High school Kirchoff’s voltage law question.

[Ten Over]

Loop C: Point F to point A has a rise of 6V, A to B has a drop of 3 x i1, and C to D has a rise of 4V.
+6-(3 x i1)+4=0
+6+4=3 x i1
10=3 x i1
i1=10/3
i1=3.33A

Point B: Currents coming toward a point are considered positive while currents going away from a point are considered negative. The algebraic sum of all currents at a point must equal zero.
i1+i2+i3=0
+3.33+2.0+i3=0
i3= -3.33-2.0
i3= -5.33
This time the negative value indicates that the current is going away from Point B. The problem asked for the value of i3 in the direction shown, so the answer is 5.33A without the negative sign.

[Ten Over]

Loop B Opposite Direction: Point C to point D has a rise of 4V and E to B has a rise of 2 x i2.
+4+(2 x i2)=0
2 x i2 = -4
i2= -4/2
i2= -2
So we get the same answer with either loop direction.

[Mark]

Loop B Opposite Direction: Point C to point D has a rise of 4V and E to B has a rise of 2 x i2.
+4+(2 x i2)=0
2 x i2 = -4
i2= -4/2
i2= -2
So we get the same answer with either loop direction.

Thanks for the explanation. It was thorough and concise, I could see how you handled my error and explained the right way to approach the problem.

What is embarrassing is, I use to know this stuff like the back of my hand. I will have to read up and revise.