Calculate absolute values with unknown constant

Question

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?

Answer 1

HINT: You can assume three cases:

• $x \geq \dfrac 32$
• $-1<x< \dfrac 32$
• $x \leq -1$

Answer 2

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.

Answer 3

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
$I:$
$-(2x+2) + (3-2x) = x+14$
$-2x -2 + 3 -2x = x+14$
$-4x+1 = x+14$
$5x=-13$
$x=-13/5 \in I\checkmark$
$II:$
$(2x+2) + (3-2x) = x+14$
$+2x +2 + 3 -2x = x+14$
$5 = x+14$
$x=-9$
$x=-9 \notin II$
$III:$
$(2x+2)+ [-(3-2x)] = x+14$
$2x+2-3 +2x = x+14$
$4x-1 = x+14$
$3x=15$
$x=5 \in III\checkmark$

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

Answer 4

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 $-\frac{13}{5}$.