algebra

The value of b is 44/7.

a + b + c ==99.......................(1)

[a+5] = [a-5] = [c*5].............(2), solve for a, b, c

Use substitution to get:

a==40,  b==50,   c==9

a + b + c = 99

a + 5 = b - 5 ===> a = b + 10

b - 5 = 5c  ===> c = \({b - 5} \over 5\)

so b + 10 + b + \({b - 5} \over 5\) = 99

so 5b + 50 + 5b + b - 5 = 495

so 11b - 45 = 495

so 11b = 540

so b = \(540 \over 11\)