A, B, and C are 3 collinear point. (3 points on the same line.)
AB=3y-1, and BC=7y
Since A, B, and C then AC = AB+BC = (3y-1)+7y = 10y-1
AC is 29 units, so we write the equation:
(AC=) 10y-1 =29
adding 1 to both sides, we have:
10y=30,
dividing both sides by 10:
y=3.
AB = 3y-1, substituting y with 3, we have AB=3*3-1=9-1=8