Question on Thevnin theorm
The Thévenin voltage is obtained by calculating the voltage across terminals x and x'. There are many ways to get there. The easiest and fastest way is to apply superposition. You have two sources that you will alternately turn off during an intermediate calculation. First, turn the 88-V off. Turning a voltage source off or reducing its voltage to 0 V is identical to replacing it with a short circuit: replace the 88-V source by a wire and calculate the voltage between x and x' while the 1-A source biases the network. To get there, calculate the voltage at y and y' then involve a resistive divider to determine the voltage between x and x'. No line of algebra, just inspect the circuit. Once this is done, turn the 1-A current source off which is similar to reducing it to 0 A or open-circuiting it: remove it from the circuit and the 88-V source remains alone. Calculate the voltage across x and x', again using a resistive divider. Now, sum up the two voltages obtained during these intermediate steps, you have the voltage between x and x'. This is \$V_{th}\$. Without disclosing the calculation steps - this is your job : ) a quick sim will let you know if you have it right, \$V_{th}=69.3\;V\$:



For the output resistance, you must turn all sources off to calculate it. This is one way of determining \$R_{th}\$ but there are other ways. This one is simple. You see that there is already a resistance across x and x'. This is the \$50-\Omega\$ resistor. Therefore, remove this resistance in your head and determine the resistance you "see" between the connecting terminals. This is \$R?\$ in the below sketch. Once \$R?\$ is determined (it is easy, it is a series-parallel association), the Thévenin resistance is simply \$R_{th}=50 || R?\$. If you bias x and x' with a 1-A test source \$I_T\$ as shown in the sketch, the voltage developed across the current source is the resistance you want (\$7.32\;\Omega\$).



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