Question

How do I solve this absolute value equation for x:

|x-5| + |2-2x| = 11

Answer 1

|x-5|=11-|2-2x|
\[11-|2-2x|=
\begin{cases}
x-5, &\text{if }x\geq5\\
5-x &\text{if }x<5
\end{cases}\]
Now separately solve 11-|2-2x|=x-5 and 11-|2-2x|=5-x and determine whether the solutions are in their correct intervals.

Answer 2

There are 4 cases, two cases for | x -5| and two cases for | 2 - 2x| :
( x - 5 )+ ( 2 - 2x) = 11
( x - 5 ) - ( 2 - 2x) = 11

- ( x - 5) + ( 2 - 2x) = 11
- ( x - 5) - ( 2 - 2x) = 11

Answer 3

@blockcolder Just expand your instruction, hopefully the asker won't be confused with absolute value :)

Answer 4

You can also consider 3 cases.
\[ x \le 1 \\
1 \le x \le 5\\
\ 5 \le x

\]

Answer 5

For the first case
\[
|x-5|=11-|2-2x| \\
-x +5 = 11 -(2-2x)\\
-x+5 = 11 -2 + 2x\\
-3 x =4 \\
x= -\frac 4 3
\]
Do the remaining two cases in a similar way/

Answer 6

Thank you for your replies. I did as you suggest and get four solutions:

A. ( x - 5 )+ ( 2 - 2x) = 11, giving x = 14, which does not satisfy the equation.

B. ( x - 5 ) - ( 2 - 2x) = 11, giving x = 6, which satisfies the equation.

C. - ( x - 5) + ( 2 - 2x) = 11, giving x = -4/3, which satisfies the equation

D. - ( x - 5) - ( 2 - 2x) = 11, giving x = 8, which does not satisfy the equation.

Am I missing something?

Answer 7

For the second case
\[
|x-5|=11-|2-2x| \\
-x +5 = 11 -(-2+2x)\\
-x+5 = 11+2 -2x\\
x =8 \\

\]
But 8 is not in the interval [1,5] so this solution is to be rejected.

Answer 8

For the third case
\[|x-5|=11-|2-2x| \\
x -5 = 11 -(-2+2x)\\
x-5 = 11+2 -2x\\
3x =18 \\
x = 6
\]
So you are right. But, you only needed to consider three cases and not 4

Answer 9

Thanks, but am still slightly confused. How did you derive the three cases x<1, x between 1 and 5, and x >5 in your reply above? Also, what about the case x = 14? Both 8 and 14 are greater than 5. Why do you exclude them on the grounds that 8 and 14 are not in the interval [1,5]?

Answer 10

you choose 1, because (2 -2x) changes sign at at x =1.
you choose 5, because (x-5) changes sign at at x =5.

Since |u|= u if u >=0
and
|u| =-u if u<=0.

Answer 11

thank you so much.

Answer 12

yw