Absolute Value Equations: Setting Restrictions

[vertciel]

Hello everyone,

For the following absolute value equations, I have no trouble solving them and finding the valid $$x$$ 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.

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1. $$| 2|x + 3| - 5 | = 7$$

2. $$| x - | 2x + 1 || = 3$$

 

[HallsofIvy]

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.

 

[matticus]

halls i think you misinterpreted the first equation, it's a little ambiguous. he probably meant |2(|x+3|)-5|.