Re: Inequality's

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: : : : 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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