### 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??

for current and prospective graduate students in mathematics

https://mathematicsgre.com/

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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??

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} ]

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} ]

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?

Posted: **Thu Aug 08, 2024 1:21 pm**

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.copilot wrote: ↑Fri Jun 28, 2024 11:01 amTry 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} ]

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

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

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????

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????

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:

https://youtube.com/playlist?list=PL81I ... X9KLdVwXUx

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.

https://www.youtube.com/playlist?list=P ... 9BWnfPxbFu

I made solutions for the previous test here:

https://youtube.com/playlist?list=PL81I ... X9KLdVwXUx

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.

https://www.youtube.com/playlist?list=P ... 9BWnfPxbFu

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

Ive been typing up a solutions manual too

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

Posted: **Wed Sep 11, 2024 6:24 pm**

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)juicetrainwreck wrote: ↑Fri Sep 06, 2024 12:33 amFor 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????

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...

Posted: **Fri Sep 13, 2024 9:47 pm**

Hi Wasnt the 3768 an already released old test?

Posted: **Fri Sep 13, 2024 10:31 pm**

No the 3768 is the most recent practice examination

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

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