1. Let the price of petrol in 2007 be $x$ dollars per litre.
2. The number of litres of petrol used by Mr Ong in 2007 is $\frac{320}{x}$.
3. The price of petrol in 2008 is $x + 0.20$ dollars per litre (since it increased by 20 cents).
4. The number of litres of petrol used by Mr Ong in 2008 is $\frac{320}{x + 0.20}$.
5. According to the problem, the number of litres used in 2008 is 20 less than that used in 2007.
6. Therefore, the equation is: $\frac{320}{x} - \frac{320}{x + 0.20} = 20$.
7. To solve this equation, first find a common denominator: $x(x + 0.20)$.
8. Multiply both sides by $x(x + 0.20)$ to eliminate the denominators: $320(x + 0.20) - 320x = 20x(x + 0.20)$.
9. Expand the terms: $320x + 64 - 320x = 20x^2 + 4x$.
10. Simplify: $64 = 20x^2 + 4x$.
11. Rearrange the equation: $20x^2 + 4x - 64 = 0$.
12. Divide the entire equation by 4 to simplify: $5x^2 + x - 16 = 0$.
13. The equation reduces to $5x^2 + x - 16 = 0$, as required.
Parent Tip: Review the logic above to help your child master the concept of algebra 2 quadratic word problems worksheet.
