1. **Problem:** Solve the first pair of simultaneous equations: $$\frac{1}{2}x + \frac{2}{3}y = 6 \frac{1}{5}$$ $$\frac{4}{3}x - \frac{1}{5}y = 13 \frac{3}{5}$$ 2. **Convert mixed numbers to improper fractions:** $$6 \frac{1}{5} = \frac{31}{5}, \quad 13 \frac{3}{5} = \frac{68}{5}$$ 3. **Rewrite the system:** $$\frac{1}{2}x + \frac{2}{3}y = \frac{31}{5}$$ $$\frac{4}{3}x - \frac{1}{5}y = \frac{68}{5}$$ 4. **Eliminate fractions by multiplying each equation by the least common denominator (LCD):** - For the first equation, LCD is 30: $$30 \times \left(\frac{1}{2}x + \frac{2}{3}y\right) = 30 \times \frac{31}{5}$$ $$15x + 20y = 186$$ - For the second equation, LCD is 15: $$15 \times \left(\frac{4}{3}x - \frac{1}{5}y\right) = 15 \times \frac{68}{5}$$ $$20x - 3y = 204$$ 5. **System now:** $$15x + 20y = 186$$ $$20x - 3y = 204$$ 6. **Multiply first equation by 3 and second by 20 to align coefficients of $y$ for elimination:** $$3 \times (15x + 20y) = 3 \times 186 \Rightarrow 45x + 60y = 558$$ $$20 \times (20x - 3y) = 20 \times 204 \Rightarrow 400x - 60y = 4080$$ 7. **Add the two equations to eliminate $y$:** $$45x + 60y + 400x - 60y = 558 + 4080$$ $$445x = 4638$$ 8. **Solve for $x$:** $$x = \frac{4638}{445}$$ 9. **Substitute $x$ back into one of the original equations, e.g., $15x + 20y = 186$:** $$15 \times \frac{4638}{445} + 20y = 186$$ 10. **Calculate:** $$\frac{15 \times 4638}{445} + 20y = 186$$ $$\frac{69570}{445} + 20y = 186$$ 11. **Isolate $y$:** $$20y = 186 - \frac{69570}{445}$$ 12. **Convert 186 to fraction with denominator 445:** $$186 = \frac{186 \times 445}{445} = \frac{82770}{445}$$ 13. **Calculate:** $$20y = \frac{82770}{445} - \frac{69570}{445} = \frac{13200}{445}$$ 14. **Solve for $y$:** $$y = \frac{13200}{445 \times 20} = \frac{13200}{8900} = \frac{132}{89}$$ 15. **Final answers:** $$x = \frac{4638}{445} \approx 10.42, \quad y = \frac{132}{89} \approx 1.48$$