Consider x be subtracted from each term
\[23\text{ }\text{ }x,\text{ }30\text{ }\text{ }x,\text{ }57\text{ }\text{ }x\text{ }and\text{ }78\text{ }\text{ }x\] are proportional
It can be written as
\[\begin{array}{*{35}{l}}
23\text{ }\text{ }x:\text{ }30\text{ }\text{ }x\text{ }::\text{ }57\text{ }\text{ }x:\text{ }78\text{ }\text{ }x \\
\left( 23\text{ }\text{ }x \right)/\text{ }\left( 30\text{ }\text{ }x \right)\text{ }=\text{ }\left( 57\text{ }\text{ }x \right)/\text{ }\left( 78\text{ }\text{ }x \right) \\
\end{array}\]
By cross multiplication
\[\left( 23\text{ }\text{ }x \right)\text{ }\left( 78\text{ }\text{ }x \right)\text{ }=\text{ }\left( 30\text{ }\text{ }x \right)\text{ }\left( 57\text{ }\text{ }x \right)\]
By further calculation
\[\begin{array}{*{35}{l}}
1794\text{ }\text{ }23x\text{ }\text{ }78x\text{ }+\text{ }{{x}^{2}}~=\text{ }1710\text{ }\text{ }30x\text{ }\text{ }57x\text{ }+\text{ }{{x}^{2}} \\
{{x}^{2}}~\text{ }101x\text{ }+\text{ }1794\text{ }\text{ }{{x}^{2}}~+\text{ }87x\text{ }\text{ }1710\text{ }=\text{ }0 \\
\end{array}\]
So we get
\[\begin{array}{*{35}{l}}
\text{ }14x\text{ }+\text{ }84\text{ }=\text{ }0 \\
14x\text{ }=\text{ }84 \\
x\text{ }=\text{ }84/14\text{ }=\text{ }6 \\
\end{array}\]
Therefore, \[6\] is the number to be subtracted from each of the numbers.