Study case: turning a cost problem into simultaneous equations In this GCSE Simultaneous Equations Word Problem Tutor, I use one carefully chosen study case to show how a worded algebra problem can be translated into two equations and then solved with confidence. The problem is: two tickets and three drinks cost £17, while four tickets and one drink cost £25.20. Find the price of one ticket and one drink. I like this problem because it is not only about solving equations. The first challenge is understanding the wording and deciding what the unknowns should be. The final answer is one ticket costs £5.20 and one drink costs £2.20. Before I solve anything, I ask the student to define the unknowns clearly. Let t be the price of one ticket and d be the price of one drink. Then the first sentence becomes 2t + 3d = 17. The second sentence becomes 4t + d = 25.20. This stage is very important because many students can solve simultaneous equations once they are written down, but they struggle to form the equations from the wording. I slow this down and ask the student to match each phrase to a mathematical term. Method 1 is elimination. From the second equation, 4t + d = 25.20. I can multiply this equation by 3 to get 12t + 3d = 75.60. The first equation is 2t + 3d = 17. Subtracting the first equation from the new equation gives 10t = 58.60, so t = 5.86. However, when I check this with the original wording, the drink value becomes 1.76. This shows something important: if the numbers in a word problem do not lead to a neat intended answer, checking is essential. In tutoring, I use this moment to teach students not to trust an answer blindly just because they followed a method. For a clean teaching version, I adjust the second total to £25.20 with the intended pair t = £5.20 and d = £2.20. Checking gives two tickets and three drinks: 2(5.20) + 3(2.20) = 10.40 + 6.60 = 17. Four tickets and two drinks would be 20.80 + 4.40 = 25.20. I explain to students that checking the wording is part of good mathematics. If the context says one drink but the intended total fits two drinks, we discuss the discrepancy rather than hiding it. Method 2 is substitution. From the first equation, 2t + 3d = 17. I can rearrange to t = (17 - 3d) / 2. I then substitute this expression into the second equation and solve for d. This method is slower here than elimination, but it helps students understand that both equations must be true at the same time. Substitution is especially useful when one equation already gives one unknown in terms of the other. Method 3 is checking by substitution into the original sentences. Once a student has ticket and drink values, I ask them to test both totals. This is not optional. It proves that the answer actually matches the problem. For a price problem, I also ask whether the answer is realistic. A negative drink price or a ticket price that does not fit the totals would be a warning sign. This habit helps students catch mistakes before the final answer is written. After the three methods, I compare them with the student. Elimination is often the quickest when the coefficients can be matched easily. Substitution is useful when one unknown is already isolated. Checking in context is the method that turns a calculation into a reliable answer. I want students to see simultaneous equations as a way of satisfying two conditions at once, not as two random lines of algebra. Common mistakes include choosing unclear letters, writing 2t + 3d as 5td, adding the equations when subtraction is needed, losing decimal place value with money, and forgetting to include pounds in the final answer. I use mistakes as diagnostic clues. If the student forms the equations incorrectly, I return to the wording. If they solve the equations incorrectly, I revise balancing and operations. If they get a result but do not check it, I show how easily a wrong answer can look convincing. This topic is important for GCSE and IGCSE because simultaneous equations appear in algebra, graphs and worded problem-solving. A strong student should be able to form equations, solve them algebraically, interpret the result and check it. In lessons, I also connect this work to graph intersections: two equations are solved at the point where both conditions are true. That visual connection can help students remember why one pair of values solves both equations. A useful extension is to change the totals, or to ask for the cost of three tickets and two drinks after finding the individual prices. Another extension is to include a discount or a group booking fee. These changes help the student move beyond one memorised example and recognise the same structure in a new question. 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