1. We are given two problems involving simultaneous equations related to costs and numbers. **Problem 1:** The cost of 2 rackets and 3 squash balls is 21.63. The cost of 5 rackets and 7 squash balls is 52.90. Let $r$ represent the cost of one racket and $s$ the cost of one squash ball. 2. Set up the equations: $$2r + 3s = 21.63$$ $$5r + 7s = 52.90$$ 3. To solve, multiply the first equation by 5 and the second by 2 to eliminate $r$: $$ (2r + 3s) \times 5 \implies 10r + 15s = 108.15 $$ $$ (5r + 7s) \times 2 \implies 10r + 14s = 105.80 $$ 4. Subtract the second from the first: $$ (10r + 15s) - (10r + 14s) = 108.15 - 105.80 $$ $$ s = 2.35 $$ 5. Substitute $s = 2.35$ into the first equation: $$ 2r + 3(2.35) = 21.63 $$ $$ 2r + 7.05 = 21.63 $$ $$ 2r = 21.63 - 7.05 = 14.58 $$ $$ r = \frac{14.58}{2} = 7.29 $$ **Answer for problem 1:** - Cost of one racket $= 7.29$ - Cost of one squash ball $= 2.35$ **Problem 2:** The cost of 7 potted plants and 2 bags of fertiliser is 43. The cost of 5 potted plants and 4 bags of fertiliser is 41. Let $p$ be the cost of one potted plant and $f$ be the cost of one bag of fertiliser. Set up the equations: $$7p + 2f = 43$$ $$5p + 4f = 41$$ Multiply the first by 2 and the second by 1 (to aid elimination of $f$): $$14p + 4f = 86$$ $$5p + 4f = 41$$ Subtract the second from the first: $$ (14p + 4f) - (5p + 4f) = 86 - 41 $$ $$ 9p = 45 $$ $$ p = 5 $$ Substitute $p = 5$ into the second equation: $$ 5(5) + 4f = 41 $$ $$ 25 + 4f = 41 $$ $$ 4f = 16 $$ $$ f = 4 $$ Calculate the total cost of 2 potted plants and 5 bags of fertiliser: $$ 2p + 5f = 2(5) + 5(4) = 10 + 20 = 30 $$ **Answer for problem 2:** - Total cost of 2 potted plants and 5 bags of fertiliser $= 30$