## Graphing |7 - 3x| - 10 = 4

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diaste
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Joined: Sat Jan 17, 2009 1:54 pm
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### Graphing |7 - 3x| - 10 = 4

I have a question on negative numbers in absolute value equations.
For instance:
$|7-3x|-10=4$
$|-3x+7|-10=4$
$-3x+7=14$
$-3x = 7$
$x = - \frac 73$
This is fine for me. I'm assuming that the result is a negative based on the division by -3. The book says that's correct.
The other solution is what's giving me trouble. I get -7 and the book says 7.
$|-7-3x|-10=4$
$|-7-3x|=14$
$-3x=21$
$x= - \frac {21}{3}$
$x=-7$

Am I supposed to think that $- \frac{21}{3}$ is the same as a negative divided by a negative? That would work out to a positive. If that's true then $- \frac 73$ would be 2.33 in decimal. I'm going to convert it to a graphing equation and check out the vectors but I really would like to hear an explanation of the negative number thing in absolute value equations. I'm thinking that because -3x is an absolute value that it's either positive or negative and so when I can simplify to a whole number it's a positive, and then again, maybe not.

Thanks! Hope you are having a great weekend!

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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### Re: Graphing |7 - 3x| - 10 = 4

I have a question on negative numbers in absolute value equations. For instance:
$|7-3x|-10=4$
$|-3x+7|-10=4$
$-3x+7=14$
The error is in the last step. To take the absolute-value bars off, you must assume that "-3x + 7" is positive. But it need not be!

Instead, consider the two cases. If -3x + 7 > 0, then 7/3 > x (or, as I prefer, x < 7/3). So, assuming x to be in this range, we then get:

. . . . .|7 - 3x| - 10 = 4

. . . . .(7 - 3x) - 10 = 4

. . . . .-3x -3 = 4

. . . . .-3x = 7

. . . . .x = -7/3

Since -7/3 is indeed less than 7/3, this solution works.

Now consider the other case. If -3x + 7 < 0, then |-3x + 7| = -(-3x + 7) = 3x - 7, since the absolute value changes the sign. (And if (-3x + 7) is negative, then obviously -(-3x + 7) must be positive.) The range of values will then be x > 7/3. Plugging in the bar-free form, we get:

. . . . .|7 - 3x| - 10 = 4

. . . . .|-3x + 7| - 10 = 4

. . . . .-(-3x + 7) - 10 = 4

. . . . .3x - 7 - 10 = 4

. . . . .3x - 17 = 4

. . . . .3x = 21

...and so forth.

Hope that helps!

Eliz.

diaste
Posts: 14
Joined: Sat Jan 17, 2009 1:54 pm
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### Re: Graphing |7 - 3x| - 10 = 4

I see that now. I got the idea after doing the graph and doing a little more reading but your explanation helps me even more.

Thanks Elizabeth!

diaste
Posts: 14
Joined: Sat Jan 17, 2009 1:54 pm
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### Re: Graphing |7 - 3x| - 10 = 4

Hmmm, a question just came up.

You wrote;

Instead, consider the two cases. If -3x + 7 > 0, then 7/3 > x (or, as I prefer, x < 7/3). So, assuming x to be in this range, we then get:

. . . . .|7 - 3x| - 10 = 4

. . . . .(7 - 3x) - 10 = 4 Here the sign didn't change. Is that because we are assuming that we need a value in that range?

. . . . .-3x -3 = 4

. . . . .-3x = 7

. . . . .x = -7/3

And you also wrote:

Now consider the other case. If -3x + 7 < 0, then |-3x + 7| = -(-3x + 7) = 3x - 7, since the absolute value changes the sign. (And if (-3x + 7) is negative, then obviously -(-3x + 7) must be positive.) The range of values will then be x > 7/3. Plugging in the bar-free form, we get:

. . . . .|7 - 3x| - 10 = 4

. . . . .|-3x + 7| - 10 = 4

. . . . .-(-3x + 7) - 10 = 4 Here you said the removal of the absolute value changes the sign. No problem. But above it didn't.

. . . . .3x - 7 - 10 = 4

. . . . .3x - 17 = 4

. . . . .3x = 21

I'm still baffled about that part......

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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The sign changes if the argument (the stuff between the bars) is negative; otherwise, it doesn't.

. . . . .-5 < 0, and |-5| = -(-5) = 5

. . . . .5 > 0, and |5| = (5) = 5

When you have variables, you don't know the sign, so you consider both options:

. . . . .if -3x + 7 < 0, then |-3x + 7| = -(-3x + 7) = 3x - 7

. . . . .if -3x + 7 > 0, then |-3x + 7| = (-3x + 7) = -3x + 7

Does that make a little more sense?

Eliz.