Math Help for Section 3.7, Page 154

Solving
Equations Involving Absolute Value
If two algebraic expressions are equal in absolute value, they must
either be
equal to each other or be the opposites
of each other. So, you can
solve equations
of the form $|ax+b|=|cx+d|$ by forming the two linear equations

    $\color{red}\overbrace{{\color{black}
ax+b=cx+d}}
^{\text{Expressions equal}}$  
 and  
 $\color{red}\overbrace{{\color{black} ax+b=-(cx+d)}}
^{\text{Expressions opposite}}$.

Example
4: Check

$x=3$:

    $\eqalign{|3x-4|=&|7x-16|
&{\small\color{red}\quad\quad\text{Write original equation.}}
\cr |3({\color{red}3})-4|\overset{?}{=}&|7({\color{red}3})-16|
&{\small\color{red}\quad\quad\text{Substitute 3 for }x.} \cr
|9-4|\overset{?}{=}&|21-16|
&{\small\color{red}\quad\quad\text{Multiply.}}
\cr |5|\overset{?}{=}&|5|
&{\small\color{red}\quad\quad\text{Subtract.}} \cr 5=&5
&{\small\color{red}\quad\quad\text{Solution checks.
}\checkmark}
}$

$x=2$:

    $\eqalign{|3x-4|=&|7x-16|
&{\small\color{red}\quad\quad\text{Write original equation.}}
\cr |3({\color{red}2})-4|\overset{?}{=}&|7({\color{red}2})-16|
&{\small\color{red}\quad\quad\text{Substitute 2 for }x.} \cr
|6-4|\overset{?}{=}&|14-16|
&{\small\color{red}\quad\quad\text{Multiply.}}
\cr |2|\overset{?}{=}&|-2|
&{\small\color{red}\quad\quad\text{Subtract.}} \cr 2=&2
&{\small\color{red}\quad\quad\text{Solution checks.
}\checkmark}
}$

Example
5: Check

$x=-8$:

    $\eqalign{|x+5|=&|x+11|
&{\small\color{red}\quad\quad\text{Write original equation.}}
\cr |{\color{red}-8}+5|\overset{?}{=}&|{\color{red}-8}+11|
&{\small\color{red}\quad\quad\text{Substitute }-8\text{ for
}x.} \cr
|-3|\overset{?}{=}&|3|
&{\small\color{red}\quad\quad\text{Add.}}
\cr  3=&3
&{\small\color{red}\quad\quad\text{Solution checks.
}\checkmark}
}$