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← Level 6 – Problem solver

7 / 30 ⏱ 18 minutes

Word problems with simultaneous equations

From a text, two equations: tickets, prices, counts of things.

From a text, two equations

You have the methods. Now the main reason simultaneous equations exist: word problems. Cinema tickets, snack prices, animals in a yard — anywhere there are two unknown things and two pieces of information about them.

The recipe stays the same. Choose the unknowns and write what they mean. For example d = price of an adult ticket, s = price of a student ticket. That line is not decoration — without it you get lost halfway through the calculation.

Then translate the sentences into equations. Each sentence with a number usually gives one equation. “Two adult and three student tickets cost 570 Kč” → 2d + 3s = 570.

You solve by the method you like. And at the end answer in a sentence — for a word problem the answer is not “x = 90”, but “a student ticket costs 90 Kč”.

This type of question is almost guaranteed in entrance exams. Come and make it automatic.

Filip and Ema were at the cinema. Filip paid for himself and dad — one adult, one student, 330 Kč in total. Ema paid for the family — two adult and one student for 530 Kč. “So how much is an adult?” Filip wonders. Ema pulls out the receipts: “Your purchase: d + s = 330. Mine: 2d + s = 530. The difference between us is exactly one extra adult — and it is 200 Kč more. An adult costs 200 Kč and a student 130.” Two receipts, two equations, تم.

💡

Always write a key: what the first unknown is, what the second is, including units. Half the mistakes in word problems come from the solver forgetting halfway what the letters mean.

From text to answer in five steps

Step 1: Key. Read the question twice and write: what do I not know? For example: r = price of a roll in Kč, k = price of a doughnut in Kč.

Step 2: Translate the sentences into equations. Look for sentences with numbers. “Three rolls and two doughnuts cost 54 Kč” → 3r + 2k = 54. “A doughnut is 12 Kč more expensive than a roll” → k = r + 12.

Watch the translations: “12 more” means +12. “Twice as expensive” means ·2. “Together” means a sum.

Step 3: Pick a method. When one unknown is the subject (k = r + 12), take substitution. When you see opposite signs, take elimination.

Step 4: Solve. 3r + 2 · (r + 12) = 54, so 3r + 2r + 24 = 54, so 5r = 30, so r = 6. And k = 6 + 12 = 18.

Step 5: Check against the question and answer in a sentence. Three rolls (18 Kč) and two doughnuts (36 Kč) is 54 Kč together — it matches. Answer: a roll costs 6 Kč, a doughnut 18 Kč.

Do the check against the text of the question, not against your equations — if the mistake was already in the translation, the equations would not confess it.

Example 1: tickets

Question: An adult and a student ticket together cost 290 Kč. The adult is 70 Kč more expensive. How much does each cost?

Key: d = adult, s = student (in Kč).

Equations: d + s = 290 and d = s + 70.

Solution by substituting: (s + 70) + s = 290, so 2s + 70 = 290, so 2s = 220, so s = 110. And d = 110 + 70 = 180.

Check against the question: 180 + 110 = 290 — it matches. Difference 70 — it matches.

Answer: An adult costs 180 Kč, a student 110 Kč.

Example 2: a phone and a case

Question: A phone with a case costs 5 500 Kč. The phone is ten times as expensive as the case. How much does the case cost?

Key: m = phone, p = case (in Kč).

Equations: m + p = 5500 and m = 10p.

Solution: 10p + p = 5500, so 11p = 5500, so p = 500. And m = 5 000.

Check: 5000 + 500 = 5500 — it matches. The phone is ten times as expensive — it matches.

Answer: The case costs 500 Kč. Watch: “ten times as expensive” is multiplying, not +10. That mix-up in translation is the most common trap.

Example 3: hens and rabbits

Question: In a yard there are hens and rabbits, together 10 heads and 28 legs. How many hens are there?

Key: s = number of hens, k = number of rabbits.

Equations: heads: s + k = 10. Legs: a hen has 2, a rabbit 4, so 2s + 4k = 28.

Solution by elimination: I multiply the first equation by −2: −2s − 2k = −20. I add with the second: 2k = 8, so k = 4. Then s = 10 − 4 = 6.

Check: 6 + 4 = 10 heads — it matches. 6 · 2 + 4 · 4 = 12 + 16 = 28 legs — it matches.

Answer: There are 6 hens in the yard.

⚠️

“How much more” is not “how many times”. The sentence “70 Kč more expensive” gives d = s + 70. The sentence “twice as expensive” gives d = 2s. If you swap them, the equations will come out — but they describe a different problem and a check against the question spots it at once. That is why you check against the text, not against your own equations.

Practice: now you do it 🎲

Now try it for real. You will see all the questions at once, like on paper. البداية with ten. When it goes well, add more.

A warm-up on simultaneous equations — you can picture each as a mini word problem about a sum and a difference. Answer with the value of x.

Practise the questions

On paper

Two word problems from start to finish: key, equations, solution, check against the question, answer in a sentence.

  1. An ice cream and a wafer together cost 62 Kč. The ice cream is 38 Kč more expensive. How much does each cost?

  2. There are 28 children in the class. There are 4 fewer boys than girls. How many girls and how many boys?

  3. Two T-shirts and one pair of shorts cost 700 Kč. A T-shirt and shorts cost 550 Kč. How much does a T-shirt cost? (Compare both purchases — what changed?)

  4. In each answer underline the units. Kč are Kč, children are children.

Now you 💪

  1. Every problem has a key: what each unknown means, with a unit.
  2. You translate “how much more” sentences to +, “how many times” sentences to multiplying.
  3. You answer in a full sentence and do the check against the text of the question.
✅

Done when: You have two word problems complete: key, two equations, solution, check against the question and an answer in a sentence.

What to take from this lesson

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