1. **Problem A.i:** Form an equation for the total amount of money Mrs. James spent. Mrs. James spent $x$ in the first shop, twice that amount in the second shop ($2x$), $3$ in the third shop, and $8$ in the last shop. The total spent was $26$. The equation is: $$x + 2x + 3 + 8 = 26$$ 2. **Problem A.ii:** Solve the equation to find the amount spent at the first shop. Start with the equation: $$x + 2x + 3 + 8 = 26$$ Combine like terms: $$3x + 11 = 26$$ Subtract 11 from both sides: $$3x + \cancel{11} - \cancel{11} = 26 - 11$$ $$3x = 15$$ Divide both sides by 3: $$\frac{3x}{\cancel{3}} = \frac{15}{\cancel{3}}$$ $$x = 5$$ So, Mrs. James spent $5$ at the first shop. 3. **Problem B:** Calculate the cost of each type of book. Let $x$ be the cost of a Mathematics book and $y$ be the cost of an English book. From the problem: - 25 Mathematics books and 10 English books cost $855$: $$25x + 10y = 855$$ - 10 Mathematics books and 40 English books cost $990$: $$10x + 40y = 990$$ Solve the system: Multiply the second equation by 2.5 to align $x$ terms: $$2.5 \times (10x + 40y) = 2.5 \times 990$$ $$25x + 100y = 2475$$ Subtract the first equation from this: $$(25x + 100y) - (25x + 10y) = 2475 - 855$$ $$25x - 25x + 100y - 10y = 1620$$ $$90y = 1620$$ Divide both sides by 90: $$\frac{90y}{\cancel{90}} = \frac{1620}{\cancel{90}}$$ $$y = 18$$ Substitute $y=18$ into the first equation: $$25x + 10(18) = 855$$ $$25x + 180 = 855$$ Subtract 180: $$25x = 675$$ Divide by 25: $$x = \frac{675}{25} = 27$$ **Final answers:** - Cost of Mathematics book $x = 27$ - Cost of English book $y = 18$