(i) 3x – y = 3
9x – 3y = 9
(ii) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(i) Given,
\[3x\text{ }-y\text{ }=\text{ }3\text{ }and\text{ }9x\text{ }-3y\text{ }=\text{ }9\] are the two conditions.
From first condition, we get,
\[x\text{ }=\text{ }\left( 3+y \right)/3\]
Presently, substitute the worth of x in the given second condition to get,
\[9\left( 3+y \right)/3\text{ }-3y\text{ }=\text{ }9\]
\[\Rightarrow 9\text{ }+3y\text{ }-\text{ }3y\text{ }=\text{ }9\]
\[\Rightarrow 9\text{ }=\text{ }9\]
Hence, y has limitless qualities and since, \[x\text{ }=\text{ }\left( 3+y \right)/3\] , so x additionally has boundless qualities.
(ii) Given,
\[0.2x\text{ }+\text{ }0.3y\text{ }=\text{ }1.3\text{ }and\text{ }0.4x\text{ }+\text{ }0.5y\text{ }=\text{ }2.3\] are the two conditions.
From first condition, we get,
\[x\text{ }=\text{ }\left( 1.3-0.3y \right)/0.2\text{ }\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left( 1 \right)\]
Presently, substitute the worth of x in the given second condition to get,
\[0.4\left( 1.3-0.3y \right)/0.2\text{ }+\text{ }0.5y\text{ }=\text{ }2.3\]
\[\Rightarrow 2(1.3-0.3y)\text{ }+\text{ }0.5y\text{ }=\text{ }2.3\]
\[\Rightarrow 2.6-0.6y\text{ }+\text{ }0.5y\text{ }=\text{ }2.3\]
\[\Rightarrow 2.6-0.1\text{ }y\text{ }=\text{ }2.3\]
\[\Rightarrow 0.1\text{ }y\text{ }=\text{ }0.3\]
\[\Rightarrow y\text{ }=\text{ }3\]
Presently, substitute the worth of y in condition (1), we get,
\[x\text{ }=\text{ }\left( 1.3-0.3\left( 3 \right) \right)/0.2\text{ }=\text{ }\left( 1.3-0.9 \right)/0.2\text{ }=\text{ }0.4/0.2\text{ }=\text{ }2\] Hence, \[x\text{ }=\text{ }2\text{ }and\text{ }y\text{ }=\text{ }3\] .