Inequalities and Interval Notation.

healingduck

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Aug 2, 2011
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So the problem is -1-x<5 The way the teacher told it was kind of hard to understand and if I can't figure it out online I plan to ask tomorrow. Its just kind of confusing and its a 10 week course in 5 weeks so I'm basically learning new stuff everyday.

Thanks for any help.
 
Did you first solve for 'x' in the usual way?
 
That's kind of the part I'm confused about. Common sense would tell me to move the one to the 5 so it would read -x<6
 
healingduck said:
That's kind of the part I'm confused about. Common sense would tell
me to move the one to the 5 so it would read -x<6

That is *one* of the correct routes.

You need to get x on one side by itself.

\(\displaystyle -x \ < \ 6\)

When you multiply or divide each side of an inequality
by a negative number, then the sign changes in *that*
step.

The following is wrong:

\(\displaystyle (-1)(-x) \ < \ (-1)(6)\)

\(\displaystyle x \ > \ -6\)


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And so, too, is the following wrong:

\(\displaystyle \frac{-x}{-1} \ < \ \frac{6}{-1}\)

\(\displaystyle x \ > \ -6\)


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Using the division option from the original problem,
it would be this:

\(\displaystyle -x \ < \ 6\)

\(\displaystyle \frac{-x}{-1} \ > \ \frac{6}{-1}\)

\(\displaystyle x \ > \ -6\)

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Or, you might want to avoid multiplying or
dividing by a negative number as in:


\(\displaystyle -x \ < \ 6\)

\(\displaystyle -x + (x - 6) \ < \ 6 + (x - 6)\)

\(\displaystyle -6 \ < \ x\)


Turning it around:

\(\displaystyle x \ > \ -6\)
 
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