: 781
Joined: Tue Jun 08, 2010 4:32 pm

Not true in many cases. Let me settle this and be clear with what I was saying:

@ptblazer, to write it in that different format for other equations (And this one) would be theoretically inequivalent. In the above, the distinction is not meaningful, but regardless, OP always needs to check for conditions in which x cannot exist.

@stratocophic: No, you cannot just simplify by multiplying things through, while not accounting for "can't divide by zeros": Here's a simple example that clarifies:

Problem: (2x^2 - x - 1) / (x - 1) =3
If we simplified all the way, then we'd get:
--> (2x +1)(x-1) / (x-1) = 3
--> 2x + 1 = 3
--> 2x = 2
--> x = 1
BUT x can't equal one because in the problem above, (x - 1) can't exist. So the answer to the problem is: "No Solution"

Posts: 98
Joined: Thu Nov 18, 2010 1:57 pm

Ugh... I hate numbers.

Renzo

Posts: 4254
Joined: Tue Dec 02, 2008 3:23 am

stratocophic

Posts: 2204
Joined: Tue Dec 22, 2009 6:24 pm

glewz wrote:Not true in many cases. Let me settle this and be clear with what I was saying:

@ptblazer, to write it in that different format for other equations (And this one) would be theoretically inequivalent. In the above, the distinction is not meaningful, but regardless, OP always needs to check for conditions in which x cannot exist.

@stratocophic: No, you cannot just simplify by multiplying things through, while not accounting for "can't divide by zeros": Here's a simple example that clarifies:

Problem: (2x^2 - x - 1) / (x - 1) =3
If we simplified all the way, then we'd get:
--> (2x +1)(x-1) / (x-1) = 3
--> 2x + 1 = 3
--> 2x = 2
--> x = 1
BUT x can't equal one because in the problem above, (x - 1) can't exist. So the answer to the problem is: "No Solution"
Hence why I said "first order." That's a quadratic equation, which is a second order equation. I don't discount the likelihood that I'm wrong. I'm not the one to be giving mathematical principles, given my sterling 2.0 GPA in college mathematics.

ptblazer

Posts: 376
Joined: Sun Oct 31, 2010 11:27 pm

glewz wrote:Not true in many cases. Let me settle this and be clear with what I was saying:

Problem: (2x^2 - x - 1) / (x - 1) =3
If we simplified all the way, then we'd get:
--> (2x +1)(x-1) / (x-1) = 3
--> 2x + 1 = 3
--> 2x = 2
--> x = 1
BUT x can't equal one because in the problem above, (x - 1) can't exist. So the answer to the problem is: "No Solution"

I see what you are saying now. I really hate stirring things up, but you're wrong here. The answer isn't no solution, because x = 1 isn't the only solution. This is a quadratic equation (x^2), meaning there are two answers. In this case what you said in your original post is appropriate and would apply. The answer (X=?, don't care to find it) is correct, only because x=1 is not a valid answer. Doh! My bad
Last edited by ptblazer on Fri Mar 25, 2011 1:10 pm, edited 1 time in total.

stratocophic

Posts: 2204
Joined: Tue Dec 22, 2009 6:24 pm

ptblazer wrote:
glewz wrote:Not true in many cases. Let me settle this and be clear with what I was saying:

Problem: (2x^2 - x - 1) / (x - 1) =3
If we simplified all the way, then we'd get:
--> (2x +1)(x-1) / (x-1) = 3
--> 2x + 1 = 3
--> 2x = 2
--> x = 1
BUT x can't equal one because in the problem above, (x - 1) can't exist. So the answer to the problem is: "No Solution"

I see what you are saying now. I really hate stirring things up, but you're wrong here. The answer isn't no solution, because x = 1 isn't the only solution. This is a quadratic equation (x^2), meaning there are two answers. In this case what you said in your original post is appropriate and would apply. The answer (X=?, don't care to find it) is correct, only because x=1 is not a valid answer.
Not sure, but I think you can use DiffEQ to solve for the second solution, because the quadratic formula's obviously not going to work

rinkrat19

Posts: 13923
Joined: Sat Sep 25, 2010 5:35 am

Wow, flashbacks to 8th grade math.

ptblazer

Posts: 376
Joined: Sun Oct 31, 2010 11:27 pm

stratocophic wrote:
ptblazer wrote:I see what you are saying now. I really hate stirring things up, but you're wrong here. The answer isn't no solution, because x = 1 isn't the only solution. This is a quadratic equation (x^2), meaning there are two answers. In this case what you said in your original post is appropriate and would apply. The answer (X=?, don't care to find it) is correct, only because x=1 is not a valid answer.

Not sure, but I think you can use DiffEQ to solve for the second solution, because the quadratic formula's obviously not going to work

I meant to address this before. You're right that you can make most first order equation into a "divide by zero" issue, which is why you need to simplfy first. Also, I'm pretty sure the quadratic equation would work, but you need to make the equation look like XXXXX = 0 first.

I actually like math quite a bit (although not claiming to be good at it) so if it seems like I'm "too" engaged in this topic, its because I am. Opportunities to discuss math rarely come up in the pursuit of a law degree, so this brief distraction is nice.

stratocophic

Posts: 2204
Joined: Tue Dec 22, 2009 6:24 pm

ptblazer wrote:
stratocophic wrote:
ptblazer wrote:I see what you are saying now. I really hate stirring things up, but you're wrong here. The answer isn't no solution, because x = 1 isn't the only solution. This is a quadratic equation (x^2), meaning there are two answers. In this case what you said in your original post is appropriate and would apply. The answer (X=?, don't care to find it) is correct, only because x=1 is not a valid answer.

Not sure, but I think you can use DiffEQ to solve for the second solution, because the quadratic formula's obviously not going to work

I meant to address this before. You're right that you can make most first order equation into a "divide by zero" issue, which is why you need to simplfy first. Also, I'm pretty sure the quadratic equation would work, but you need to make the equation look like XXXXX = 0 first.

I actually like math quite a bit (although not claiming to be good at it) so if it seems like I'm "too" engaged in this topic, its because I am. Opportunities to discuss math rarely come up in the pursuit of a law degree, so this brief distraction is nice.
Right, but in doing so don't you make it first order anyway in this case since the x-1 cancels? It took me one semester to forget almost all of the calculus I knew, now that it's been over 2 years I've got nothing left but bits and pieces of everything

ptblazer

Posts: 376
Joined: Sun Oct 31, 2010 11:27 pm

stratocophic wrote:
Right, but in doing so don't you make it first order anyway in this case since the x-1 cancels? It took me one semester to forget almost all of the calculus I knew, now that it's been over 2 years I've got nothing left but bits and pieces of everything

(2x^2 - x - 1) / (x - 1) =3
(2x^2 - x - 1) = 3x -3
2x^2 - 4x + 2 = 0
(2x - 2)(x - 1) = 0

well shit, haha... I owe glewz an apology. Turns out both answers are x=1 and the answer to this problem is no solution. Should have solved it in the first place.