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### The solutions to GR3768

Posted: Tue Jun 25, 2024 6:49 pm
The newly released practice math Subject Test is very difficult.Is there any solution manual to this available??

### Re: The solutions to GR3768

Posted: Fri Jun 28, 2024 11:01 am
Try here. https://www.mathsub.com/solutions/

I also found this answer on the internet, not sure if its relevant:

We have real-valued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:

[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]

where (f) and (g) are real-valued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).

The correct answer is:

[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u} - \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]

### Re: The solutions to GR3768

Posted: Sun Jun 30, 2024 4:29 pm
There is no solution manual as far as I know. However, if there are specific questions you're having trouble with, maybe you can post them here and people can explain their solutions?

### Re: The solutions to GR3768

Posted: Thu Aug 08, 2024 1:21 pm
copilot wrote:
Fri Jun 28, 2024 11:01 am
Try here. https://www.mathsub.com/solutions/

I also found this answer on the internet, not sure if its relevant:

We have real-valued differentiable functions (u(x, y)) and (v(x, y)) implicitly defined by the equations:

[ x = f(u, v) \quad \text{and} \quad y = g(u, v) ]

where (f) and (g) are real-valued differentiable functions. We want to find an expression for (\frac{\partial u}{\partial x}).

The correct answer is:

[ \frac{\partial u}{\partial x} = \frac{\partial g}{\partial v} \frac{\partial f}{\partial u} - \frac{\partial f}{\partial v} \frac{\partial g}{\partial u} ]
Actually he has all previous GRE math solutions, except for the newly released ones .Also he is not responding to emails..This newly released sample test seems like it is a more realistic review of undergrad math , although I am not sure the actual test I will be taking will resemble anything like this.

### Re: The solutions to GR3768

Posted: Mon Sep 02, 2024 8:56 pm
I'm sure the test will be similar to 3768.

If you study all the previous exams there are some repeat/similar questions. For example there's always a question about dim(u+v) = dim(u) + dim(v) - dim(u intersect v)

there's usually a problem about isomorphic graphs

Etc

### Re: The solutions to GR3768

Posted: Fri Sep 06, 2024 12:33 am
For number 21, is this valid reasoning?

Add the two congruences together to get $4x + 9y \equiv 6 \mod 13$. Add this new congruence to the second one to get $5x + 16y \equiv 7 \mod 13$. Since $16y \equiv 3y$ we have that $5x + 3y \equiv 7 \mod 13$.

p.s. how to get LaTeX to render here????

### Re: The solutions to GR3768

Posted: Tue Sep 10, 2024 5:58 pm
I'm making video solutions. It'll likely take a couple months, but I'll be posting them relatively regularly on youtube.

I made solutions for the previous test here:

Hopefully I've learned a little and these will be better. Good luck with the studies!

Here's the playlist for 3768. I'll add videos as I go. Depending on how busy I am with work, it may take a couple months to do the entire exam.

### Re: The solutions to GR3768

Posted: Wed Sep 11, 2024 12:47 am
Thanks Ill check it out!

Ive been typing up a solutions manual too

there is another guy on youtube with many of the solutions to gr3768

### Re: The solutions to GR3768

Posted: Wed Sep 11, 2024 6:24 pm
juicetrainwreck wrote:
Fri Sep 06, 2024 12:33 am
For number 21, is this valid reasoning?

Add the two congruences together to get $4x + 9y \equiv 6 \mod 13$. Add this new congruence to the second one to get $5x + 16y \equiv 7 \mod 13$. Since $16y \equiv 3y$ we have that $5x + 3y \equiv 7 \mod 13$.

p.s. how to get LaTeX to render here????
I'm not sure if that's a valid reason or not. But in case it helps what I did was multiplying by 2 [x+7y \equiv 1 (mod 13)] getting 2x+14y \equiv 2 (mod 13). Since we're in mod 13 the 14y is the same as y, so you get 2x+y \equiv 2 (mod 13). And then sum this to the other equation and you'll get 5x +3y \equiv 7 (mod 13)

About the solutions I wish someone share them too I have my exam at the end of September and I'm more and more anxious So far, the only ones that I've found are the ones from the JoshMathWizz YT chanel but only a few problems are solved...

### Re: The solutions to GR3768

Posted: Fri Sep 13, 2024 9:47 pm
Hi Wasnt the 3768 an already released old test?

### Re: The solutions to GR3768

Posted: Fri Sep 13, 2024 10:31 pm
No the 3768 is the most recent practice examination

### Re: The solutions to GR3768

Posted: Mon Sep 16, 2024 1:13 am
I made anki decks for many of the exams, with the solutions provided by IACOLEY and RAMBO. msg if youd like a copy

they are basic cards (front -> back) with timers