1. **Stating the problem:** We have two lorries transporting sand: a 7-tonne lorry making $x$ trips and a 14-tonne lorry making $y$ trips. The total sand transported is 133 tonnes, and the total cost incurred is 47000. 2. **Writing the equations:** From the problem, the total sand transported is: $$7x + 14y = 133$$ The total cost incurred is: $$3000x + 4000y = 47000$$ 3. **Using substitution method to solve:** From the first equation, solve for $x$: $$7x + 14y = 133$$ $$7x = 133 - 14y$$ $$x = \frac{133 - 14y}{7}$$ 4. **Substitute $x$ into the second equation:** $$3000\left(\frac{133 - 14y}{7}\right) + 4000y = 47000$$ Simplify: $$\frac{3000}{7}(133 - 14y) + 4000y = 47000$$ $$\frac{3000 \times 133}{7} - \frac{3000 \times 14y}{7} + 4000y = 47000$$ $$\frac{399000}{7} - 6000y + 4000y = 47000$$ $$\frac{399000}{7} - 2000y = 47000$$ 5. **Isolate $y$:** $$-2000y = 47000 - \frac{399000}{7}$$ Calculate $\frac{399000}{7} = 57000$: $$-2000y = 47000 - 57000$$ $$-2000y = -10000$$ Divide both sides by $-2000$: $$y = \frac{-10000}{\cancel{-2000}} \cancel{-1} = 5$$ 6. **Find $x$ using $y=5$:** $$x = \frac{133 - 14 \times 5}{7} = \frac{133 - 70}{7} = \frac{63}{7} = 9$$ 7. **Calculate profit:** Total sand transported = 133 tonnes. Payment received = $500 \times 133 = 66500$ Total cost = 47000 Profit = Payment - Cost = $66500 - 47000 = 19500$ **Final answers:** $$x = 9, \quad y = 5, \quad \text{Profit} = 19500$$