Solving $3|2-x| + |2x-4| = 8$

Question

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. >_<

Consider the cases $x <\frac 1 2, \frac 1 2 \leq x \leq 2, x>2$ separately. — Kavi Rama Murthy, Jan 14 '20 at 06:28
Well, the obvious way is to consider when the expressions within the parentheses change their signs. So break up the equation at $2-x = 0$ and $2x-4= 0$. This time, it turns out that they're the same point. — Matti P., Jan 14 '20 at 06:28
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Jan 14 '20 at 06:32
In this case, noticing $|2x-4|=2|x-2|$ simplifies things substantially. In connection with solving equations with absolute values, this question and other posts linked there might be useful. — Martin Sleziak, Jan 14 '20 at 06:57
BTW I am not really sure why this is tagged (inequality) , when the question is about solving an equation. — Martin Sleziak, Jan 14 '20 at 06:58
$|2-x|$ and $|4-2x|$ both equal $0$ at $x = 2.$ You have two cases to consider $x \le 2$ and $x > 2$ — Doug M, Jan 14 '20 at 07:26

score 3 · Answer 1 · answered Jan 14 '20 at 06:38

3

$ 3|2-x| + |2x-4| = 8 \iff 5|x-2|=8 \iff 5(x-2)= \pm 8.$

Can you proceed ?

answered Jan 14 '20 at 06:38

Fred

77,394

Almost, so this suggests |2-x| = |x-2| ? – once again Jan 14 '20 at 06:53
1

$|a-b| = |b-a|$. Always. – fleablood Jan 14 '20 at 07:25
Because $b-a = -(a -b)$ and $|M| = |-M|$. So $|a-b| = |-(b-a)| = |b-a|$. Always. – fleablood Jan 14 '20 at 07:27

lab bhattacharjee · Answer 2 · 2020-01-14T06:49:24.510

1

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 $|y|=b,\sqrt{p^2+q^2}=b$

edited Jan 14 '20 at 06:49

answered Jan 14 '20 at 06:27

lab bhattacharjee

274,582

1

Seriously, could the downvoter explain their action? That would help the non-specialist understand what's the problem in an answer from a high-rep user (my guess is $|a|=||-a$ ?) – Déjà vu Jan 14 '20 at 06:47
@RingØ, Sorry for the obvious typo. – lab bhattacharjee Jan 14 '20 at 06:50
I'm not the downvoter, but explaining a dv would help, my $0.02c$... – Déjà vu Jan 14 '20 at 06:51
@RingØ, Wish somebody take the pain to share the mistake – lab bhattacharjee Jan 14 '20 at 06:52
Well, I'd never down-vote this but I'd pretty sure its more abstract and advanced than the OP is prepared for. – fleablood Jan 14 '20 at 07:28

score 1 · Answer 3 · answered Jan 14 '20 at 07:24

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

$3|2-x| + |2x-4| = 8$

$3(x-2) + (2x-4) = 8$

$5x - 10 = 8$

$x = \frac {18}5 = 3\frac 35$.

Solving $3|2-x| + |2x-4| = 8$

3 Answers3