# Difference between revisions of "MIE 2016/Day 1/Problem 2"

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− | + | ==Problem 2== | |

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\begin{cases}x^2-5x-14>0\\x>0\end{cases}</math> | ||

+ | |||

+ | <math>\left.\begin{array}{l}x > 0\\x > 7\\x < -2\end{array}\right\}x\in(-2,0)\cup(7,+\infty)</math> | ||

+ | |||

+ | |||

+ | 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 .