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Absolute Value Equations: Setting Restrictions

  1. Jan 9, 2008 #1
    Hello everyone,

    For the following absolute value equations, I have no trouble solving them and finding the valid [tex] x [/tex] solutions by plugging all the x solutions into the original equation.

    However, I am just wondering if could someone please show me how to set restrictions for the following equations? I want to know how to solve these equations in two ways instead of just one.

    Thank you very much.


    1. [tex] | 2|x + 3| - 5 | = 7 [/tex]

    2. [tex] | x - | 2x + 1 || = 3 [/tex]
  2. jcsd
  3. Jan 10, 2008 #2


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    The first one isn't an "absolute value" equation at all since the variable, x, does not appear in the absolute vaues: |2|= 2 and |-5|= -5 so that equation is just 2x+ 15= 7.

    For the second one, |2x+ 1| "changes" when 2x+ 1= 0 or x= -1/2. Now separate it into cases:

    If x< -1/2, then 2x+ 1< 0 so |2x+1|= -(2x+1)= -2x-1. Your equation is now |x+ 2x+ 1|= |3x+1|= 3. |3x+1| "changes" when 3x+ 1= 0 or x= -1/3. Since -1/3< -1/2, x< -1/2 immediately gives x< -1/3 so we have -(3x-1)= -3x-1= 3. Solve that equation. If the solution is less than -1/2, that is a solution to the original problem. If not, this gives no solution.

    If x>= -1/2, then 2x+ 1>= 0 so |2x+1|= 2x+ 1. Your equation is now |x- 2x- 1|= |-3x-1|= 3. |-3x-1| "changes" when -3x- 1= 0 or x= -1/3. Now we have two possibilities: -1/2<= x< -1/3 or -1/3<= x.
    If -1/2<= x< -1/3, then |-3x-1|= -(-3x-1)= 3x+ 1 and the equation becomes 3x+ 1= 3. Solve that for x and check if the answer is between -1/2 and -1/3. If it is, that is a solution to the original equation, if not, this gives no solution.

    Finally, if x>= -1/3 then |-3x-1|= -3x-1 and the equation become -3x-1= 3. Solve that for x and check if the answer is >= -1/3.
  4. Jan 11, 2008 #3
    halls i think you misinterpreted the first equation, it's a little ambiguous. he probably meant |2(|x+3|)-5|.
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