Systems of Equations • Topic 4 of 4

Word-Problem Systems

Most GRE systems arrive dressed as stories, and the whole difficulty is translation. Name each unknown with a letter, then convert every sentence into an equation — usually one for a total count and one for a total value or a relationship. Two adult and three child tickets costing a certain amount, alongside a different mix at a different total, is a classic two-equation setup. Watch the units and make sure both equations describe the same quantities in the same terms. Once the pair is written, solve with whichever method is cleaner — substitution if a variable is isolated, elimination if coefficients align. Finally, re-read the question: it may ask for one unknown, their sum, or a derived quantity, so answer what was actually asked rather than stopping at the first value you find.

✅ Solved examples

1. Two coffees and three teas cost $13; one coffee and two teas cost $8. Find each price.
Let c, t. 2c + 3t = 13 and c + 2t = 8. From the second, c = 8 − 2t; substitute: 2(8 − 2t) + 3t = 13 ⇒ 16 − t = 13 ⇒ t = 3, c = 2. Coffee $2, tea $3.
2. A number and its reverse: the tens plus units digit is 9, and the tens digit is twice the units. Find the digits.
Let t, u. t + u = 9 and t = 2u. Substitute: 2u + u = 9 ⇒ u = 3, t = 6. Digits 6 and 3.
3. Adult tickets are $12 and child $8; 30 tickets sold for $308. How many of each?
Let a, c. a + c = 30 and 12a + 8c = 308. From the first c = 30 − a; substitute: 12a + 8(30 − a) = 308 ⇒ 4a + 240 = 308 ⇒ a = 17, c = 13.
4. The sum of two numbers is 40 and their difference is 8. Find the larger.
x + y = 40, x − y = 8. Add: 2x = 48 ⇒ x = 24 (the larger). y = 16.

✏️ Practice — try these, take hints as needed

1. Three pens and two notebooks cost $16; one pen and two notebooks cost $10. Find each price.
Let p, n. Write both equations.
Subtract to cancel n.
2p = 6.
Pen $3, notebook $3.50
2. Two numbers sum to 50; one is 10 more than the other. Find both.
x + y = 50, x = y + 10.
Substitute.
2y + 10 = 50.
30 and 20
3. A farmer has chickens and cows totalling 20 heads and 56 legs. How many cows?
Heads: c + h = 20. Legs: 4c + 2h = 56.
Substitute h = 20 − c.
2c + 40 = 56.
8 cows (12 chickens)
4. 4 kg of apples and 3 kg of oranges cost $18; 2 kg apples and 3 kg oranges cost $12. Price per kg of apples?
Subtract the equations.
2·(apple) = 6.
Solve for apple price.
Apples $3/kg (oranges $2/kg)
5. Two numbers: sum 24, difference 6. Find the smaller.
x + y = 24, x − y = 6.
Add for x, subtract for y.
Smaller is y.
9 (larger is 15)

📝 Topic test — 8 questions

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