#### Answer

$x=\dfrac{31}{14}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
0.9(2x-8)=4-(x+5)
,$ remove first the decimal numbers by multiplying both sides by a power of $10.$ Then use the properties of equality to isolate the variable. Do checking of the solution.
$\bf{\text{Solution Details:}}$
Since the largest number of decimal places a term has is $1,$ multiply both sides by $10.$ This results to
\begin{array}{l}\require{cancel}
10[0.9(2x-8)]=10[4-(x+5)]
\\\\
9(2x-8)=40-10(x+5)
.\end{array}
Using the Distributive Property and the properties of equality to isolate the variable, the equation above is equivalent to
\begin{array}{l}\require{cancel}
9(2x-8)=40-10(x+5)
\\\\
9(2x)+9(-8)=40-10(x)-10(5)
\\\\
18x-72=40-10x-50
\\\\
18x+10x=40-50+72
\\\\
28x=62
\\\\
x=\dfrac{62}{28}
\\\\
x=\dfrac{31}{14}
.\end{array}
Checking: If $x=\dfrac{31}{14},$ then
\begin{array}{l}\require{cancel}
0.9(2x-8)=4-(x+5)
\\\\
0.9\left( 2\cdot\dfrac{31}{14}-8\right)=4-\left(\dfrac{31}{14}+5\right)
\\\\
\dfrac{9}{10}\left( \dfrac{62}{14}-8\right)=4-\left(\dfrac{31}{14}+5\right)
\\\\
\dfrac{9}{10}\left( \dfrac{62}{14}-\dfrac{112}{14}\right)=\dfrac{56}{14}-\left(\dfrac{31}{14}+\dfrac{70}{14}\right)
\\\\
\dfrac{9}{10}\left( \dfrac{50}{14}\right)=\dfrac{56}{14}-\dfrac{101}{14}
\\\\
\dfrac{9}{\cancel{10}}\left( \dfrac{\cancel{10}(5)}{14}\right)=\dfrac{56}{14}-\dfrac{101}{14}
\\\\
\dfrac{45}{14}=\dfrac{45}{14}
\text{ (TRUE) }
.\end{array}
Hence, the solution is $
x=\dfrac{31}{14}
.$