1. ## Absolute Values

all x ∈ R that satisfy |x+1|-2|x-4|= 3

help?

2. In this problem, use cases.
$\begin{array}{l}
x \in ( - \infty , - 1]\quad \Rightarrow \quad \left( { - x - 1} \right) - 2\left( { - x + 4} \right) = 3 \\
x \in ( - 1,4)\quad \Rightarrow \quad \left( {x + 1} \right) - 2\left( { - x + 4} \right) = 3 \\
x \in [4,\infty )\quad \Rightarrow \quad (x + 1) - 2(x - 4) = 3 \\
\end{array}$

Now solve each of the three cases.
But make sure the solution fits the case.

3. Originally Posted by Plato
In this problem, use cases.
$\begin{array}{l}
x \in ( - \infty , - 1]\quad \Rightarrow \quad \left( { - x - 1} \right) - 2\left( { - x + 4} \right) = 3 \\
x \in ( - 1,4)\quad \Rightarrow \quad \left( {x + 1} \right) - 2\left( { - x + 4} \right) = 3 \\
x \in [4,\infty )\quad \Rightarrow \quad (x + 1) - 2(x - 4) = 3 \\
\end{array}$

Now solve each of the three cases.
But make sure the solution fits the case.

Thanks, but how do we know this? I mean, how do we know that it's -infintiy, -1, and -1,4 etc?

4. Originally Posted by weasley74
Thanks, but how do we know this? I mean, how do we know that it's -infintiy, -1, and -1,4 etc?
Well you are studying absolute value.
Are you not expected to understand how it all works?
You must have been given examples.

5. Originally Posted by Plato
$\begin{array}{l}
x \in ( - \infty , - 1]\\
x \in ( - 1,4)\\
x \in [4,\infty )\\\
\end{array}$
is it supposed to be ( and not ] ?

6. Originally Posted by weasley74
is it supposed to be ( and not ] ?
Because $
\Re = ( - \infty , - 1] \cup ( - 1,4) \cup [4,\infty )$
we have considered every real number.
That is the whole point. But there are other ways this can be done.

7. Originally Posted by Plato
In this problem, use cases.
$\begin{array}{l}
x \in ( - \infty , - 1]\quad \Rightarrow \quad \left( { - x - 1} \right) - 2\left( { - x + 4} \right) = 3 \\
x \in ( - 1,4)\quad \Rightarrow \quad \left( {x + 1} \right) - 2\left( { - x + 4} \right) = 3 \\
x \in [4,\infty )\quad \Rightarrow \quad (x + 1) - 2(x - 4) = 3 \\
\end{array}$

Now solve each of the three cases.
But make sure the solution fits the case.
so for the first one, for which x=12, the solution doesn't fit the case?
the second one, x=10/3, fits and the same for the third case when x=6 ?

8. Originally Posted by weasley74
so for the first one, for which x=12, the solution doesn't fit the case?
the second one, x=10/3, fits and the same for the third case when x=6 ?
Seems correct

Here's the graph for $y=|x+1| - 2 |x-4|-3$, do you see where the lines crosses the x axis?