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
: Then treat the inequality like you would an equal sign and start moving your terms until you have the variable isolated like this:
Oops, it becomes +2 when moved, Brandy.
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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