Lesson HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 3

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<H2>How to solve equations containing Linear Terms under the Absolute Value sign. Lesson 3</H2>

In this lesson you will see how to solve equations containing Linear Terms under the Absolute Value sign.
We consider, one after the other, the solution procedure for the following examples:

a) &nbsp;{{{x}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= {{{abs(x+1)}}} - {{{abs(x-1)}}};
b) &nbsp;{{{x+1}}} &nbsp;= {{{abs(x+1)}}} - {{{abs(x-1)}}};
c) &nbsp;{{{3x-4}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}};
d) &nbsp;{{{2x}}} &nbsp;&nbsp;&nbsp;&nbsp;= {{{abs(x+1)}}} - {{{abs(x-1)}}};
e) &nbsp;{{{2}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= {{{abs(x+1)}}} - {{{abs(x-1)}}};
f) &nbsp;{{{3.5}}} &nbsp;&nbsp;= {{{abs(x+1)}}} - {{{abs(x-1)}}}.

The starting lessons in this series are  
&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=http://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-1.lesson>How to solve equations containing Linear Terms under the Absolute Value sign. Lesson 1</A> 
&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=http://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-2.lesson>How to solve equations containing Linear Terms under the Absolute Value sign. Lesson 2</A> 
where simple cases are considered.

The goal is to get you acquainted with the different typical situations you can meet solving this kind of problems. 
When you learn how to solve these equations, you will easily proceed in many other cases.

<B>Problem 1</B>

Solve absolute value equation &nbsp;a) &nbsp;{{{x}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= {{{abs(x+1)}}} -{{{abs(x-1)}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1);
Solve absolute value equation &nbsp;b) &nbsp;{{{x+1}}} &nbsp;&nbsp;= {{{abs(x+1)}}} -{{{abs(x-1)}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2);
Solve absolute value equation &nbsp;c) &nbsp;{{{3x-4}}}  = {{{abs(x+1)}}} -{{{abs(x-1)}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3);
Solve absolute value equation &nbsp;d) &nbsp;{{{2x}}} &nbsp;&nbsp;&nbsp;&nbsp;= {{{abs(x+1)}}} -{{{abs(x-1)}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(4);
Solve absolute value equation &nbsp;e) &nbsp;{{{2}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= {{{abs(x+1)}}} -{{{abs(x-1)}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5);
Solve absolute value equation &nbsp;f) &nbsp;{{{3.5}}} &nbsp;&nbsp;&nbsp;= {{{abs(x+1)}}} -{{{abs(x-1)}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(6).

<B>Solution to &nbsp;1a)</B>

We have two groups of the terms under the Modulus sign. The first group is &nbsp;{{{abs(x+1)}}}, &nbsp;the second group is &nbsp;{{{abs(x-1)}}}.

For the first group, according to the definition of Absolute Value,

&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(x+1)}}} = &nbsp;&nbsp;&nbsp;{{{x+1}}}, &nbsp;&nbsp;&nbsp;if &nbsp;{{{x+1}}} >= 0, &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(x+1)}}} = &nbsp;{{{-(x+1)}}}, &nbsp;if &nbsp;{{{x+1}}} < 0.

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So, the Modulus &nbsp;{{{abs(x+1)}}}&nbsp; is the linear function  &nbsp;{{{x+1}}}&nbsp; over the range&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
{{{x}}} >= {{{-1}}}, &nbsp;and the linear function &nbsp;{{{-(x+1)}}}&nbsp; over the range &nbsp;{{{x}}} < {{{-1}}}.

Similarly, the Modulus &nbsp;{{{abs(x-1)}}}&nbsp; is the linear function  &nbsp;{{{x-1}}}&nbsp; over the range 
{{{x}}} >= {{{1}}}, &nbsp;and the linear function &nbsp;{{{-(x-1)}}}&nbsp; over the range &nbsp;{{{x}}} < {{{1}}}.

Now, we can divide the entire set of real numbers into adjacent ranges 
{{{x<=-1}}}, &nbsp;{{{-1<x<=1}}} &nbsp;and&nbsp; {{{1<x}}}&nbsp; and combine the respective linear 
functions over these ranges.

In this way, you will get

{{{abs(x+1)}}} - {{{abs(x-1)}}} = {{{-(x+1)}}} - [{{{-(x-1)}}}] = {{{-2}}} &nbsp;over &nbsp;{{{x}}} <= {{{-1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;&nbsp;&nbsp;&nbsp;{{{x+1)}}} - [{{{-(x-1)}}}] &nbsp;= {{{2x}}} &nbsp;over &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{-1<x<=1}}}, &nbsp;and&nbsp;

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;&nbsp;&nbsp;&nbsp;{{{x+1)}}} - {{{(x-1)}}} &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=  &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{1<x}}}.

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<B>Figure 1a</B>. &nbsp;&nbsp;Plot of the function&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
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line(  1.0,  2.0,   4.2,  2.0),
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red(line( -4.2, -4.2,   4.2,  4.2))
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&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1a'</B>. &nbsp;&nbsp;To the solution  
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
  </TD>
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The resulting plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; is shown in &nbsp;<B>Figure 1a</B>. &nbsp;It consists of three piecewise straight lines.

Thus, the equation &nbsp;{{{x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; which we should solve is equivalent to the three equations over the respective ranges:

{{{x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;which is &nbsp;{{{x}}} = {{{-2}}}&nbsp; over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(7)

{{{x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; 1 < {{{x}}} <= {{{1}}}, &nbsp;which is &nbsp;{{{x}}} = {{{2x}}} &nbsp;over &nbsp; {{{-1}}} < {{{x}}} <= {{{1}}}, &nbsp;&nbsp;&nbsp;&nbsp;(8)

{{{x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{1}}} < {{{x}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;which is {{{x}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp; {{{1}}} < {{{x}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(9)

Equation &nbsp;(7)&nbsp; has the solution &nbsp;{{{x}}} = {{{-2}}}. &nbsp;This solution belongs to the range &nbsp;{{{x}}} <= {{{-1}}} &nbsp;and satisfies all requirements. 
Equation &nbsp;(8)&nbsp; has the solution &nbsp;{{{x}}} = &nbsp;&nbsp;{{{0}}}. &nbsp;This solution belongs to the range &nbsp;{{{-1}}} < {{{x}}} <= {{{1}}} &nbsp;and satisfies all requirements. 
Equation &nbsp;(9)&nbsp; has the solution &nbsp;{{{x}}} = &nbsp;&nbsp;{{{2}}}. &nbsp;This solution belongs to the range &nbsp;{{{1}}} < {{{x}}} &nbsp;and satisfies all requirements.

So, the equation &nbsp;(1)&nbsp; has  three solutions &nbsp;{{{x}}} = {{{-2}}}, &nbsp;{{{x}}} = {{{0}}} &nbsp;and&nbsp; {{{x}}} = {{{2}}}.

In &nbsp;<B>Figure 1a'</B>&nbsp; the plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} is presented by the black lines. &nbsp;The red line represents the linear function {{{y}}} = {{{x}}}. 
The three intersection points represent the three solutions we just found.

<B>Solution &nbsp;1b)</B>

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We just derived in the solution to &nbsp;<B>Problem 1a)</B>&nbsp; above  that&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

{{{abs(x+1)}}} - {{{abs(x-1)}}} = {{{-2}}} &nbsp;over &nbsp;{{{x}}} <= {{{-1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;{{{2x}}} &nbsp;over &nbsp;{{{-1<x<=1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp;{{{1<x}}}.

Hence, the equation &nbsp;{{{x+1}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; which we should solve 
is equivalent to the three equations over the respective ranges:

{{{x+1}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;which is &nbsp;{{{x+1}}} = {{{-2}}}&nbsp; over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(10)

{{{x+1}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; 1 < {{{x}}} <= {{{1}}}, &nbsp;which is &nbsp;{{{x+1}}} = {{{2x}}} &nbsp;over &nbsp; {{{-1}}} < {{{x}}} <= {{{1}}}, &nbsp;&nbsp;&nbsp;&nbsp;(11)

{{{x+1}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{1}}} < {{{x}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;which is {{{x+1}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp; {{{1}}} < {{{x}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(12)

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red(line( -4.2, -3.2,   4.2,  5.2))
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&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1b</B>. &nbsp;&nbsp;To the solution  
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{x+1}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
  </TD>
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Equation &nbsp;(10)&nbsp; has the solution &nbsp;{{{x}}} = {{{-3}}}. &nbsp;This solution belongs to the range &nbsp;{{{x}}} <= {{{-1}}} &nbsp;and satisfies all requirements. 
Equation &nbsp;(11)&nbsp; has the solution &nbsp;{{{x}}} = &nbsp;&nbsp;&nbsp;{{{1}}}. &nbsp;This solution belongs to the range &nbsp;{{{-1}}} < {{{x}}} <= {{{1}}}, &nbsp;and satisfies all requirements. 
Equation &nbsp;(12)&nbsp; has the root &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{x}}} = &nbsp;&nbsp;&nbsp;&nbsp;{{{1}}}. &nbsp;This root does not belong to the range &nbsp;{{{1}}} < {{{x}}}, &nbsp;hence, does not satisfy all requirements.

So, the equation &nbsp;(2)&nbsp; has  two solutions &nbsp;{{{x}}} = {{{-3}}} &nbsp;and&nbsp; {{{x}}} = {{{1}}}.

In &nbsp;<B>Figure 1b</B>&nbsp; the plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} is presented by the black lines. &nbsp;The red line represents the linear function {{{y}}} = {{{x+1}}}. 
The two intersection points represent the two solutions we just found.

<B>Solution &nbsp;1c)</B>

{{{abs(x+1)}}} - {{{abs(x-1)}}} = {{{-2}}} &nbsp;over &nbsp;{{{x}}} <= {{{-1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;{{{2x}}} &nbsp;over &nbsp;{{{-1<x<=1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp;{{{1<x}}}.

Hence, the equation &nbsp;{{{3x-4}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; which we should solve 
is equivalent to the three equations over the respective ranges:

{{{3x-4}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;which is &nbsp;{{{3x-4}}} = {{{-2}}}&nbsp; over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(13)

{{{3x-4}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; 1 < {{{x}}} <= {{{1}}}, &nbsp;which is &nbsp;{{{3x-4}}} = {{{2x}}} &nbsp;over &nbsp; {{{-1}}} < {{{x}}} <= {{{1}}}, &nbsp;&nbsp;&nbsp;&nbsp;(14)

{{{3x-4}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{1}}} < {{{x}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;which is {{{3x-4}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp; {{{1}}} < {{{x}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(15)

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            line(  1.0,  1.97,  4.2,  1.97),

red(line( -4.2, -16.6,  4.2,  8.6))
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1c</B>. &nbsp;&nbsp;To the solution  
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{3x-4}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
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Equation &nbsp;(13)&nbsp; has the root &nbsp;{{{x}}} = {{{2/3}}}. &nbsp;This root does not belong to the range &nbsp;{{{x}}} <= {{{-1}}}, &nbsp;and, hence, does not satisfy all requirements. 
Equation &nbsp;(14)&nbsp; has the root &nbsp;{{{x}}} = {{{4}}}. &nbsp;This root does not belong to the range &nbsp;{{{-1}}} < {{{x}}} <= {{{1}}}, &nbsp;and, hence, does not satisfy all requirements. 
Equation &nbsp;(15)&nbsp; has the root &nbsp;{{{x}}} = {{{2}}}. &nbsp;This root belongs to the range &nbsp;{{{1}}} < {{{x}}}, &nbsp;and, hence, satisfy all requirements.

So, the equation &nbsp;(3)&nbsp; has  the unique solution &nbsp;{{{x}}} = {{{2}}}.

In &nbsp;<B>Figure 1c</B>&nbsp; the plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} is presented by the black lines. &nbsp;The red line represents the linear function {{{y}}} = {{{3x-4}}}. 
The two intersection points represent the two solutions we just found.

<B>Solution &nbsp;1d)</B>

{{{abs(x+1)}}} - {{{abs(x-1)}}} = {{{-2}}} &nbsp;over &nbsp;{{{x}}} <= {{{-1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;{{{2x}}} &nbsp;over &nbsp;{{{-1<x<=1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp;{{{1<x}}}.

Hence, the equation &nbsp;{{{2x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; which we should solve 
is equivalent to the three equations over the respective ranges:

{{{2x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;which is &nbsp;{{{2x}}} = {{{-2}}}&nbsp; over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(16)

{{{2x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; 1 < {{{x}}} <= {{{1}}}, &nbsp;which is &nbsp;{{{2x}}} = {{{2x}}} &nbsp;over &nbsp; {{{-1}}} < {{{x}}} <= {{{1}}}, &nbsp;&nbsp;&nbsp;&nbsp;(17)

{{{2x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{1}}} < {{{x}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;which is {{{2x}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp; {{{1}}} < {{{x}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(18)

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red(line( -4.2, -8.3,   4.2,  8.5))
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&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1d</B>. &nbsp;&nbsp;To the solution  
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{2x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
  </TD>
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</TABLE>
Equation &nbsp;(16)&nbsp; has the root &nbsp;{{{x}}} = {{{-1}}}. &nbsp;This root belongs to the range &nbsp;{{{x}}} <= {{{-1}}}, &nbsp;and, hence, satisfies all requirements. 
Equation &nbsp;(17)&nbsp; is the identity. &nbsp;Every real number &nbsp;{{{x}}}&nbsp; of the range &nbsp;{{{-1}}} < {{{x}}} <= {{{1}}}&nbsp; is the solution. &nbsp;So, the equation &nbsp;(17)&nbsp; has infinitely many solutions.  
Equation &nbsp;(18)&nbsp; has the root &nbsp;{{{x}}} = {{{1}}}. &nbsp;This root does not belong to the range &nbsp;{{{1}}} < {{{x}}}, &nbsp;and, hence, does not satisfy all requirements.

So, the equation &nbsp;(4)&nbsp; has  infinitely many solutions.

In &nbsp;<B>Figure 1d</B>&nbsp; the plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} is presented by the black lines. &nbsp;The red line represents the linear function {{{y}}} = {{{2x}}}. 
The straight line {{{y}}} = {{{2x}}} coincides with the plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; over the range &nbsp;{{{-1}}} <= {{{x}}} <= {{{1}}}.

<B>Solution &nbsp;1e)</B>

{{{abs(x+1)}}} - {{{abs(x-1)}}} = {{{-2}}} &nbsp;over &nbsp;{{{x}}} <= {{{-1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;{{{2x}}} &nbsp;over &nbsp;{{{-1<x<=1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp;{{{1<x}}}.

Hence, the equation &nbsp;{{{2x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; which we should solve 
is equivalent to the three equations over the respective ranges:

{{{2}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;which is &nbsp;{{{2}}} = {{{-2}}}&nbsp; over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(19)

{{{2}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; 1 < {{{x}}} <= {{{1}}}, &nbsp;which is &nbsp;{{{2}}} = {{{2x}}} &nbsp;over &nbsp; {{{-1}}} < {{{x}}} <= {{{1}}}, &nbsp;&nbsp;&nbsp;&nbsp;(20)

{{{2}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{1}}} < {{{x}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;which is {{{2}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp; {{{1}}} < {{{x}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(21)

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&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1e</B>. &nbsp;&nbsp;To the solution  
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{2}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
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Equation &nbsp;(19)&nbsp; has no solutions.

Equation &nbsp;(20)&nbsp; has the root &nbsp;{{{x}}} = {{{1}}}. &nbsp;This root belongs to the range &nbsp;{{{-1}}} < {{{x}}} <= {{{1}}}&nbsp; and, hence, satisfies all requirements.

Equation &nbsp;(21)&nbsp; is the identity. &nbsp;Every real number &nbsp;{{{x}}}&nbsp; of the range &nbsp;{{{1}}} < {{{x}}}&nbsp; is the solution. &nbsp;So, the equation &nbsp;(21)&nbsp; has infinitely many solutions.

Thus, the equation &nbsp;(5)&nbsp; has  infinitely many solutions.

In &nbsp;<B>Figure 1e</B>&nbsp; the plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} is presented by the black lines. &nbsp;The red line represents the linear function {{{y}}} = {{{2}}}. 
The straight line {{{y}}} = {{{2}}} coincides with the plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; over the range &nbsp;{{{1}}} <= {{{x}}}.

<B>Solution &nbsp;1f)</B>

{{{abs(x+1)}}} - {{{abs(x-1)}}} = {{{-2}}} &nbsp;over &nbsp;{{{x}}} <= {{{-1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;{{{2x}}} &nbsp;over &nbsp;{{{-1<x<=1}}},

{{{abs(x+1)}}} - {{{abs(x-1)}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp;{{{1<x}}}.

Hence, the equation &nbsp;{{{3.5}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; which we should solve 
is equivalent to the three equations over the respective ranges:

{{{3.5}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;which is &nbsp;{{{3.5}}} = {{{-2}}}&nbsp; over&nbsp; {{{x}}} <= {{{-1}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(22)

{{{3.5}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; 1 < {{{x}}} <= {{{1}}}, &nbsp;which is &nbsp;{{{3.5}}} = {{{2x}}} &nbsp;over &nbsp; {{{-1}}} < {{{x}}} <= {{{1}}}, &nbsp;&nbsp;&nbsp;&nbsp;(23)

{{{3.5}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} &nbsp;over&nbsp; {{{1}}} < {{{x}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;which is {{{3.5}}} = &nbsp;&nbsp;{{{2}}} &nbsp;over &nbsp; {{{1}}} < {{{x}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(24)

</TD>
  <TD> 
{{{drawing( 250, 250, -4.5, 4.5, -3.5, 5.5,
            grid(1),
            line (-4.2, -2.0,  -1.0, -2.0),
            line (-4.2, -1.98, -1.0, -1.98),

line (-1.0, -2.0,   1.0,  2.0),

line(  1.0,  2.0,   4.2,  2.0),
            line(  1.0,  1.97,  4.2,  1.97),

red(line( -4.2,  3.5,   4.2,  3.5))
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1f</B>. &nbsp;&nbsp;To the solution  
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{3.5}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
  </TD>
  </TR>
</TABLE>
Equation &nbsp;(22)&nbsp; has no solutions.

Equation &nbsp;(23)&nbsp; has no solutions.

Equation &nbsp;(24)&nbsp; has no solutions.

Thus, the equation &nbsp;(6)&nbsp; has no solutions.

In &nbsp;<B>Figure 1f</B>&nbsp; the plot of the function &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}} is presented by the black lines. &nbsp;The red line represents the constant function {{{y}}} = {{{3.5}}}.

There is no intersection point of the plot  &nbsp;{{{y}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp; and the plot &nbsp;{{{y}}} = {{{3.5}}}.

Below the plots of the &nbsp;<B>Figure 1a</B> - <B>Figure 1f</B>&nbsp; are presented again for your convenience.

<TABLE> 
  <TR>
  <TD> 
{{{drawing( 250, 250, -4.5, 4.5, -3.5, 5.5,
            grid(1),
            line (-4.2, -2.0,  -1.0, -2.0),
            line (-4.2, -1.98, -1.0, -1.98),

line (-1.0, -2.0,   1.0,  2.0),

line(  1.0,  2.0,   4.2,  2.0),
            line(  1.0,  1.97,  4.2,  1.97),

red(line( -4.2, -4.2,   4.2,  4.2))
)}}}

<B>Figure 1a</B>. To the solution &nbsp;{{{x}}} = {{{abs(x+1)-abs(x-1)}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(three roots)
  </TD>
   <TD>
{{{drawing( 250, 250, -4.5, 4.5, -3.5, 5.5,
            grid(1),
            line (-4.2, -2.0,  -1.0, -2.0),
            line (-4.2, -1.98, -1.0, -1.98),

line (-1.0, -2.0,   1.0,  2.0),

line(  1.0,  2.0,   4.2,  2.0),
            line(  1.0,  1.97,  4.2,  1.97),

red(line( -4.2, -3.2,   4.2,  5.2))
)}}}

<B>Figure 1b</B>. To the solution &nbsp;{{{x+1}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp;&nbsp;&nbsp;&nbsp; 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(two roots)
  </TD>
   <TD>
{{{drawing( 250, 250, -4.5, 4.5, -3.5, 5.5,
            grid(1),
            line (-4.2, -2.0,  -1.0, -2.0),
            line (-4.2, -1.98, -1.0, -1.98),

line (-1.0, -2.0,   1.0,  2.0),

line(  1.0,  2.0,   4.2,  2.0),
            line(  1.0,  1.97,  4.2,  1.97),

red(line( -4.2, -16.6,  4.2,  8.6))
)}}}

<B>Figure 1c</B>. To the solution  &nbsp;{{{3x-4}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(one root)
  </TD>
  </TR>
</TABLE><TABLE> 
  <TR>
  <TD> 
{{{drawing( 250, 250, -4.5, 4.5, -3.5, 5.5,
            grid(1),
            line (-4.2, -2.0,  -1.0, -2.0),
            line (-4.2, -1.98, -1.0, -1.98),

line (-1.0, -2.0,   1.0,  2.0),

line(  1.0,  2.0,   4.2,  2.0),
            line(  1.0,  1.97,  4.2,  1.97),

red(line( -4.2, -8.3,   4.2,  8.5))
)}}}

<B>Figure 1d</B>. To the solution  &nbsp;{{{2x}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(infinitely many roots)
  </TD>
   <TD>
{{{drawing( 250, 250, -4.5, 4.5, -3.5, 5.5,
            grid(1),
            line (-4.2, -2.0,  -1.0, -2.0),
            line (-4.2, -1.98, -1.0, -1.98),

line (-1.0, -2.0,   1.0,  2.0),

line(  1.0,  2.0,   4.2,  2.0),
            line(  1.0,  1.97,  4.2,  1.97),

red(line( -4.2,  2.03,  4.2,  2.03))
)}}}

<B>Figure 1e</B>. To the solution  &nbsp;{{{2}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(infinitely many roots)
  </TD>
   <TD>
{{{drawing( 250, 250, -4.5, 4.5, -3.5, 5.5,
            grid(1),
            line (-4.2, -2.0,  -1.0, -2.0),
            line (-4.2, -1.98, -1.0, -1.98),

line (-1.0, -2.0,   1.0,  2.0),

line(  1.0,  2.0,   4.2,  2.0),
            line(  1.0,  1.97,  4.2,  1.97),

red(line( -4.2,  3.5,   4.2,  3.5))
)}}}

<B>Figure 1f</B>. To the solution &nbsp;{{{3.5}}} = {{{abs(x+1)}}} - {{{abs(x-1)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(no roots)
  </TD>
  </TR>
</TABLE>

Note that the strategy of the solution is the same in all examples. &nbsp;It is to break up the entire set of real numbers into sub-domains (ranges) 
where the absolute value of linear term is a linear function, &nbsp;and then to solve the corresponding linear equations in each sub-domain (range).

My other lessons on &nbsp;<B>Absolute Value equations</B>&nbsp; in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/-Absolute-Value-equations.lesson>Absolute Value equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-1.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 1</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-2.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 2</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-eqns-containing-Lin-Terms-under-Abs-value-L4.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 4</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Quadratic-Terms-under-Abs-Value-sign-L-1.lesson>HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 1</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Quadratic-Terms-under-Abs-Value-sign-L-2.lesson>HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2</A>

See also the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/absolute-value/Review-of-lessons-on-Absolute-Value-equations.lesson>OVERVIEW of lessons on Absolute Value equations</A>&nbsp; under the current topic in this site.

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-I.