2006 iTest Problems/Problem 37
Solution
To begin, let's number the equations as follows
In order to solve for , we begin by subtracting the first equation from the second equation:
Similarily, subtracting the second eqaution from the third equation:
Subtracting the factored equations we see that
Therefore, we can conclude that either , or that . As we are told that has a unique value, we can use either option.
Continuing with the second option, we can plug into the first equation:
Multiplying each side of the resulting equation by four:
We can then subtract twice our original second equation, , from the equation from the previous step, yielding:
Rearranging the new equation, we see that:
We can ignore the negative root of the RHS because both y and z are positive, and because which is a result of (3) being greater than (1). Continuing to manipulate the newest equation yields:
Plugging this back into (2)
Solving for yields that
Therefore, the solution is
See Also
2006 iTest (Problems) | ||
Preceded by: Problem 36 |
Followed by: Problem 38 | |
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