Question:

### (i) \(x + y = 14\)
1. Solve the given equation for one variable, e.g. \(x = 14 - y\).
2. Since there is no second independent equation, the solution set is expressed as \(\{(x, y) \mid x = 14 - y\}\).
Solution: \(x = 14 - y\) (infinitely many ordered pairs).

### (ii) \(\begin{cases}s - t = 3\\2s + t = 2\end{cases}\)
1. From the first equation, \(s = t + 3\).
2. Substitute \(s = t + 3\) into the second equation:
\[2(t + 3) + t = 2 \Rightarrow 2t + 6 + t = 2 \Rightarrow 3t = -4 \Rightarrow t = -\frac{4}{3}.\]
3. Substitute \(t\) back into \(s = t + 3\):
\[s = -\frac{4}{3} + 3 = \frac{5}{3}.\]
Solution: \(s = \frac{5}{3},\; t = -\frac{4}{3}.\)

### (iii) \(\begin{cases}3x - y = 3\\9x - 3y = 9\end{cases}\)
1. Notice that the second equation is exactly three times the first; therefore both equations represent the same straight line.
2. Solve the first equation for \(y\):
\[y = 3x - 3.\]
3. The system has infinitely many solutions lying on the line \(y = 3x - 3\).
Solution: \(\{(x, y) \mid y = 3x - 3\}\).

### (iv) \(\begin{cases}0.2x + 0.3y = 1.3\\0.4x + 0.5y = 2.3\end{cases}\)
1. Multiply the first equation by 2 to make the coefficient of \(x\) the same as in the second equation:
\[0.4x + 0.6y = 2.6.\]
2. Subtract the second original equation from this new equation:
\[(0.4x + 0.6y) - (0.4x + 0.5y) = 2.6 - 2.3\]
\[0.1y = 0.3 \Rightarrow y = 3.\]
3. Substitute \(y = 3\) into the first original equation:
\[0.2x + 0.3(3) = 1.3 \Rightarrow 0.2x + 0.9 = 1.3 \Rightarrow 0.2x = 0.4 \Rightarrow x = 2.\]
Solution: \(x = 2,\; y = 3.\)

### (v) \(\begin{cases}\frac{2}{3}x + y = 0\\x - 2y = 5\end{cases}\)
1. From the first equation, express \(y\) in terms of \(x\):
\[y = -\frac{2}{3}x.\]
2. Substitute this expression for \(y\) into the second equation:
\[x - 2\left(-\frac{2}{3}x\right) = 5 \Rightarrow x + \frac{4}{3}x = 5 \Rightarrow \frac{7}{3}x = 5 \Rightarrow x = \frac{15}{7}.\]
3. Find \(y\) using \(y = -\frac{2}{3}x\):
\[y = -\frac{2}{3}\times \frac{15}{7} = -\frac{30}{21} = -\frac{10}{7}.\]
Solution: \(x = \frac{15}{7},\; y = -\frac{10}{7}.\)

### (vi) \(\begin{cases}\frac{3}{5}x - 2y = -3\\\frac{8}{3}x - y = 13\\\frac{3}{2}x + 6y = 0\end{cases}\)
1. First pair (equations 1 & 2):
- From equation 1, solve for \(x\):
\[\frac{3}{5}x = 2y - 3 \Rightarrow x = \frac{5}{3}(2y - 3) = \frac{10}{3}y - 5.\]
- Substitute this \(x\) into equation 2:
\[\frac{8}{3}\left(\frac{10}{3}y - 5\right) - y = 13\]
\[\frac{80}{9}y - \frac{40}{3} - y = 13\]
\[\left(\frac{80}{9} - \frac{9}{9}\right) y = 13 + \frac{40}{3}\]
\[\frac{71}{9}y = \frac{79}{3}\]
\[y = \frac{79}{3}\times \frac{9}{71} = \frac{237}{71}.\]
- Then \(x = \frac{10}{3}y - 5 = \frac{10}{3}\times \frac{237}{71} - 5 = \frac{2370}{213} - 5 = \frac{2370 - 1065}{213} = \frac{1305}{213} = \frac{435}{71}.\]
2. Check with the third equation:
Substitute \(x = \frac{435}{71}\) and \(y = \frac{237}{71}\) into \(\frac{3}{2}x + 6y = 0\):
\[\frac{3}{2}\times \frac{435}{71} + 6\times \frac{237}{71} = \frac{1305}{142} + \frac{1422}{71} = \frac{1305}{142} + \frac{2844}{142} = \frac{4149}{142}
eq 0.\]
Hence the three equations are inconsistent; no common solution exists for all three simultaneously.
3. Therefore we solve any two of them (as done above) and state that the third equation is not compatible with the first two.
Solution (using equations 1 & 2): \(x = \frac{435}{71},\; y = \frac{237}{71}.\) The system of three equations has no solution.

Summary of Answers
- (i) \(x = 14 - y\) (infinitely many solutions)
- (ii) \(s = \frac{5}{3},\; t = -\frac{4}{3}\)
- (iii) \(y = 3x - 3\) (infinitely many solutions)
- (iv) \(x = 2,\; y = 3\)
- (v) \(x = \frac{15}{7},\; y = -\frac{10}{7}\)
- (vi) The first two equations give \(x = \frac{435}{71},\; y = \frac{237}{71}\); the third equation is inconsistent, so the three‑equation system has no solution.