#### Answer

$x=-\dfrac{3}{2}$

#### Work Step by Step

The given equation, $
|x-1|=5x+10
,$ is equivalent to
\begin{array}{l}\require{cancel}
x-1=5x+10
\\\\\text{OR}\\\\
x-1=-(5x+10)
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-1=5x+10
\\
x-5x=10+1
\\
-4x=11
\\
\dfrac{-4x}{-4}=\dfrac{11}{-4}
\\
x=-\dfrac{11}{4}
\\\\\text{OR}\\\\
x-1=-(5x+10)
\\
x-1=-5x-10
\\
x+5x=-10+1
\\
6x=-9
\\
\dfrac{6x}{6}=-\dfrac{9}{6}
\\
x=-\dfrac{3}{2}
.\end{array}
Since the right side of the given equation is not a constant, then checking of solution/s is required.
Substituting $
x=-\dfrac{11}{4}
$ in the original equation results to
\begin{array}{l}\require{cancel}
|x-1|=5x+10
\\
\left|-\dfrac{11}{4}-1\right|=5\left(-\dfrac{11}{4}\right)+10
\\
\left|-\dfrac{11}{4}-\dfrac{4}{4}\right|=-\dfrac{55}{4}+10
\\
\left|-\dfrac{15}{4}\right|=-\dfrac{55}{4}+\dfrac{40}{4}
\\
\left|-\dfrac{15}{4}\right|=-\dfrac{15}{4}
\\
\dfrac{15}{4}=-\dfrac{15}{4}
\text{ (FALSE)}
.\end{array}
Since the substitution above resulted in a FALSE statement, then $
x=-\dfrac{11}{4}
,$ is not a solution (an extraneous solution).
Substituting $
x=-\dfrac{3}{2}
$ in the original equation results to
\begin{array}{l}\require{cancel}
|x-1|=5x+10
\\
\left|-\dfrac{3}{2}-1\right|=5\left(-\dfrac{3}{2}\right)+10
\\
\left|-\dfrac{3}{2}-\dfrac{2}{2}\right|=-\dfrac{15}{2}+10
\\
\left|-\dfrac{5}{2}\right|=-\dfrac{15}{2}+\dfrac{20}{2}
\\
\left|-\dfrac{5}{2}\right|=\dfrac{5}{2}
\\
\dfrac{5}{2}=\dfrac{5}{2}
\text{ (TRUE)}
.\end{array}
Hence, $
x=-\dfrac{3}{2}
.$