Answer:
We have the problem:
|3r−15| if r<5
First we see the equality, if r = 5 we have:
I3r - 15I = I3*5 - 15I = I0I = 0.
Then the only restriction that we have is:
I3r - 15I > 0.
now, we could simplify it a bit further:
if r < 5, then the thing inside the absolute value will always be negative:
Then we can write:
I3*r - 15I = -(3*r -15) > 0
multiplying by -1 in both sides
(3r - 15) < 0.
if we keep simplifying this, we will get our initial restriction:
3r - 15 < 0
3r < 15
r < 15/3 = 5
r < 5
