Three-part Problem: Thevenin + Dependent current source

[s3a]

Homework Statement

The diagram of the electric circuit, the problem statement (and an attempt) are included in the file below.:
https://www.docdroid.net/RD0ZEgA/question-3.pdf

Homework Equations

Dependent Current Source

Nodal analysis

Short Circuit Current

Thevenin Voltage

Thevenin Resistance

The Attempt at a Solution

See the (same) included file.:
https://www.docdroid.net/RD0ZEgA/question-3.pdf

Could someone please help point out the flaws in the work? If you show me another way to do the problem's parts that would also be nice, but I'm largely interested in knowing what's wrong with the work in the attachment.

Any input would be greatly appreciated!

P.S.
I didn't attach the file because, when I tried to, the website said it's too large.

 

[diredragon]

The problem i immediately saw was with the part b). You get $V_1=0.10$. Shouldn't $V_1$ be $V_1=0$ since it's short-circuited to the ground?
Using that i get: $\frac{0.5V}{0.25}+\frac{V_x}{3}=I_0$, where $V_x = 0.5$. Then $I_0=2.16666$.

 

[s3a]

The problem i immediately saw was with the part b). You get $V_1=0.10$. Shouldn't $V_1$ be $V_1=0$ since it's short-circuited to the ground?
Using that i get: $\frac{0.5V}{0.25}+\frac{V_x}{3}=I_0$, where $V_x = 0.5$. Then $I_0=2.16666$.

Oh, yes!

Shouldn't it be like this, though (instead of what you typed)?:
(V_1 - 0.5)/0.25 + V_x/3 - I_0 = 0

(V_1 - 0.5)/0.25 + (V_1 - 0.5)/3 - I_0 = 0

(0 - 0.5)/0.25 + (0 - 0.5)/3 - I_0 = 0

(-0.5)/0.25 + (-0.5)/3 - I_0 = 0

I_0 = (-0.5)/0.25 + (-0.5)/3

I_0 = -2.16666666666666666667 A (meaning that the current is pointing upwards, instead of downwards)

 

[diredragon]

Oh, yes!

Shouldn't it be like this, though (instead of what you typed)?:
(V_1 - 0.5)/0.25 + V_x/3 - I_0 = 0

(V_1 - 0.5)/0.25 + (V_1 - 0.5)/3 - I_0 = 0

(0 - 0.5)/0.25 + (0 - 0.5)/3 - I_0 = 0

(-0.5)/0.25 + (-0.5)/3 - I_0 = 0

I_0 = (-0.5)/0.25 + (-0.5)/3

I_0 = -2.16666666666666666667 A (meaning that the current is pointing upwards, instead of downwards)

Using these equations the $-I_0$ suggest that (following your sign rule) the $I_0$ enters the node. Since you get a negative value it means that the current flow in the opposite direction than you speculated it would. Meaning it goes downward since you draw it going upward.
See that in my equations i immediately concluded that it would go downward and i get a positive result proving that it does.

 

[s3a]

Ah, yes. I get it / part b now. Thanks. :)

About the rest, do you have an idea as to why grades are deducted? Is anything wrong there?

Edit:
I still get it, but I think you meant upwards instead of downwards and downwards instead of upwards.

 

[diredragon]

Ah, yes. I get it / part b now. Thanks. :)

About the rest, do you have an idea as to why grades are deducted? Is anything wrong there?

Edit:
I still get it, but I think you meant upwards instead of downwards and downwards instead of upwards.

The a) part seems correct, and the c) uses the previous results so it could be because of the incorrect b) part.

 

[s3a]

For what it's worth, in response to the what you seem to have edited away about the "+" for the last part of the first equation in part a, it does seem to have a "+" omitted, but the work continues as if it were there and as if the -0.5 + V_1 (including the "-" in front of the 0.5) were all on the numerator of the fraction. (You probably noticed that, but I just wanted to state it, just in case.)

Having said that, I get it all now, and thanks again! :)