Putting the value of y1 = 4 – x1 in equation 3, we get
\[\begin{array}{*{35}{l}}
4{{x}_{1}}~+\text{ }3\left( 4\text{ }-\text{ }{{x}_{1}} \right)\text{ }=\text{ }5 \\
\Rightarrow ~4{{x}_{1}}~+\text{ }12\text{ }-\text{ }3{{x}_{1}}~=\text{ }5 \\
\Rightarrow ~{{x}_{1}}~=\text{ }5\text{ }-\text{ }12 \\
\Rightarrow ~{{x}_{1}}~=\text{ }-\text{ }7 \\
\end{array}\]
Putting the value of x1 in equation 4, we get
\[\begin{array}{*{35}{l}}
{{y}_{1}}~=\text{ }4\text{ }-\text{ }\left( -7 \right) \\
\Rightarrow ~{{y}_{1}}~=\text{ }4\text{ }+\text{ }7 \\
\Rightarrow ~{{y}_{1}}~=\text{ }11 \\
\end{array}\]
Hence, the required points on the given line are (3, 1) and (-7, 11)