# Re: Inequality's

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Posted by Soroban on August 27, 2002 at 16:20:44:

In Reply to: Re: Inequality's posted by Brandy on August 27, 2002 at 15:20:35:

: : 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)

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