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How to solve complex absolute value inequalities like \[\left| x \right| + \left| {x - 2} \right| < 5\] ?

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Last updated date: 12th Jul 2024
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Answer
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Hint:
Here, we have to find the solution set for the given complex absolute value inequalities. We will use the absolute value of the function, to find the interval of the absolute value and the sign of the absolute value of the function. We will use the sign of the value obtained by the interval of the absolute value, we will find the solution set for the complex value inequalities. Thus, the solution for the complex absolute value inequalities is the required answer.

Complete step by step solution:
We are given with a complex absolute value inequality \[\left| x \right| + \left| {x - 2} \right| < 5\].
Now, by rewriting the complex absolute value inequalities, we get
\[ \Rightarrow \left| x \right| + \left| {x - 2} \right| - 5 < 0\]
Let us consider the given function \[f\left( x \right)\] .
\[f\left( x \right) = \left| x \right| + \left| {x - 2} \right| - 5\]
Since the given function is an absolute value, the value should always be a positive sign, we get
\[\begin{array}{l}x \ge 0\\x - 2 \ge 0\end{array}\]
Thus, by rewriting, we get
\[\begin{array}{l}x \ge 0\\x \ge 2\end{array}\]
Thus, \[x\] lies between \[0 < x < 2\] .
Now, by considering the interval from the number line, we get
\[ - \infty < x < 0\], \[0 < x < 2\] and \[2 < x < \infty \].
Now, by using the interval of \[x\] for the absolute value \[\left| x \right|\] , we get
\[\ln \left| x \right|\], \[x\] is negative in the interval \[\left( { - \infty ,0} \right)\]
\[\ln \left| x \right|\], \[x\] is positive in the interval \[\left( {0,2} \right)\]
\[\ln \left| x \right|\], \[x\] is positive in the interval \[\left( {2,\infty } \right)\]
Now, by using the interval of \[x\] for the absolute value \[\left| {x - 2} \right|\], we get
\[\ln \left| {x - 2} \right|\], \[x\] is negative in the interval \[\left( { - \infty ,0} \right)\]
\[\ln \left| {x - 2} \right|\], \[x\] is negative in the interval \[\left( {0,2} \right)\]
\[\ln \left| {x - 2} \right|\], \[x\] is positive in the interval \[\left( {2,\infty } \right)\]
By considering the function \[f\left( x \right)\] in the interval \[\left( { - \infty ,0} \right)\], we get
\[ - x - \left( {x - 2} \right) - 5 < 0\]
\[ \Rightarrow - x - x + 2 - 5 < 0\]
Simplifying the function, we get
\[ \Rightarrow - 2x - 3 < 0\]
Now, by rewriting the function, we get
\[ \Rightarrow - 2x < 3\]
\[ \Rightarrow x < - \dfrac{3}{2}\]
Now, by considering the function \[f\left( x \right)\] in the interval \[\left( {0,2} \right)\], we get
\[x - \left( {x - 2} \right) - 5 < 0\]
\[ \Rightarrow x - x + 2 - 5 < 0\]
Simplifying the function, we get
\[ \Rightarrow 0 - 3 < 0\]
There exists no solution in the interval \[\left( {0,2} \right)\].
Now, by considering the function \[f\left( x \right)\] in the interval \[\left( {0,\infty } \right)\], we get
\[x + x - 2 - 5 < 0\]
Now, by simplifying the function, we get
\[ \Rightarrow 2x - 7 < 0\]
Now, by rewriting the function, we get
\[ \Rightarrow 2x < 7\]
\[ \Rightarrow x < \dfrac{7}{2}\]
Thus, \[ - \dfrac{3}{2} < x < \dfrac{7}{2}\] for the given complex absolute value inequalities.

Therefore, the solution set \[ - \dfrac{3}{2} < x < \dfrac{7}{2}\] for the complex absolute value inequalities \[\left| x \right| + \left| {x - 2} \right| < 5\].

Note:
We know that the absolute value is defined as the non-negative integer without regard to its sign, it can be either positive or negative integer. We will follow these steps to solve an absolute value equality which contains two absolute values. We will write both the equations without the absolute values. The first set of equations will set the quantity inside the bars on the left side of the equation equal to the quantity inside the bars on the right side of the equation and then solve the linear equation formed. The second set of equations will set the quantity inside the bars on the left side of the equation opposite to the quantity inside the bars on the right side of the equation and then solve the linear equation formed. Thus, the solution set for the linear equation.