Question 1064087
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The key to this problem is the idea that if the average of *[tex \Large n] numbers is *[tex \Large A], then the sum of the *[tex \Large n] numbers must be *[tex \Large n\ \times\ A].


So if the average of the first 2 tests is 8, the sum of the scores of the first two tests must be 16.  So let *[tex \Large x] represent the score of the first test and *[tex \Large y] represent the score of the second test.  Now you can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 16]


Likewise, if the average of the first three tests is 12, then the sum of the first three scores must be 3 times 12, or 36.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ +\ z\ =\ 36]


Finally, we know that the 3rd test score is 13 points more than the first test score, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ x\ +\ 13]


Now you have three linear equations in three variables, so solve for *[tex \Large x], *[tex \Large y], and *[tex \Large z] by any convenient means.  Given the configuration of the of the third equation, I would first make a substitution.  Then I would simplify and solve for *[tex \Large x] and *[tex \Large y] by elimination.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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