GRE Help

nikkimckelvy

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I'm an adult who is looking to take the GRE after years of being out of school. I need help on the math section. Does anyone know of any good websites? I've contacted the one that came with my GRE study book and a couple of sites similar to this one. So far no luck.

I need a website that explains the logic behind the different formulas. Without that, I'm lost. I need to know when to use the formulas and why to use them. The knowledge of why they were invented and the logic motivating the people who discovered it would also be helpful. If I can understand that, I can usually understand when to apply the problem.

For example, if x^2-y^2= 28 and x-y=8 what is the average of x and y? If I apply FOIL then 28= (x-y) (x+y). That means that 28=8(x+y). From here what do I do? Do I do 28=8x+8y or do I do 28/8=x+y? That would mean that 7/2=x+y, and none of those answers the question of what is the average of x and y.

I got that question from Barrons GRE. Understanding the concepts behind this stuff would really help. So would understanding when this is applicable in life. Any help would be appreciated.

Sincerely,
Nikki
 

galactus

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I hate to admit it, but I don't know what they're talking about either. You found that x-y=8, then x+y=7/2

I suppose the average would be \(\displaystyle \frac{x+y}{2}=\frac{7}{4}\)

If you google it, you can probably find some sample tests. I have an old one on file. I will mail it to you as an attachment.

If I can find it. Let me check.
 

stapel

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nikkimckelvy said:
I need to know when to use the formulas and why to use them.
It sounds like you might need a semester or two of algebra. You might want to enroll at your local community college for this. :idea:

nikkimckelvy said:
The knowledge of why they were invented and the logic motivating the people who discovered it would also be helpful.
Unless the GRE has added a section on the history of mathematics or on its logical foundations, this information would almost certainly never apply to GRE test questions (even where such information is available; much is lost or unknown). The ability to "see" what tools to apply, and when, is something that comes from experience (and, sometimes, trial and error). Reading biographies of mathematicians would not, in general, be helpful. :shock:

nikkimckelvy said:
For example, if x^2-y^2= 28 and x-y=8 what is the average of x and y?
The "average of x and y" is their sum, divided by 2. This might explain why "7/2 = x + y" is not one of the answer options; "x + y" is just the sum, not the average. :oops:

nikkimckelvy said:
If I apply FOIL then 28= (x-y) (x+y).
Actually, "FOIL" is a special algorithm for exactly one case of polynomial multiplication. I think you mean that you "factored", which is the opposite process to multiplying together. :?:

nikkimckelvy said:
That means that 28=8(x+y).
Now divide through by 16 (so the right-hand side ends up as "(x + y)/2"), and simplify the left-hand side. :wink:

Eliz.
 

Denis

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nikkimckelvy said:
For example, if x^2-y^2= 28 and x-y=8 what is the average of x and y? If I apply FOIL then 28= (x-y) (x+y). That means that 28=8(x+y). From here what do I do? Do I do 28=8x+8y or do I do 28/8=x+y? That would mean that 7/2=x+y, and none of those answers the question of what is the average of x and y.
I must be missing something; why can't it be this simple:

x^2 - y^2 = 28 [1]
x - y = 8, so y = x - 8 [2]
Substitute [2] in [1]:
x^2 - (x - 8)^2 = 28
solve to get x = 23/4 ; then y = -9/4

Average = [23/4 + (-9/4)] = 7/2
 

stapel

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Denis said:
...solve to get x = 23/4 ; then y = -9/4

Average = [23/4 + (-9/4)] = 7/2
Wouldn't the "average" be the sum divided by 2?

Eliz.
 

Loren

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x[sup:i46o58nu]2[/sup:i46o58nu]-y[sup:i46o58nu]2[/sup:i46o58nu]=28
x-y=8

If equals are divided by equals, the quotients are equal.

\(\displaystyle \frac{(x+y)(x-y)}{x-y}=\frac{28}{8}\)

x+y = 7/2

But the average of two quantities is their sum divided by 2.

\(\displaystyle \frac{x+y}{2}=\frac{\frac{7}{2}}{2}= \frac{7}{4}\)
 

Denis

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Re:

stapel said:
Denis said:
...solve to get x = 23/4 ; then y = -9/4
Average = [23/4 + (-9/4)] = 7/2
Wouldn't the "average" be the sum divided by 2?
Eliz.
Duh....
 
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