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November 7th, 2017, 09:03 AM   #1
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absolute equations

Hi All,

what would be the best way to solve for x:

13|x-7| + 11 = 7|x-7| + 11

I started subtracting 11 from both sides and then dividing both sides by 13, but then got stuck with how to isolate the x.

Thanks in advance.
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November 7th, 2017, 09:37 AM   #2
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Originally Posted by allo95 View Post
Hi All,

what would be the best way to solve for x:

13|x-7| + 11 = 7|x-7| + 11

I started subtracting 11 from both sides and then dividing both sides by 13, but then got stuck with how to isolate the x.

Thanks in advance.
Let y = x - 7. Then your equation is
13y + 11 = 7y + 11

Cancel the 11.
13y = 7y

What value can y take? Once you have that you know that you can use the definition of |p|:
$\displaystyle |p| = \begin{cases} -p & p \leq 0 \\ p & 0 < p \end{cases}$

Can you finish from here?

-Dan
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November 7th, 2017, 06:23 PM   #3
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Yes, I think I can. Thank you.
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November 8th, 2017, 05:24 AM   #4
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Quote:
Originally Posted by topsquark View Post
Let y = x - 7.
I presume you meant "Let y= |x- 7|."

Quote:
Then your equation is
13y + 11 = 7y + 11

Cancel the 11.
13y = 7y

What value can y take? Once you have that you know that you can use the definition of |p|:
$\displaystyle |p| = \begin{cases} -p & p \leq 0 \\ p & 0 < p \end{cases}$

Can you finish from here?

-Dan
I always thought the definition of |p| was
$\displaystyle |p| = \begin{cases} p & p \geq 0 \\ -p & p < 0 \end{cases}$

Last edited by Country Boy; November 8th, 2017 at 05:28 AM.
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November 8th, 2017, 08:14 AM   #5
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Originally Posted by Country Boy View Post
I presume you meant "Let y= |x- 7|."
Oops! Thanks for the catch.

Quote:
Originally Posted by Country Boy View Post
I always thought the definition of |p| was
$\displaystyle |p| = \begin{cases} p & p \geq 0 \\ -p & p < 0 \end{cases}$
Smart*ss.

-Dan
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