Solve problems with two unknowns.
This challenge is all about using algebraic expressions and equations to solve real-life problems. Think smart, work symbolically, and let expressions guide your way to the answers.

🙂 = Correct, ☹️ = Incorrect

📌 Rules

• Use substitution when one value is known: Replace the known variable with its given value to reduce the problem to a single unknown. For example, if 3 muffins + 2 cupcakes = £12 and 1 muffin = £2, calculate 3 × 2 = £6 for muffins, leaving 12 – 6 = £6 for 2 cupcakes (so 1 cupcake = £3).

• Use multiplicative relationships (Ratios & Multipliers): Express one unknown in terms of the other using multiplication. If a field’s length is 3 times its width (l = 3w), the perimeter formula 2l + 2w = 96 becomes 2(3w) + 2w = 96, which simplifies to 8w = 96 (so width w = 12m and length l = 36m).

• Use additive differences (Sums & Differences): When two values add to a total and one is greater by a fixed amount, subtract the difference from the total first, then divide by 2 to find the smaller quantity. For 240 trees where one row has 40 more: subtract 240 – 40 = 200, then divide 200 ÷ 2 = 100 trees (shorter row) and 100 + 40 = 140 trees (longer row).

💡 Tips

• Distance, Speed, and Time with two speeds: If two objects travel in opposite directions for 2 hours to reach 300 miles apart, their combined distance in 1 hour is 300 ÷ 2 = 150 mph. If one is 60 mph faster, use the difference method: 150 – 60 = 90, so 90 ÷ 2 = 45 mph (slower car) and 45 + 60 = 105 mph (faster car).

• Quick typing on our platform: Typing your answer is designed to be super quick and simple! Just type plain numbers like 3, 16, or 105 without adding any currency symbols, unit labels, or spaces.

⚠️ Warnings

• Do not confuse perimeter parts with single sides: A rectangle has 2 lengths and 2 widths! For a field where length is 3 × width, the total perimeter is made of 8 equal width units (3 + 3 + 1 + 1 = 8), not 4.

• Check that both values satisfy all conditions: Always double-check your final numbers by adding them together or substituting them back into the problem to verify they meet both the total and the relationship required.