.
Let x be Fred's normal walking speed, in feet per second (the value under the question.
Then Fred' speed moving with walkway is (x+ 4.4) ft/s,
while Fred' speed moving with against walkway is (x- 4.4) ft/s.
The time moving 110 ft with walkway is
seconds,
while the time moving 110 ft against walkway is
seconds.
The total two-ways time moving forward and back is
+
= 50 seconds.
It is your time equation. All its terms are time periods.
To solve it, multiply both sides by (x-4.4)*(x+4.4) =
=
. You will get then
110*(x + 4.4) + 110*(x - 4.4) = 50*(x^2 - 19.36).
220x = 50x^2 - 968
50x^2 - 220x - 968 = 0
=
=
.
Only positive root x =
is meaningful, which gives you the
ANSWER. Fred's normal walking speed is about 7.12 ft/s, according to the given data.
CHECK.
+
= 50 seconds (approximately).
My comment. My solution and my answer are correct - it was CHECKED.
But the answer of 7.12 ft/s does not seem to be likelihood,
which means that the input data in the post are not realistic.