Tough Absolute Value Equation?

Lemon

New member
Joined
Sep 22, 2011
Messages
6
Hey guys need help with this

equation:

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

I should be able to do the rest, i just need to know how to do the first step

Please help will update with final answer :)

Thanks!
 
Hey guys need help with this

equation:

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

I should be able to do the rest, i just need to know how to do the first step

Please help will update with final answer :)

Thanks!

I suggest you learn how to draw modulus graphs. This problem can be solved by using the graphs of y = |x-5| and y = |2-2x| and seeing where they might add to give 11. When you have an idea of the lines involved to get 11, then you take either the positive value of the graphs, or the negative values, or a mix of the two.

For example, I did a graph and saw that they meet somewhere before the y-axis, which is where both graphs have a negative slope, that is, I have y = 5-x and y = 2-2x.

Putting that into the equation gives me the first answer.

5-x + 2-2x = 11

The other is somewhere after the x=coordinate 5, where both graphs have a positive slope.
 
equation:
|x-5| + |2-2x| = 11
If I were you, I would first rewrite it.
\(\displaystyle |x-5|+2|x-1|=11.\)
Now you have three areas to investigate.

On \(\displaystyle (\infty,1]\) solve \(\displaystyle -x+5-2x+2=11\).

On \(\displaystyle (1,5]\) solve \(\displaystyle -x+5+2x-2=11\).

On \(\displaystyle (5,\infty)\) solve \(\displaystyle x-5+2x-2=11\).

Intersect the solution sets.
 
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