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January 31, 2015

January 31, 2015

Posted by **Telli** on Thursday, August 16, 2007 at 2:27pm.

1/3x > 2 and 1/4x > 2

First should I solve for x. How would you handle this problem?

yes, solve for x in each..

x>6; x>8

so if the second condition is met, that is x is g reater than eight, the first is not applicable.

first solve each of the inequalities

here is the first:

1/3x > 2

1 > 6x

6x<1

x<1/6

in the same way from the second, x<1/8

Also it should be obvious that x cannot be zero or a negative, since 1/(a negative) is not greater than 2

so we have x<1/6 AND x<1/8

so the only values that would work are

0 < x < 1/8

See how important it is to use proper notation?

if you meant (1/3)x > 2 and (1/4)x > 2

then bobpursley's solution is correct

if you meant 1/(3x) > 2 and 1/(4x) > 2

then my solution is correct.

5N+3-4NGRAETER THEN -5-3N

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