Determine the current in each branch of the network shown . . .

Question : Determine the current in each branch of the network shown in Figure.

Doubt by Gurdev

Solution : 

Applying Kirchhoffs's Loop Rule in closed Loop 1 

1I₁+4I₂+2(I₂+I₃)=10

I₁+4I₂+2I₂+2I₃=10

I₁+6I₂+2I₃=10 — (1)

Applying Kirchhoffs's Loop Rule in closed Loop 2 

1I₁+4(I₁-I₂)-2(I₂+I₃-I₁)=10

I₁+4I₁-4I₂-2I₂-2I₃+2I₁=10

7I₁-6I₂-2I₃=10 — (2)

Applying Kirchhoffs's Loop Rule in closed Loop 3 

2(I₂+I₃)+2(I₂+I₃-I₁)=5

2I₂+2I₃+2I₂+2I₃-2I₁=5

-2I₁+4I₂+4I₃=5 — (3)

Adding equations (1) and (2) to eliminate I2 and I3

(I₁ + 6I₂ + 2I₃) + (7I₁ - 6I₂ - 2I₃) = 10 + 10

I₁ + 6I₂ + 2I₃+ 7I₁ - 6I₂ - 2I₃ = 20

8I₁ = 20
I₁ = 20/8
I₁=5/2 A
I₁=2.5 A

Substituting I₁=2.5 A in equation (1) 

2.5+6I₂+2I₃=10
6I₂+2I₃=10-2.5
6I₂+2I₃=7.5 — (A)

Substituting I₁=2.5 A in equation (3)

-2(2.5)+4I₂+4I₃=5 
-5+4I₂+4I₃=5
4I₂+4I₃=5+5
4I₂+4I₃=10 — (B)

Now Solving Equation (A)  and (B)

[6I₂+2I₃=7.5 ]×2

4I₂+4I₃=10

12I₂+4I₃=15

  4I₂+4I₃=10
-      -         -
-----------------
8I₂+0=5
8I₂=5
I₂=5/8 A
I₂=0.625 A

Putting I₂=5/8 in equation (A)

6(5/8)+2I₃=7.5
15/4+2I₃=7.5
2I₃=7.5-(15/4)
2I₃=(15/2)-(15/4)
2I₃=(30/4)-(15/4)
2I₃=15/4
I₃=15/8 A
OR 

I₃=1.875 A

The currents in the various branches of the network are
AB = I₂ = 5/8 A = 0.625 A
AC = I₁ = 2.5 A or 5/2 A
AD = I₁-I₂ = 2.5 - 0.625 = 1.875 A or 15/8 A
BC = I₂+I₃ = 5/8 + 15/8 = 20/8 = 5/2 = 2.5 A

CD = I₂+I₃-I1 
= 5/8 + 15/8 - 5/2

= 5/8 + 15/8 - 20/8
= 20/8 - 20/8
= 0 
DEB = I₃ = 15/8  = 1.875 A