Question 349235
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Let *[tex \Large x] represent the circumference of the smaller wheel.  Then *[tex \Large x\ + 3] is the circumference of the larger wheel.  Divide the circumference into the distance traveled to get the number of revolutions.  Let *[tex \Large y] be the number of revolutions of the front wheel.  Then the number of revolutions of the back wheel must be *[tex \Large y\ -\ 100]


Now we can write two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{6000}{x}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 100\ =\ \frac{6000}{x\ +\ 3}]


Solve the second equation for *[tex \Large y]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{6000}{x\ +\ 3}\ +\ 100]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{6300\ +\ 100x}{x\ +\ 3}]


Verification of the addition in the last step is left as an exercise for the student.


Now that we have two different expressions both equal to *[tex \Large y], set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{6000}{x}\ =\ \frac{6300\ +\ 100x}{x\ +\ 3}]


Cross-multiply, collect like terms, and put the equation into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 100x^2\ +\ 300x\ -\ 18000\ =\ 0]


Divide both sides by 100:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 3x\ -\ 180\ =\ 0]


Factor, solve the quadratic, discard the negative root.  The positive root will be the circumference of the smaller wheel.  The positive root plus three will be the circumference of the larger wheel.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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