Great. Thanks.

solve({(1+R_4/R_3)((-5)((R_2^-1+R_5^-1)^-1/(R_1+(R_2^-1+R_5^-1)^-1))+2.5((R_1^-1+R_5^-1)^-1/(R_2+(R_1^-1+R_5^-1)^-1)))=0,(1+R_4/R_3)((5)((R_2^-1+R_5^-1)^-1/(R_1+(R_2^-1+R_5^-1)^-1))+2.5((R_1^-1+R_5^-1)^-1/(R_2+(R_1^-1+R_5^-1)^-1)))=2.5,R_1=10,R_3=10,R_5=4.99})

1. I chose R1, R3 and R5 so that the system of equations is easy to solve. As you noticed, the remaining unknowns are R4 and R2, each one in its own product term. So, If we divide one equation to another, R4 disappears and we get one equation with one unknown, R2.
2. No. If you look into the circuit from V1 point of view, you will notice that V1 “sees” R1 in series with R2 in parallel with R5. So the input impedance from V1 point of view is R1 + (R2 || R5).

Firstly, in your example you chose to fix R1, R3 and R5 to leave two unknowns for your two equations. How do you decide which Rx values you can fix, and which two you will solve for? I tried choosing different values and got non-sense results when I tried to solve…again, rough algebra skills at this point.

Secondly, if you choose R1 = 10k, is the V1 input impedance 10k? If the positive input on the op amp was being driven to zero I would be comfortable in saying this, but I am not sure what is happening in this case…again, rough circuit skills.

Use the bipolar to unipolar calculator
http://masteringelectronicsdesign.com/summing-amplifier-calculator-java/
to view the Vin value.