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Assuming ideal op-amplifiers, I did the following steps:

\$v_{out1}=v_1\frac{R_2}{R_1}=2v_1\$

\$v_{out2}=v_o=\left(\frac{v_{out1}}{R_3}+\frac{v_2}{R_4}\right)R_5=4v_1+2v_2\$

My textbook claims \$4v_1-2v_2\$

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  • \$\begingroup\$ There are 2 things you need to know here. The correct transfer function for an inverting opamp, and how to apply superposition. \$\endgroup\$ – efox29 Oct 9 '15 at 0:03
  • \$\begingroup\$ I know the transfer function, and I am aware of the superposition theorem, but I'm not sure if I've done that step correctly, considering my answer comes out right except for a minus sign on the second term. \$\endgroup\$ – B. Lee Oct 9 '15 at 0:05
  • \$\begingroup\$ Vout1 is not 2Vi. Transfer function is wrong. \$\endgroup\$ – efox29 Oct 9 '15 at 0:07
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These stages are inverting. Your equations are both missing a - in the transfer function.

V1 gets inverted twice, that's why it has a positive coefficient. V2 gets inverted once, that's why it has a negative coefficient.

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