<PRE><B>Question 1050565</B>
let f = circumference of the front wheel.
let r = circumference of the rear wheel.


you are given that the circumference of the rear wheel is 3 feet more than the circumference of the front wheel.


this means that r = f + 3.


when the bicycle travels 6000 feet, the number of rotations of the front wheel is equal to 6000 / f and the number of rotations of the rear wheel is equal to 6000 / r.


you are given that the number of rotations of the rear wheel is 100 less than the number of rotations of the front wheel.


this means that (6000 / r) = (6000 / f) - 100.


since you already know that r = f + 3, then this equation becomes:


(6000 / (f + 3) = 6000 / f - 100


if we multiply both sides of this equation by f * (f + 3), we get:


6000 * f = 6000 * (f + 3) - 100 * f * (f + 3).


we simplify this equation to get:


6000 * f = 6000 * f + 18000 - 100 * f^2 - 300 * f


if we subtract 6000 * f from both sides of the equation, we get:


0 = 18000 - 100 * f^2 - 300 * f


if we divide both sides of the equation by 100, we get:


0 = 180 - f^2 - 3 * f


if we multiply both sides of the equation by -1, we get:


0 = -180 + f^2 + 3f


if we reorder the terms of the equation in descending order of degree, we get:


0 = f^2 + 3f - 180.


this is the same as f^2 + 3f - 180 = 0


this is a quadratic equation.


we can factor this quadratic equation to get:


(f + 15) * (f - 12) = 0


we solve for x to get:


x = -15 or x = 12.


x can&#39;t be negative, so x = 12 should be our solution.


x = 12 is the circumference of the front wheel.
x = 15 is the circumference of the rear wheel.


6000 / 12 = 500
6000 / 15 = 400


the difference is 100 with the rear wheel rotating 100 less times than the front wheel.


the solution looks good.






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