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 November 7th, 2017, 09:03 AM #1 Newbie   Joined: Nov 2017 From: New Jersey Posts: 2 Thanks: 0 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.
November 7th, 2017, 09:37 AM   #2
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Quote:
 Originally Posted by allo95 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

 November 7th, 2017, 06:23 PM #3 Newbie   Joined: Nov 2017 From: New Jersey Posts: 2 Thanks: 0 Yes, I think I can. Thank you.
November 8th, 2017, 05:24 AM   #4
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Quote:
 Originally Posted by topsquark 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.

November 8th, 2017, 08:14 AM   #5
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Quote:
 Originally Posted by Country Boy I presume you meant "Let y= |x- 7|."
Oops! Thanks for the catch.

Quote:
 Originally Posted by Country Boy 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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