I get:
$$\begin{align*}0\:\text{V}+V_1 - R_1\cdot I_1 - R_4\cdot\left(I_1-I_2\right)&=0\:\text{V}\tag{mesh 1}\\\\0\:\text{V}- R_4\cdot\left(I_2-I_1\right)-R_2\cdot I_2-R_5\cdot\left(I_2-I_3\right)&=0\:\text{V}\tag{mesh 2}\\\\0\:\text{V}-R_5\cdot\left(I_3-I_2\right) -R_3\cdot I_3-V_{I_s}&=0\:\text{V}\tag{mesh 3}\\\\I_3&=-I_s\tag{given}\end{align*}$$
Note that you must account for the current source's voltage drop.
The above is four equations and four unknowns. Also, since I kept all this as variables and not specific given values it allows me to make changes to answer questions.
Using SymPy (freely available) I find:
values = { R1:3, R2:4, R3:12, R4:8, R5:6 }mesh1 = Eq( 0 + V1 - R1*I1 - R4*(I1-I2), 0 )mesh2 = Eq(0 - R4*(I2-I1) - R2*I2 - R5*(I2-I3), 0 )mesh3 = Eq(0 - R5*(I3-I2) - R3*I3 - VIs, 0 )given = Eq( I3, -Is )for i,j in solve( [mesh1, mesh2, mesh3, given], [I1, I2, I3, VIs] ).items(): i,j.subs(values | { V1:42, Is:2 })(I1, 330/67)(I2, 102/67)(I3, -2)(VIs, 3024/67)for i,j in solve( [mesh1, mesh2, mesh3, given], [I1, I2, I3, VIs] ).items(): i,j.subs(values | { V1:0, Is:2 })(I1, -48/67)(I2, -66/67)(I3, -2)(VIs, 2016/67)You know that \$V_{\small{X}}=V_{I_s}+I_3\cdot R_3\$.
So in the first case with both sources active: \$V_{\small{X}}=\frac{3024}{67}\:\text{V}+\left(-2\:\text{A}\right)\cdot 12\:\Omega\approx 21.134\:\text{V}\$.
In the second case with the voltage source off: \$V_{\small{X}}=\frac{2016}{67}\:\text{V}+\left(-2\:\text{A}\right)\cdot 12\:\Omega\approx 6.090\:\text{V}\$.