\begin{align} a &= 7\\ b &= 4\\ x &= a-b-N_i+1\\ &= 7-4-1+1\\ &= 3 \end{align} \begin{align} 4I_4 &+& 10I_3&-&5I_2 && &=0\\ 240 &+& 40 I_1&+& 5I_2&+&10I_3 &= 0\\ 4I_4&+&240&+&40I_1 && &=0 \end{align}
In this exersice, it uses the mesh analysis method. Why, it takes the KVL for the J2 and the 2 outer loops, instead of using KVL in J2 with the inner loops J3 and J4?
Your question could have been better with a little extra time added to improve its readability.
I'll call your loop currents \$J_1\dots J_4\$ and accept your use of \$I\$ for the currents through two-terminal devices, as shown. I'll also assign \$V_1\$ to be the voltage across the \$19\:\text{A}\$ current source and \$V_2\$ to be the voltage across the dependent current source. Finally, I'll use SymPy (freely available and free to use) and write out the equations:
j1,j2,j3,j4,v1,v2 = symbols('j1,j2,j3,j4,v1,v2', real=True)
i1 = j1 - j3
i2 = j2 - j3
i3 = j2 - j4
i4 = j4
mesh1 = Eq( 0 + v1 - 40*i1, 0 )
mesh2 = Eq( 0 + 4*i4 - 10*i3 - 5*i2, 0 )
mesh3 = Eq( 0 + 40*i1 + 5*i2 - v2, 0 )
mesh4 = Eq( 0 + v2 + 10*i3 + 240, 0 )
eqj1 = Eq( j1, 19 )
eqdep = Eq( 2*i2, j4 - j3 )
solve( [ mesh1, mesh2, mesh3, mesh4, eqj1, eqdep ], [ j1, j2, j3, j4, v1, v2 ] )
{j1: 19, j2: 18, j3: 26, j4: 10, v1: -280, v2: -320}
Your writing is not entirely clear to me, so I'll let you look over the above for answers.
To confirm, LTspice says:
If I understand your question correctly, you want to ask why, for writing KVL, the G3 and G4 loops are combined, as well as the G2, G3, and G4 loops.
In general, we cannot directly calculate the voltage across the current source and the current through the voltage source on paper. Therefore, in the mesh analysis method, in loops that include a current source, we increase the loop boundary so that the current source falls inside the loop. For example, in your circuit, we have a current source for the branch between the G3 and G4 loops, so I consider these two loops together so that there is no need for the voltage of the current source.
On the other hand, we obtain the hypothetical current that we considered for the meshes with the help of the current of these current sources so that the number of equations is not less than the number of unknowns (because by doing this, it seems that we have one equation instead of two equations, and don't forget, we need an equal number of unknowns and equations.), such as in your circuit, the current of the G1 loop is 19 amperes.
We do the same for the node analysis method for the voltage source.
Let me know if you need more clarification!
