Solving $3|2-x| + |2x-4| = 8$ I'm not sure how to solve $3|2-x| + |2x-4| = 8$. I don't exactly need an answer, the method is what I'm interested in. I've done equations with one absolute value element. >_<
 A: $ 3|2-x| + |2x-4| = 8 \iff 5|x-2|=8 \iff 5(x-2)= \pm 8.$
Can you proceed ?
A: Hint:
As for real $a,|a|=|-a|$
$|2(x-2)|=2|x-2|=2|2-x|$
Finally if $|y|=b\ge0,y=\pm b$ for real $y$
If $y$ is complex, we can set $y=p+iq$ where $p,q$ are real
If $|y|=b,\sqrt{p^2+q^2}=b$
A: Well, it THIS case,  $|2x-4| = 2\cdot |x-2|= 2\cdot |2-x|$ so $3|2-x| + |2x-4| = 3|2-x|+2|2-x| = 5|2-x| =8$ so 
$|2-x| =|x-2|= \frac 85$ and so $x-2 =\pm \frac 85$ and $x = \frac 85 + 2=\frac {18}5=3\frac 35$ or $x = 2-\frac 85 = \frac 25$.
But in general do cases.
Do cases:
$2-x \ge 0$ if $x \le 2$ and $2-x < 0$ if $x > 2$.
And $2x-4 \ge 0$ if $x \ge 2$ and $2x-4 < 0$ if $x < 2$.
So do cases:
Case 1:  $x < 2$ and $2-x >0$ so $|2-x| = 2-x$ and $2x-4< 0$ so $|2x-4| = 4-2x$ and we have
$3|2-x| + |4-2x| = 8$
$3(2-x) + (4-2x) = 8$
$10 - 5x = 8$ 
$5x =2$ and $x =\frac 25$.
Case 2:  $x = 2$ and $2-x=0$ and $2x-4=0$ and so we have
$3|2-x| + |4-2x| = 8$
$3*0 + 0 = 8$
$0 = 8$ which is impossible.
Case 3: $x > 2$ and $2-x < 0$ so $|2-x| = x-2$ and $2x -4 > 0$ so $|2x-4| = 2x -4$
So 
$3|2-x| + |2x-4| = 8$
$3(x-2) + (2x-4) = 8$
$5x - 10 = 8$
$x = \frac {18}5 = 3\frac 35$.
