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August 29, 2014

August 29, 2014

Posted by **Wenny** on Saturday, January 26, 2008 at 8:04pm.

a>3

3a>3(3)

3a-a^2>9-a^2

a(3-a)>(3-a)(3+a)

a>3+a

0>3

the problem with the proof is that the fourth step needs the sign to be reverse...but why?

i looked at step 3 and there is nothing negative to reverse the sigh for...

- Math -
**drwls**, Saturday, January 26, 2008 at 9:37pmFrom the first statement, a must be a positive number. In fact, it must exceed 3. (This is very important later). The second and third statements are correct, since you are either multiplying by positive numbers or just factoring. The fourth statement is incorrect because 3-a is a negative number.The directon of the > sign must change.

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