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Solve |x - 1| - 2|x| = 3|x + 1|.

 May 26, 2020
 #1
avatar+25237 
+3

Solve

\(|x - 1| - 2|x| = 3|x + 1|\).

 

\(\begin{array}{|rcll|} \hline \text{zero in }~ |x - 1|: \\ \hline x - 1 &=& 0 \\ \hline \mathbf{x} &=&\mathbf{ 1 } \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{zero in }~ |x|: \\ \hline \mathbf{x} &=&\mathbf{0} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{zero in }~ |x + 1|: \\ \hline \hline x + 1 &=& 0 \\ \hline \mathbf{x} &=&\mathbf{ -1 } \\ \hline \end{array}\)

 

We create a sign table:

\(\begin{array}{|l|c|c\|\|c|c|c|c|c|} \hline \text{Interval or position} : & \left(-\infty,-1\right) & -1 & \left(-1,0\right) & 0 & \left(0,1\right) & 1 & \left(1,\infty\right) \\ \hline \text{sign of } (x+1): & - & 0 & + & + & + & + & + \\ \hline \text{sign of } (x): & - & - & - & 0 & +& +& + \\ \hline \text{sign of } (x-1): &- & - & - & - & - & 0 & + \\ \hline \end{array} \)

 

For the 4 cases  \((-\infty, -1] ,\ (-1 , 0] ,\ (0, 1] ,\ (1, \infty)\)
we can therefore always resolve all amounts and solve the corresponding equation without amounts,
taking into account the respective case condition.
The negative terms from the amounts are highlighted in green:

 

\(\begin{array}{|rcll|} \hline (-\infty, -1] \\ \hline {\color{green}-(x - 1)} - 2({\color{green}-x}) &=& 3({\color{green}-(x + 1)}) \\ -x+1 +2x &=& -3(x + 1) \\ -x+1 +2x &=& -3x -3 \\ 4x &=& -4 \\ \mathbf{x} &=& \mathbf{-1}\ \checkmark \qquad -\infty < x \le -1 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline (-1 , 0] \\ \hline {\color{green}-(x - 1)} - 2({\color{green}-x}) &=& 3(x + 1) \\ -x+1 +2x &=& 3x+3 \\ 2x &=& -2 \\ x &=& -1 \quad \mathbf{\text{no}} \qquad -1 < x \le 0 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (0, 1] \\ \hline {\color{green}-(x - 1)} - 2(x) &=& 3(x + 1) \\ -x+1 -2x &=& 3x+3 \\ 6x &= & -2 \\ x &=& -\dfrac{1}{3} \quad \mathbf{\text{no}} \qquad 0 < x \le 1 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (1, \infty) \\ \hline (x - 1) - 2(x) &=& 3(x + 1) \\ x-1 -2x &=& 3x+3 \\ 4x &= & -4 \\ x &=& -1 \quad \mathbf{\text{no}} \qquad 1 < x < \infty \\ \hline \end{array}\)

 

\(\mathbf{x = -1}\)

 

laugh

 May 26, 2020
edited by heureka  May 26, 2020
 #2
avatar+30322 
+2

I prefer to draw a graph first, for these sort of questions:

 

 May 26, 2020

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