Question 1209274
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A tractor has a rear wheel with a circumference of x m. The circumference of the front wheel is 5 m less. 
If the front wheel makes 15 more revolutions than the rear wheel every 450 m, 
what is the circumference, in m, of the front wheel?
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<pre>
The rear wheel has the circumference of x meters.

Hence, the front wheel has the circumference of (x-5) meters, according to the problem.


The rear wheel  makes  {{{450/x}}}  rotations every 450 m.

The front wheel makes  {{{450/(x-5)}}}  rotations every 450 m.


The problem says

    {{{450/(x-5)}}} - {{{450/x}}} = 15.


It is your setup equation.


To solve it, first multiply both sides by x*(x-5);  then simplify

    450x - 450(x-5) = 15x(x-5)

    450x - 450x + 2250 = 15x(x-5)

    2250 = 15x(x-5)

     150 = x*(x-5).


At this point, you may guess that x = 15.


Indeed,  15*(15-5) = 15*10 = 150.

In addition, the function x*(x-5) is monotonic at x > 5,  
so, this guessed solution is UNIQUE in the domain x > 5.


Alternatively, you may solve equation x^2 - 5x - 150 = 0  factoring or by using
the quadratic formula.


So, the circumstance of the rear wheel is 15 meters.
Hence, the circumstance if the front wheel is 15-5 = 10 meters.    <U>ANSWER</U>
</pre>

Solved.


You may check the answer on your own, by substituting the numbers into the problem.