Difference between revisions of "2006 iTest Problems/Problem 37"
(Created page with "==Solution== To begin, let's number the equations as follows <cmath>\begin{align} x^2 + xy + y^2 &= 1 \\ y^2 + yz +z^2 &= 2 \\ z^2 + zx +x^2 &= 3 \end{align}</cmath> In or...") |
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Therefore, the solution is <math> 6+2+6+9 = \boxed{23}</math> | Therefore, the solution is <math> 6+2+6+9 = \boxed{23}</math> | ||
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+ | ==See Also== | ||
+ | {{iTest box|year=2006|num-b=36|num-a=38|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}} |
Latest revision as of 18:27, 14 January 2020
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, Answer Key) | ||
Preceded by: Problem 36 |
Followed by: Problem 38 | |
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