Thevenin Equivalent Problem
I see in a comment that you are looking for nodal analysis. Here's my own approach to it:


  
  
  Equation rules:
  
  
  Keep all outgoing currents on the left side
  Keep all incoming currents on the right side
  If not already chosen by the problem, pick a convenient node and call
  it "ground." This allows you to assign \$0\:\text{V}\$ to that node and
  this will simplify some of the equations. Which node to select is up to
  you. But I will pick one that either helps make the schematic easier to
  understand, or else will simplify the equations more than other choices.
  
  
  I like to imagine a 3D chessboard -- those funny chessboards with tiny
  sub-chessboard areas that sit at different heights. Each node in the
  circuit is represented as one of these sub-areas (or "platforms"), sitting at some unknown
  elevation and held there on a long pole beneath it. This unknown elevation
  is the voltage you must find. One of these can be placed at "0 height." But
  that's just arbitrary, since then there still may be platforms above and/or
  below. So it's just a way of conveniently locating the "ground floor," so
  to speak.
  
  I treat the currents as water. Water flows onto and off of each platform.
  The left side of the equation, the outgoing currents, is represented by the
  water that is flowing off of the perimeter of the platform. The right side
  of the equation, the incoming currents, is represented by the water that is
  flowing onto the platform from other platforms above it. The amount of water
  falling onto such a platform must equal the water flowing off of it via the
  perimeter edge. (Of course, otherwise water would magically build up on the platform -- or the reverse, get created out of thin air. Both are impossible.)
  
  (No water is ever lost and all of it
  remains in play and the water cannot accumulate on the floors -- the net
  flow onto a floor must equal the net flow off of it -- KCL! -- and hence
  the justification for setting the left side equal to the right side.)
  
  


Your schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

\$N_3\$ is free, since it has a voltage source holding its value. So \$V_{N_3}=V_1=100\:\text{V}\$. This only leaves two other nodes to worry over.

I'm going to provide the general solution here, because the approach can be used broadly. Your circuit could actually be "cut short" by the fact that you already know that \$R_1\$ is to be removed and/or shorted out. But this would be a special-case solution. I'd rather teach you the general solution, out of which all other questions can be answered.

The nodal equations are:

$$\begin{align*}
\frac{V_{N_1}}{R_4}+\frac{V_{N_1}}{R_5}&amp;=10\:\text{mA}+\frac{V_{N_2}}{R_4}+\frac{0\:\text{V}}{R_5}\\\\
\frac{V_{N_2}}{R_1}+\frac{V_{N_2}}{R_3}+\frac{V_{N_2}}{R_4}&amp;=\frac{0\:\text{V}}{R_1}+\frac{V_{N_3}=100\:\text{V}}{R_3}+\frac{V_{N_1}}{R_4}
\end{align*}$$

There are two unknowns and two variables to solve for, here.

$$\begin{align*}
V_{N_1}&amp;=R_5\frac{V_1\:R_1+I_1\left(R_1\:R_3+R_1\:R_4+R_3\:R_4\right)}{R_1\:R_3+R_1\:R_4+R_1\:R_5+R_3\:R_4+R_3\:R_5}\tag{$R_1\ne\infty$}\\\\
&amp;=R_5\frac{V_1+I_1\left(R_3+R_4\right)}{R_3+R_4+R_5}\tag{$R_1=\infty$}\\\\
V_{N_2}&amp;=R_1\frac{V_1\left(R_4+R_5\right)+I_1\:R_3\:R_5}{R_1\:R_3+R_1\:R_4+R_1\:R_5+R_3\:R_4+R_3\:R_5}\tag{$R_1\ne\infty$}\\\\
&amp;=\frac{V_1\left(R_4+R_5\right)+I_1\:R_3\:R_5}{R_3+R_4+R_5}\tag{$R_1=\infty$}
\end{align*}$$



Note that if you plug in the actual value for \$R_1=100\:\Omega\$ in the above equations (the ones where \$R_1\ne\infty\$), then you will get the voltages you already worked out:

$$\begin{align*}
V_{N_1}\approx 15.25641\:\text{V}\\\\
V_{N_2}\approx 23.46154\:\text{V}
\end{align*}$$

These solved voltages do not really help you answer the final solution, though.



Once you solve these equations, you can work out the fact that if you short \$R_1\$ (just set \$R_1=0\:\Omega\$ in the above equations) then the current through that short is \$\frac{61}{180}\:\text{A}\$ (the sum of the currents coming in from the left and the right.) And if you open \$R_1\$ (just set \$R_1=\infty\:\Omega\$ in the above equations) then the voltage across the vacant location for \$R_1\$ is \$V_{N_2}=76\frac{1}{4}\:\text{V}\$.

The above equations allow you to set \$R_1=0\$. Here, the voltage at \$V_{N_2}\$ is going to obviously be zero volts (since you just shorted it to ground.) However, the voltage at \$V_{N_1}\$ will now be very important. It will cause a current through \$R_4\$! So compute \$V_{N_1}\$ using the above equation with \$R_1=0\$ and then work out the current in \$R_4\$ from that. Add to that, the current that must also be created now in \$R_3\$. Since you know that \$V_{N_2}=0\$ these two currents should be trivial to work out. Sum them. This is the short circuit current.

These equations also support \$R_1=\infty\$. In this case, you just solve for \$V_{N_2}\$ and you have the open circuit voltage.

With both of these, you have your answer.



P.S. If you are not already using Sage and/or Sympy, you should be. These programs help you rapidly solve equations, symbolically. It took me mere seconds of time to develop the equations I showed above from the two nodal equations.

 var('vn1 vn2 vn3 r1 r2 r3 r4 r5 i1')
 ans=solve([Eq(vn1/r4+vn1/r5,i1+vn2/r4),Eq(vn2/r1+vn2/r3+vn2/r4,vn3/r3+vn1/r4)],[vn1,vn2])


From that, the short circuit current was easy:

 (ans[vn1].subs({r1:0,r2:200,r3:300,r4:400,r5:500,i1:.01,vn3:100})/400+100/300).n()


And so was the open circuit node voltage for \$V_{N_2}\$:

 y1[vn2].subs({r1:1E20,r2:200,r3:300,r4:400,r5:500,i1:.01,vn3:100}).n()




If you were to take this as a nodal analysis problem, but wanted to solve it more specifically than the above general approach, then you have the following two circuits:







simulate this circuit

The nodal equations for the top schematic is:

$$\begin{align*}
\frac{V_{N_1}}{R_4}+\frac{V_{N_1}}{R_5}&amp;=10\:\text{mA}+\frac{V_{N_2}}{R_4}+\frac{0\:\text{V}}{R_5}\\\\
\frac{V_{N_2}}{R_3}+\frac{V_{N_2}}{R_4}&amp;=\frac{V_{N_3}=100\:\text{V}}{R_3}+\frac{V_{N_1}}{R_4}
\end{align*}$$

This solves out as \$V_{N_2}=76\frac{1}{4}\:\text{V}\$.

The nodal equation for the bottom schematic is trivially:$$\begin{align*}
\frac{V_{N_1}}{R_4}+\frac{V_{N_1}}{R_5}&amp;=10\:\text{mA}+\frac{0\:\text{V}}{R_4}+\frac{0\:\text{V}}{R_5}\\\\
\end{align*}$$

This solves out as \$V_{N_1}=2\frac{2}{9}\:\text{V}\$. So it is now easy to figure out the current through \$V_2\$ as the sum, \$I_{V_2}=\frac{2\frac{2}{9}\:\text{V}}{400\:\Omega}+\frac{100\:\text{V}}{300\:\Omega}=\frac{61}{180}\:\text{A}\$. And once again you are at the same place and able to work out the Thevenin resistance by dividing the open circuit voltage by the short circuit current.