(i) x + y = 14

x – y = 4

(ii) s – t = 3

(s/3) + (t/2) = 6

Arrangements:

(I) Given,

$$ \[x\text{ }+\text{ }y\text{ }=\text{ }14\text{ }and\text{ }x\text{ }-\text{ }y\text{ }=\text{ }4\]  are the two conditions.

From first condition, we get,

\[x\text{ }=\text{ }14\text{ }-\text{ }y\]  

Presently, substitute the worth of x in second condition to get,

\[\left( 14\text{ }-\text{ }y \right)\text{ }-\text{ }y\text{ }=\text{ }4\] 

\[14\text{ }-\text{ }2y\text{ }=\text{ }4\] 

\[2y\text{ }=\text{ }10\]  

Or then again \[y\text{ }=\text{ }5\]  

By the worth of y, we would now be able to track down the specific worth of \[x\] ;

\[\because x\text{ }=\text{ }14-y\]  

\[\therefore x\text{ }=\text{ }14-5\]  

Or on the other hand \[x\text{ }=\text{ }9\]  

Henceforth, \[x\text{ }=\text{ }9\text{ }and\text{ }y\text{ }=\text{ }5\] .

(ii) Given,

\[s\text{ }-t\text{ }=\text{ }3\text{ }and\text{ }\left( s/3 \right)\text{ }+\text{ }\left( t/2 \right)\text{ }=\text{ }6\]  are the two conditions.

From first condition, we get,

\[s\text{ }=\text{ }3\text{ }+\text{ }t\text{ }\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\left( 1 \right)\]  

Presently, substitute the worth of s in second condition to get,

\[\left( 3+t \right)/3\text{ }+\text{ }\left( t/2 \right)\text{ }=\text{ }6\]

\[\Rightarrow \left( 2\left( 3+t \right)\text{ }+\text{ }3t\text{ } \right)/6\text{ }=\text{ }6\]  

\[\Rightarrow \left( 6+2t+3t \right)/6\text{ }=\text{ }6\]  

\[\Rightarrow \left( 6+5t \right)\text{ }=\text{ }36\]  

\[\Rightarrow 5t\text{ }=\text{ }30\]  

\[\Rightarrow t\text{ }=\text{ }6\]  

Presently, substitute the worth of t in condition (1)

\[s\text{ }=\text{ }3\text{ }+\text{ }6\text{ }=\text{ }9\]   In this way, s = 9 and t = 6.