Ideal Op-Amp Question
The "ideal" op-amp will manipulate the inverting-input to have the same voltage level as the non-inverting pin i.e. \$V_n\$ is made to equal \$V_p\$ and this is due to negative feedback via the 8k resistor: -



Clearly 6 volts (\$V_p\$) is present on the non-inverting input because of the 2k and 1k potential divider. Then, due to negative feedback, the op-amp (acting as an error amplifier) will make the inverting input (\$V_n\$) also become 6 volts. It does this by passing current from/into its output pin and through the 8k feedback resistor. This puts the circuit in equilibrium and Vo is a result of that equilibrium.

So, if V1 is 4 volts and, 6 volts must be present at the inverting-input, then there must be a current from V1 of -2 volts / 2k = -1 mA. Given that the inverting and non-inverting input pins are ideal (infinite impedance) then all that current comes-from/goes-to the op-amp output. 

So, it's a simple case of asking what voltage (Vout) is needed to force 1 mA through an 8 k resistor when that resistor has 6 volts at the other terminal. Simple math tells us that Vout = +14 volts.