1. **State the problem:** Bill buys 6 apples and 5 oranges for 39. The price of an apple is $\frac{5}{7}$ of that of an orange. (a) Find the price of an orange. (b) The price of a pear is 0.8 higher than that of an orange. Bill has 100 originally, find the maximum number of extra pears he can buy. 2. **Define variables:** Let the price of an orange be $x$. Then the price of an apple is $\frac{5}{7}x$. 3. **Write the total cost equation:** $$6 \times \frac{5}{7}x + 5x = 39$$ 4. **Simplify the equation:** $$\frac{30}{7}x + 5x = 39$$ Convert 5x to $\frac{35}{7}x$ to add: $$\frac{30}{7}x + \frac{35}{7}x = 39$$ $$\frac{65}{7}x = 39$$ 5. **Solve for $x$:** Multiply both sides by 7: $$65x = 39 \times 7$$ $$65x = 273$$ Divide both sides by 65: $$x = \frac{273}{65}$$ Show cancellation: $$x = \frac{\cancel{273}}{\cancel{65}}$$ Simplify fraction: $$x = 4.2$$ So, the price of an orange is 4.2. 6. **Find the price of a pear:** Price of pear = price of orange + 0.8 = $4.2 + 0.8 = 5.0$ 7. **Calculate money left after buying apples and oranges:** Total spent = 39 Money left = 100 - 39 = 61 8. **Find maximum number of pears Bill can buy with remaining money:** Number of pears = $\left\lfloor \frac{61}{5.0} \right\rfloor = 12$ **Final answers:** (a) Price of an orange = 4.2 (b) Maximum extra pears Bill can buy = 12