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For such a simple circuit we don't have to use the mesh analysis to solve such simple circuit it. Save the mash/nodal analysis for more complex circuits.

For the first case (without the current source) we can find

Ix' = 3.5V /(15Ω + 10Ω) = - 0.14A but we need to give it a minus sign because Ix' current flows in the opposite direction to the Ix current

The second case (without the voltage source)

Ix" = 2A * 10Ω/(10Ω + 15Ω) = 0.8A (current divider rule)

So finally we have:

Ix = 0.8A - 0.14A = 0.66A = 660mA

From what I see you made a mistake in your nodal equations. Please notice that node 2 is "short" via "voltage source short" to the reference point.

Exactly in the same way as shown here
https://forum.allaboutcircuits.com/threads/shorted-out-components-in-a-circuit.146251/#post-1244323

Why didn't you see this "short circuit"?

So, you only have one node, not two nodes.

V1/10Ω + V1/15Ω - 2A = 0 ---> V1 = 12V -----> Ix" = 12V/15Ω = 0.8A

 

I strongly recommend that you get in the habit of carefully checking your setup equations (in this case your node and mesh equations) carefully after you write them and before you start to solve them. All of the electrical engineering associated with the problem is captured by these equations; everything that follows is simply math. If there is an error in these equations, you probably won't be able to catch it later because you are simply solving a different problem as far as the math is concerned. So go through term by term and ensure that you agree with it.

Look at each term of your first node equation. Do you agree that each term represents the current flowing into the the node?

 

For such a simple circuit we don't have to use the mesh analysis to solve such simple circuit it. Save the mash/nodal analysis for more complex circuits.

For the first case (without the current source) we can find

Ix' = 3.5V /(15Ω + 10Ω) = - 0.14A but we need to give it a minus sign because Ix' current flows in the opposite direction to the Ix current

Anyone reviewing your work later, very possibly including you, is likely to spot the "mistake" when they see a positive numerator and a positive denominator yielding a negative quotient in the above and "fix" it, thus messing things up. Let the math do its work.

You shouldn't have to go throwing magical mystery minus signs around if you've set up the equations correctly.

Ohm's Law says that Ix' is

Ix' = (0 V - 3.5 V) / (15 Ω + 10 Ω) = -140 mA

 

MOD: Cleaned up your images, now readable.
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Hi,

You may want to note that when solving for voltages there is only ONE unknown node voltage.