Difference between revisions of "MIE 2016/Day 1/Problem 2"
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The following system has <math>k</math> integer solutions. We can say that: | The following system has <math>k</math> integer solutions. We can say that: | ||
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− | + | ==Solution 2== | |
+ | <math>\frac{x^2-2x-14}{x}>3</math> | ||
+ | <math>\frac{x^2-2x-14}{x}-\frac{3x}{x}>0</math> | ||
+ | |||
+ | <math>\frac{x^2-5x-14}{x}>0 | ||
+ | |||
+ | <math>\left. | ||
+ | |||
+ | |||
+ | Adding the other interval we get | ||
+ | |||
+ | |||
+ | <math>x\in(-2,0)\cup(7,12]</math> | ||
+ | |||
+ | |||
+ | If <math>k</math> is the number of integer solutions, then <math>k=7</math>. <math>\boxed{D}</math> | ||
===See Also=== | ===See Also=== |
Revision as of 18:29, 8 January 2018
Problem 2
The following system has integer solutions. We can say that:
(a)
(b)
(c)
(d)
(e)
Solution 2
Adding the other interval we get
If is the number of integer solutions, then .