# Distributive Property and Absolute Value?

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• Aug 27th 2009, 08:53 AM
sdudley24
Distributive Property and Absolute Value?
OK...so something doesn't seem right when I distribute a positive number versus a negative number into an absolute value. Can someone please explain this to me.

For example...it was a couple of inequality problems that caught my attention on this:

-2 abs(3x - 4) < 16

versus

3 abs(2x + 5) > 9

I don't think you can distribute a negative into absolute value...or can you? Please help! (Thinking)
• Aug 27th 2009, 09:06 AM
math84
You can distribute only nonnegative numbers into absolute values. With \$\displaystyle -2 |3x - 4|\$, you can distribute the \$\displaystyle 2\$ but the negative sign must stay outside.
• Aug 27th 2009, 09:14 AM
sdudley24
ok...
So what do you do with the negative outside of the parentheses if you leave it there? Can someone show me an example or a place on the web where it explains this. It would be nice if I could at least see the first example I posted worked out...

Anyone???
• Aug 27th 2009, 09:26 AM
Defunkt
Are you sure \$\displaystyle (-2) |3x-4| < 16 \$ is the right inequality?!

Since the absolute value is always non-negative, multiplying it by a negative value will always yield a non-positive result and thus obviously lesser than 16 (a positive value)!
• Aug 27th 2009, 09:39 AM
sdudley24
yes...
That is the problem that was given to me in an exercise to work out...is it impossible to do? I just don't get it...
• Aug 27th 2009, 09:42 AM
Defunkt
It's possible! There's just no use playing with it as it can easily be seen that it is correct for any \$\displaystyle x\$
• Aug 27th 2009, 09:59 AM
sdudley24
bad example...
OK...so I guess that was a bad example to use for me to understand exactly how to do this. Can you post a few examples or rules that I can follow and learn? Anyone?!?!

-2 abs(x + 5)

-5 abs (-x - 6)

-7 abs (x - 2)

Do the rules change if the above examples were equations or inequalities?
• Aug 27th 2009, 10:19 AM
Defunkt
Quote:

Originally Posted by sdudley24
OK...so I guess that was a bad example to use for me to understand exactly how to do this. Can you post a few examples or rules that I can follow and learn? Anyone?!?!

-2 abs(x + 5)

-5 abs (-x - 6)

-7 abs (x - 2)

Do the rules change if the above examples were equations or inequalities?

No, they don't. Consider this:

Is \$\displaystyle -2 * |-5|\$ equal to \$\displaystyle |(-2)(-5)|\$? Of course not:

\$\displaystyle -2 * |-5| = -10 \neq 10 = |(-2)(-5)|\$

If we want to solve, for example, \$\displaystyle -2|x+5| = -12\$ then we will do it like this:

Divide both sides by \$\displaystyle -2\$:

\$\displaystyle |x+5| = 6 \Rightarrow x+5 = 6\$ or \$\displaystyle x+5 = -6 \Rightarrow x_1 = 1; x_2 = -11\$
• Aug 27th 2009, 10:45 AM
sdudley24
better...
OK...that helped me out...and if anyone else wants to add to this whole concept, I would greatly appreciate it.

Thanks so much!