i) \( 2 X+3 Y=10 \) \( 4 X-Y=5 \) ii) \[ \begin{array}{c} 3 X-2 Y=7 \\ 5 X+4 Y=1 \end{array} \] iii) \[ \begin{array}{l} X+Y=7 \\ 2 X-4 Y=1 \end{array} \] iv) \[ \begin{array}{l} 4 X+5 Y=12 \\ 2 X-3 Y=-5 \end{array} \] v) \[ \begin{array}{l} 3 X+2 Y=10 \\ 2 X-4 Y=6 \end{array} \] vii) \( 2 \frac{5}{8}+2 \frac{3}{5} Y=5 \) vi) \[ \begin{array}{c} 10 X+2 Y=6 \\ 2 X+2 \frac{3}{7} Y=6 \end{array} \] viii) \( X+3 Y=8 \) \[ 7 X+5 Y=10 \]

i) We have \[ 2X+3Y=10 \quad (1) \] \[ 4X-Y=5 \quad (2) \] Solve (2) for \(Y\): \[ 4X - Y = 5 \implies Y = 4X - 5. \] Substitute into (1): \[ 2X + 3(4X-5) = 10. \] \[ 2X + 12X - 15 = 10 \implies 14X = 25. \] \[ X = \frac{25}{14}. \] Then, \[ Y = 4\left(\frac{25}{14}\right) - 5 = \frac{100}{14} - 5 = \frac{50}{7} - 5 = \frac{50-35}{7} = \frac{15}{7}. \] --- ii) The system is \[ 3X-2Y=7 \quad (1) \] \[ 5X+4Y=1 \quad (2) \] Multiply (1) by 2: \[ 6X - 4Y = 14. \] Now add (2): \[ (6X-4Y)+(5X+4Y) = 14+1 \implies 11X = 15. \] \[ X=\frac{15}{11}. \] Substitute \(X\) into (1): \[ 3\left(\frac{15}{11}\right) - 2Y = 7 \implies \frac{45}{11} - 2Y = 7. \] \[ -2Y = 7 - \frac{45}{11} = \frac{77-45}{11} = \frac{32}{11}. \] \[ Y = -\frac{16}{11}. \] --- iii) The system is \[ X+Y=7 \quad (1) \] \[ 2X-4Y=1 \quad (2) \] From (1): \[ X=7-Y. \] Substitute into (2): \[ 2(7-Y)-4Y=1 \implies 14-2Y-4Y=1. \] \[ 14-6Y=1 \implies -6Y=-13. \] \[ Y=\frac{13}{6}. \] Then, \[ X=7-\frac{13}{6}=\frac{42-13}{6}=\frac{29}{6}. \] --- iv) The system is \[ 4X+5Y=12 \quad (1) \] \[ 2X-3Y=-5 \quad (2) \] Multiply (2) by 2: \[ 4X-6Y=-10. \] Subtract this from (1): \[ (4X+5Y)-(4X-6Y)=12-(-10) \implies 11Y=22. \] \[ Y=2. \] Substitute into (1): \[ 4X+5(2)=12 \implies 4X=12-10=2. \] \[ X=\frac{1}{2}. \] --- v) The system is \[ 3X+2Y=10 \quad (1) \] \[ 2X-4Y=6 \quad (2) \] Multiply (1) by 2: \[ 6X+4Y=20. \] Now add (2): \[ (6X+4Y)+(2X-4Y)=20+6 \implies 8X=26. \] \[ X=\frac{26}{8}=\frac{13}{4}. \] Substitute into (1): \[ 3\left(\frac{13}{4}\right)+2Y=10 \implies \frac{39}{4}+2Y=10. \] \[ 2Y=10-\frac{39}{4}=\frac{40-39}{4}=\frac{1}{4}. \] \[ Y=\frac{1}{8}. \] --- vii) The equation is \[ 2\frac{5}{8}+2\frac{3}{5}Y=5. \] Convert the mixed numbers: \[ 2\frac{5}{8}=\frac{21}{8}, \quad 2\frac{3}{5}=\frac{13}{5}. \] Thus, the equation becomes \[ \frac{21}{8}+\frac{13}{5}Y=5. \] Isolate \(Y\): \[ \frac{13}{5}Y=5-\frac{21}{8}=\frac{40}{8}-\frac{21}{8}=\frac{19}{8}. \] \[ Y=\frac{19}{8}\cdot\frac{5}{13}=\frac{95}{104}. \] --- vi) The system is \[ 10X+2Y=6 \quad (1) \] \[ 2X+2\frac{3}{7}Y=6 \quad (2) \] Convert the mixed number in (2): \[ 2\frac{3}{7}=\frac{17}{7}. \] So (2) becomes \[ 2X+\frac{17}{7}Y=6. \] Solve (2) for \(X\): \[ 2X=6-\frac{17}{7}Y \implies X=3-\frac{17}{14}Y. \] Substitute into (1): \[ 10\left(3-\frac{17}{14}Y\right)+2Y=6. \] \[ 30-\frac{170}{14}Y+2Y=6. \] Write \(2Y\) as \(\frac{28}{14}Y\): \[ 30-\frac{170-28}{14}Y=6 \implies 30-\frac{142}{14}Y=6. \] \[ 30-\frac{71}{7}Y=6. \] Subtract 30: \[ -\frac{71}{7}Y=6-30=-24. \] \[ Y=24\cdot\frac{7}{71}=\frac{168}{71}. \] Then, \[ X=3-\frac{17}{14}\left(\frac{168}{71}\right). \] Simplify: \[ \frac{168}{14}=12, \text{ so } X=3-\frac{17\cdot 12}{71}=3-\frac{204}{71}. \] Write 3 as \(\frac{213}{71}\): \[ X=\frac{213-204}{71}=\frac{9}{71}. \] --- viii) The system is \[ X+3Y=8 \quad (1) \] \[ 7X+5Y=10 \quad (2) \] Solve (1) for \(X\): \[ X=8-3Y. \] Substitute into (2): \[ 7(8-3Y)+5Y=10. \] \[ 56-21Y+5Y=10 \implies 56-16Y=10. \] \[ -16Y=10-56=-46. \] \[ Y=\frac{46}{16}=\frac{23}{8}. \] Then, \[ X=8-3\left(\frac{23}{8}\right)=8-\frac{69}{8}=\frac{64-69}{8}=-\frac{5}{8}. \] --- Final Answers: i) \(\left(\frac{25}{14},\frac{15}{7}\right)\) ii) \(\left(\frac{15}{11},-\frac{16}{11}\right)\) iii) \(\left(\frac{29}{6},\frac{13}{6}\right)\) iv) \(\left(\frac{1}{2},2\right)\) v) \(\left(\frac{13}{4},\frac{1}{8}\right)\) vii) \(Y=\frac{95}{104}\)