# Re: Inequality's

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Posted by Brandy on August 28, 2002 at 15:34:52:

In Reply to: Re: Inequality's posted by Soroban on August 27, 2002 at 16:20:44:

: : : I need to solve this Inequality in Interval Notation:

: : : -(4a+6)<3a-2

: : First release the parentheses so that your problem now looks like this

: : -4a-6<3a-2

: : Then treat the inequality like you would an equal sign and start moving your terms until you have the variable isolated like this:

: : -6<(3a+4a)-2
: : -6<7a-2
: : -6-2<7a

: Oops, it becomes +2 when moved, Brandy.

: : -8<7a
: : -8/7=a

: An obvious typo: the missing "<"

: A matter of personal taste: I prefer to have the
: variable on the left.

: I agree that "x > 2" and "2 < x" are equivalent -
: no argument. But I've found that "2 < x" is a
: bit irksome. "All x such that 2 is less than x"
: doesn't give me an instant picture.

: I'll go through the problem again.

: Release the parentheses:
: -4a - 6 < 3a - 2

: Move variables to the left:
: -4a - 6 - 3a < -2

: Move constants to the right:
: -4a - 3a < -2 + 6

: Combine terms: -7a < 4

: Final step: divide by -7

: And THIS is why I wanted to go through the
: problem again - to make this important point.

: Phaedra, when dealing with inequalties,
: you can treat the statement like an equality -
: EXCEPT when multiplying or dividing by a
: *negative* quantity, REVERSE THE INEQUALITY.

: So, dividing by -7: a > -4/7

: In interval notation: (-4/7, Infinity)

ACK!!! It's always the little mistakes that kill me. Thanks for the correction!

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