## The solutions to GR3768

Forum for the GRE subject test in mathematics.
Chessplayer
Posts: 3
Joined: Tue Jun 25, 2024 6:42 pm

### The solutions to GR3768

The newly released practice math Subject Test is very difficult.Is there any solution manual to this available??

copilot
Posts: 4
Joined: Mon Jun 17, 2024 1:50 pm

### Re: The solutions to GR3768

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}).

[ \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} ]

boredmathguy
Posts: 8
Joined: Thu Mar 23, 2023 1:48 pm

### Re: The solutions to GR3768

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?

Chessplayer
Posts: 3
Joined: Tue Jun 25, 2024 6:42 pm

### Re: The solutions to GR3768

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}).

[ \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.

juicetrainwreck
Posts: 11
Joined: Fri Aug 09, 2024 1:01 am

### Re: The solutions to GR3768

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

juicetrainwreck
Posts: 11
Joined: Fri Aug 09, 2024 1:01 am

### Re: The solutions to GR3768

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????

brianstonelake
Posts: 2
Joined: Sat Jun 04, 2016 8:30 pm

### Re: The solutions to GR3768

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.

juicetrainwreck
Posts: 11
Joined: Fri Aug 09, 2024 1:01 am

### Re: The solutions to GR3768

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

estibalizmc
Posts: 1
Joined: Sun Sep 08, 2024 11:31 am

### Re: The solutions to GR3768

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...

smuth
Posts: 6
Joined: Sat Jun 19, 2021 2:41 pm

### Re: The solutions to GR3768

Hi Wasnt the 3768 an already released old test?

juicetrainwreck
Posts: 11
Joined: Fri Aug 09, 2024 1:01 am

### Re: The solutions to GR3768

No the 3768 is the most recent practice examination

juicetrainwreck
Posts: 11
Joined: Fri Aug 09, 2024 1:01 am

### Re: The solutions to GR3768

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