With port currents defined entering the two-port, V2=−20I2. The z equation gives $80I_1+100I_2=-20I_2$, so I2=−(2/3)I1.
At the input, the 20 V source and 10 Ω resistor give $20=10I_1+V_1=50I_1+60I_2$. Substitution gives I1=2 A and I2=−4/3 A.
The load power is $I_2^2R_L=(16/9)20=$ **35.5556 W**, within 34–36 W.
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