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Well, we have the following circuit (and we assume an ideal model of an OPAMP):

^{simulate this circuit – Schematic created using CircuitLab}

Using KCL, we can write:

$$ \begin{cases} \text{I}_4=\text{I}_-+\text{I}_2\\ \\ \text{I}_x=\text{I}_++\text{I}_3\\ \\ \text{I}_\text{o}=\text{I}_4+\text{I}_5 \end{cases}\tag1 $$

Using KVL, we can write:

$$ \begin{cases} \text{I}_x=\frac{\text{V}_x-\text{V}_+}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_--0}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_+-0}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{o}-\text{V}_-}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_\text{o}-0}{\text{R}_5}\\ \\ \end{cases}\tag2 $$

Notice: in the ideal OPAMP circuit we assume that \$\text{I}_+=\text{I}_-=0\$ and \$\text{V}_+=\text{V}_-\$.

Now, the gain is defined by:

$$\text{G}:=\frac{\text{V}_\text{o}}{\text{V}_x}\tag3$$

We can find an expression for the output voltage \$\text{V}_\text{o}\$, by solving the systems of equations:

$$\text{V}_\text{o}=\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag4$$

So, we get:

$$\text{G}=\frac{1}{\text{V}_x}\cdot\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}=\frac{\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag5$$

In your case we get:

 

$$\text{G}=\frac{68000\cdot\left(30000+63000\right)}{30000\cdot\left(12000+68000\right)}=\frac{527}{200}=2.635\tag6$$

Solving it, in general, gives (notice that \$\text{V}_+=\text{V}_-=\text{V}_\text{p}\$:

In your case (using your values):

I checked my solution using LTspice and I got it right.

Well, we have the following circuit (and we assume an ideal model of an OPAMP):

^{simulate this circuit – Schematic created using CircuitLab}

Using KCL, we can write:

$$ \begin{cases} \text{I}_4=\text{I}_-+\text{I}_2\\ \\ \text{I}_x=\text{I}_++\text{I}_3\\ \\ \text{I}_\text{o}=\text{I}_4+\text{I}_5 \end{cases}\tag1 $$

Using KVL, we can write:

$$ \begin{cases} \text{I}_x=\frac{\text{V}_x-\text{V}_+}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_--0}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_+-0}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{o}-\text{V}_-}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_\text{o}-0}{\text{R}_5}\\ \\ \end{cases}\tag2 $$

Notice: in the ideal OPAMP circuit we assume that \$\text{I}_+=\text{I}_-=0\$ and \$\text{V}_+=\text{V}_-\$.

Now, the gain is defined by:

$$\text{G}:=\frac{\text{V}_\text{o}}{\text{V}_x}\tag3$$

We can find an expression for the output voltage \$\text{V}_\text{o}\$, by solving the systems of equations:

$$\text{V}_\text{o}=\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag4$$

So, we get:

$$\text{G}=\frac{1}{\text{V}_x}\cdot\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}=\frac{\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag5$$

In your case we get:

$$\text{G}=\frac{68000\cdot\left(30000+63000\right)}{30000\cdot\left(12000+68000\right)}=\frac{527}{200}=2.635\tag6$$

Solving it, in general, gives (notice that \$\text{V}_+=\text{V}_-=\text{V}_\text{p}\$:

In your case (using your values):

I checked my solution using LTspice and I got it right.

Well, we have the following circuit (and we assume an ideal model of an OPAMP):

^{simulate this circuit – Schematic created using CircuitLab}

Using KCL, we can write:

$$ \begin{cases} I_4=I_-+I_2\\ \\ I_x=I_++I_3\\ \\ I_\text{o}=I_4+I_5 \end{cases}\tag1 $$

Using KVL, we can write:

$$ \begin{cases} I_x=\frac{V_x-V_+}{R_1}\\ \\ I_2=\frac{V_--0}{R_2}\\ \\ I_3=\frac{V_+-0}{R_3}\\ \\ I_4=\frac{V_\text{o}-V_-}{R_4}\\ \\ I_5=\frac{V_\text{o}-0}{R_5}\\ \\ \end{cases}\tag2 $$

Notice: in the ideal OPAMP circuit we assume that \$I_+=I_-=0\$. If (and only if) the opamp is used in a circuit with negative feedback then we can assume that \$V_+=V_-\$.

Now, the gain is defined by:

$$G:=\frac{V_\text{o}}{V_x}\tag3$$

We can find an expression for the output voltage \$V_\text{o}\$, by solving the systems of equations:

$$V_\text{o}=\frac{V_xR_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}\tag4$$

So, we get:

$$G=\frac{1}{V_x}\times\frac{V_xR_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}=\frac{R_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}\tag5$$

In your case we get:

$$G=\frac{68000\times\left(30000+63000\right)}{30000\times\left(12000+68000\right)}=\frac{527}{200}=2.635\tag6$$

Solving it, in general, gives (notice that \$V_+=V_-=V_\text{p}\$):

In your case (using your values):

I checked my solution using LTspice and I got it right.

Well, we have the following circuit (and we assume an ideal model of an OPAMP):

^{simulate this circuit – Schematic created using CircuitLab}

Using KCL, we can write:

$$ \begin{cases} \text{I}_4=\text{I}_-+\text{I}_2\\ \\ \text{I}_x=\text{I}_++\text{I}_3\\ \\ \text{I}_\text{o}=\text{I}_4+\text{I}_5 \end{cases}\tag1 $$

Using KVL, we can write:

$$ \begin{cases} \text{I}_x=\frac{\text{V}_x-\text{V}_+}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_--0}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_+-0}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{o}-\text{V}_-}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_\text{o}-0}{\text{R}_5}\\ \\ \end{cases}\tag2 $$

Notice: in the ideal OPAMP circuit we assume that \$\text{I}_+=\text{I}_-=0\$ and \$\text{V}_+=\text{V}_-\$.

Now, the gain is defined by:

$$\text{G}:=\frac{\text{V}_\text{o}}{\text{V}_x}\tag3$$

We can find an expression for the output voltage \$\text{V}_\text{o}\$, by solving the systems of equations:

$$\text{V}_\text{o}=\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag4$$

So, we get:

$$\text{G}=\frac{1}{\text{V}_x}\cdot\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}=\frac{\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag5$$

In your case we get:

$$\text{G}=\frac{68000\cdot\left(30000+63000\right)}{30000\cdot\left(12000+68000\right)}=\frac{527}{200}=2.635\tag6$$

Solving it, in general, gives (notice that \$\text{V}_+=\text{V}_-=\text{V}_\text{p}\$:

In your case (using your values):

I checked my solution using LTspice and I got it right.

Well, we have the following circuit (and we assume an ideal model of an OPAMP):

^{simulate this circuit – Schematic created using CircuitLab}

Using KCL, we can write:

$$ \begin{cases} \text{I}_4=\text{I}_-+\text{I}_2\\ \\ \text{I}_x=\text{I}_++\text{I}_3\\ \\ \text{I}_\text{o}=\text{I}_4+\text{I}_5 \end{cases}\tag1 $$

Using KVL, we can write:

$$ \begin{cases} \text{I}_x=\frac{\text{V}_x-\text{V}_+}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_--0}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_+-0}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_\text{o}-\text{V}_-}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_\text{o}-0}{\text{R}_5}\\ \\ \end{cases}\tag2 $$

Notice: in the ideal OPAMP circuit we assume that \$\text{I}_+=\text{I}_-=0\$ and \$\text{V}_+=\text{V}_-\$.

Now, the gain is defined by:

$$\text{G}:=\frac{\text{V}_\text{o}}{\text{V}_x}\tag3$$

We can find an expression for the output voltage \$\text{V}_\text{o}\$, by solving the systems of equations:

$$\text{V}_\text{o}=\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag4$$

So, we get:

$$\text{G}=\frac{1}{\text{V}_x}\cdot\frac{\text{V}_x\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}=\frac{\text{R}_3\left(\text{R}_2+\text{R}_4\right)}{\text{R}_2\left(\text{R}_1+\text{R}_3\right)}\tag5$$

In your case we get:

$$\text{G}=\frac{68000\cdot\left(30000+63000\right)}{30000\cdot\left(12000+68000\right)}=\frac{527}{200}=2.635\tag6$$

Solving it, in general, gives (notice that \$\text{V}_+=\text{V}_-=\text{V}_\text{p}\$):

In your case (using your values):

I checked my solution using LTspice and I got it right.

Well, we have the following circuit (and we assume an ideal model of an OPAMP):

^{simulate this circuit – Schematic created using CircuitLab}

Using KCL, we can write:

$$ \begin{cases} I_4=I_-+I_2\\ \\ I_x=I_++I_3\\ \\ I_\text{o}=I_4+I_5 \end{cases}\tag1 $$

Using KVL, we can write:

$$ \begin{cases} I_x=\frac{V_x-V_+}{R_1}\\ \\ I_2=\frac{V_--0}{R_2}\\ \\ I_3=\frac{V_+-0}{R_3}\\ \\ I_4=\frac{V_\text{o}-V_-}{R_4}\\ \\ I_5=\frac{V_\text{o}-0}{R_5}\\ \\ \end{cases}\tag2 $$

Notice: in the ideal OPAMP circuit we assume that \$I_+=I_-=0\$. If (and only if) the opamp is used in a circuit with negative feedback then we can assume that \$V_+=V_-\$.

Now, the gain is defined by:

$$G:=\frac{V_\text{o}}{V_x}\tag3$$

We can find an expression for the output voltage \$V_\text{o}\$, by solving the systems of equations:

$$V_\text{o}=\frac{V_xR_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}\tag4$$

So, we get:

$$G=\frac{1}{V_x}\times\frac{V_xR_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}=\frac{R_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}\tag5$$

In your case we get:

$$G=\frac{68000\times\left(30000+63000\right)}{30000\times\left(12000+68000\right)}=\frac{527}{200}=2.635\tag6$$

Solving it, in general, gives (notice that \$V_+=V_-=V_\text{p}\$):

In your case (using your values):

I checked my solution using LTspice and I got it right.