I am to calculate all $x$ if $f(x) = g(x)$ and if

$$f(x)= |2x+2| + |3-2x|$$ $$g(x)= x + 14$$

How do I mix regular numbers with absolute values in such a sense? I thought I could calculate it like this:

$$|2x+2| + |3-2x| = 3x+2 + 3+2x = 3x+5$$

But then I realised that the end value of either of the absolute values is determined by $x$. It feels like a catch 22: how do I calculate this equation without knowing the value of $x$ until I've calculated it?


HINT: You can assume three cases:

  • $x \geq \dfrac 32$
  • $-1<x< \dfrac 32$
  • $x \leq -1$
  • $\begingroup$ For $x\ge3/2$, the argument of the second absolute value, $3-2x$, definitely is not positive. $\endgroup$ – celtschk Aug 26 '12 at 8:31
  • $\begingroup$ @celtschk: I said the term comes out of the absolute like it is. $\endgroup$ – Gigili Aug 26 '12 at 8:35
  • 1
    $\begingroup$ Well, your edit doesn't mean the previous text is inaccessible. And your equations (now edited out together with the text) clearly show that you meant the argument being positive. $\endgroup$ – celtschk Aug 26 '12 at 8:42
  • $\begingroup$ @celtschk: That's why I edited it. I simply made a mistake there. $\endgroup$ – Gigili Aug 26 '12 at 9:25
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    $\begingroup$ @celtschk - I think Gigili's statement about absolute values being positive was perfectly clear, before his/her edit, even though it wasn't precise English for what he/she meant. Is it really compulsory for you to give him/her such a hard time about it? $\endgroup$ – user22805 Aug 26 '12 at 11:38

Think the definition of $\lvert x\rvert$: $$\lvert x\rvert = \cases{x & if $x\ge 0$\\-x & if $x<0$}$$ Now, in your case, you've got two absolute values with expressions in there. Of course those expressions are inserted into the definition, and give you a distinction of cases each. For each of the resulting cases, the absolute value is then replaced by the corresponding expression, and the equation can be solved normally, but of course only solutions which fulfil the conditions are valid for each case.


First, partition of range of $x$ into 3 parts:

  1. $x\geq\frac{3}{2}$, in this case, $f(x)=(2x+2)+(2x-3)=4x-1$. Then solve the eqation $4x-1=x+14$ and find that $x=5$.

  2. $-1\leq x\leq\frac{3}{2}$, then $f(x)=(2x+2)+(3-2x)=5$. Then you can find that the solution of $5=x+14$ is $x=9$, which out of the range $-1\leq x\leq\frac{3}{2}$. Therefore, this is not a solution.

  3. $x\leq-1$, then $f(x)=(-2x-2)+(3-2x)=1-4x$. The solution of $1-4x=x+14$ is $x=-\frac{13}{5}$, which is in the range.

Therefore, the solution of $f(x)=g(x)$ is $x=5$ and $-\frac{13}{5}$.


1. Find the roots of all |A| expressions(total of m):

x=-1, x=3/2  

2.Partition the $\mathcal{R}$ (if you want to solve in Real numbers set) into m+1 sub domains

I    x <- 1  
II   -1 =< x <= 3/2  
III  x>3/2 

3.Find the sign of |A| in each sub domain and solve for x
$-(2x+2) + (3-2x) = x+14$
$-2x -2 + 3 -2x = x+14$
$-4x+1 = x+14$
$x=-13/5 \in I\checkmark$
$(2x+2) + (3-2x) = x+14$
$+2x +2 + 3 -2x = x+14$
$5 = x+14$
$x=-9 \notin II$
$(2x+2)+ [-(3-2x)] = x+14$
$2x+2-3 +2x = x+14$
$4x-1 = x+14$
$x=5 \in III\checkmark$

NOTE: in each sub domain check if the answer is in its corresponding sub domain

  • $\begingroup$ I'm wondering about the down vote..... $\endgroup$ – Zeta.Investigator Aug 26 '12 at 17:14

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