Mesh analysis equations with dependent current sources
It helps to redraw the schematic, a little, and to include some other unknowns in the schematic.



simulate this circuit &ndash; Schematic created using CircuitLab
One of the things that earlier students often forget about is that current sources have applied voltages across them and that these may be required in order to work out the KVL loops.
With the above, we can write out the loops:
$$\begin{align*}
0\:\text{V}+V_{I_1}-R_1\cdot\left(I_A-I_B\right)&amp;=0\:\text{V}\\
0\:\text{V}-R_1\left(I_B-I_A\right)-R_2\cdot I_B-V_{I_2}&amp;=0\:\text{V}\\
0\:\text{V}+V_{I_2}-R_3\cdot I_C-R_4\cdot\left(I_C-I_D\right)&amp;=0\:\text{V}\\
0\:\text{V}-R_4\cdot\left(I_D-I_C\right)-V_{I_3}&amp;=0\:\text{V}
\end{align*}$$
This suggests two unknown loop currents and three unknown voltages, but only four equations.
You know a few other details:
$$\begin{align*}
I_A&amp;=2\:\text{A}\\
I_D&amp;=-5\:\text{A}\\
I_X&amp;=I_A-I_B \\
I_2&amp;=I_C-I_B=1.5\cdot I_X=1.5\cdot\left(I_A-I_B\right)\\\therefore \\I_C&amp;=\frac32\cdot I_A-\frac12\cdot I_B\\&amp;=3\:\text{A}-\frac12\cdot I_B
\end{align*}$$
Only the last equation is the important addition. Now you have five unknowns and five equations.