We Call distance of AB \(=x\left(km\right)\) and call speed of Mr.Joseph when him drives car from A to B is \(y\left(m\right)\).
The usual time to Mr.Joseph drives car from A to B is  \(\dfrac{x}{y}\) (hours).
If the speeds of the car is increased by 20% then the new speeds is \(y+20\%y=1,2y\).
When the time to Mr.Joseph drives car from A to B is : \(\dfrac{x}{1,2y}=\dfrac{5}{6}\times\dfrac{x}{y}\).
We have: \(\dfrac{5}{6}\times\dfrac{x}{y}=\dfrac{x}{y}-1\Leftrightarrow\left(1-\dfrac{5}{6}\right)\dfrac{x}{y}=1\)\(\Leftrightarrow\dfrac{x}{y}=6\)\(\Leftrightarrow x=6y\).
If the speeds of the car is increased by 20% then the new speeds is: \(y+30\%y=1,3y\).
When the time to Mr.Joseph drives car from A to B is :
\(\dfrac{100}{y}+\dfrac{x-100}{1,3y}=\dfrac{x}{y}-1\)
\(\Leftrightarrow\dfrac{100\times1,3+x-100}{1,3y}=6-1\)
\(\Leftrightarrow30+x=6,5y\).
We have the system of equation:
\(\left\{{}\begin{matrix}30+x=6,5y\\x=6y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=60\left(km\right)\\x=60\times6=360\left(km\right)\end{matrix}\right.\).
So, this distance of AB is 360km.

Condition determines: \(14x-42\ne0\Leftrightarrow x\ne3\).
We have:
\(\dfrac{7\left(x-3\right)}{14\left(x-3\right)}=\dfrac{7}{14}=\dfrac{7:7\times2}{14:7\times2}=\dfrac{2}{4}\)

Find the x satisty: \(\left|x-1\right|=2x-4\)

\(11+22+33+44+...+999\)
\(=11\left(1+2+......+111\right)\)
\(=11\times\dfrac{111+1}{2}\times\dfrac{111}{2}\)
\(=34188\)