Getting the wrong Thevenin equivalent resistor
There are lots of ways to proceed on the problem. Your circuit example can be addressed by what's usually learned early -- before mesh or nodal analysis. So I'll start with the basic Thevenin to Norton and Norton to Thevenin conversions. Let's start.
Step 1
Perform a Thevenin to Norton conversion on the left-side two components:



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I think you can see how this was achieved, if you are at all familiar with this transformation.
Step 2
Take the two resistors in parallel and combine them:



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Again, this is pretty basic and I'm sure you've already become familiar with this.
Step 3
Perform a Norton to Thevenin conversion on the left-side two components:



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Again, I think you are familiar with this conversion.
Step 4
Take the two resistors in series and combine them:



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Step 5
Perform a Thevenin to Norton conversion on the left-side two components:



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Step 6
Combine the current sources:



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Step 7
At this point, just convert back to Thevenin. Here you will get \$V_\text{TH}=125\:\text{V}\$ and \$R_\text{TH}=25\:\Omega\$.



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Another approach is to perform nodal analysis and later inject a \$1\:\text{A}\$ source. So, for example:



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The nodal equations are:
$$\begin{align*}
\frac{V_\text{X}}{2\:\Omega}+\frac{V_\text{X}}{2\:\Omega}+\frac{V_\text{X}}{24\:\Omega}&amp;=\frac{100\:\text{V}}{2\:\Omega}+\frac{0\:\text{V}}{2\:\Omega}+\frac{V_\text{TH}}{24\:\Omega}\\\\
\frac{V_\text{TH}}{24\:\Omega}&amp;=\frac{V_\text{X}}{24\:\Omega}+3\:\text{A}
\end{align*}$$
Which solve out as: \$V_\text{TH}=125\:\text{V}\$ and \$V_\text{X}=53\:\text{V}\$.
Now, inject \$1\:\text{A}\$ into the (+) node. The new nodal equations are:
$$\begin{align*}
\frac{V_\text{X}}{2\:\Omega}+\frac{V_\text{X}}{2\:\Omega}+\frac{V_\text{X}}{24\:\Omega}&amp;=\frac{100\:\text{V}}{2\:\Omega}+\frac{0\:\text{V}}{2\:\Omega}+\frac{V_\text{TH}}{24\:\Omega}\\\\
\frac{V_\text{TH}}{24\:\Omega}&amp;=\frac{V_\text{X}}{24\:\Omega}+3\:\text{A}+1\:\text{A}
\end{align*}$$
And the new results are: \$V_\text{TH}=150\:\text{V}\$ and \$V_\text{X}=54\:\text{V}\$.
This means that \$R_\text{TH}=\frac{150\:\text{V}-125\:\text{V}}{1\:\text{A}-0\:\text{A}}=25\:\Omega\$.
So, \$V_\text{TH}=125\:\text{V}\$ from the open-circuit case and then \$R_\text{TH}=25\:\Omega\$, as derived from changes due to the injection of a \$1\:\text{A}\$ current. So the results using nodal analysis confirm the multi-step process used earlier here.