Simultaneous Equations: An Algebra Extension

Sometimes two unknown prices are tied to two different purchases. Guessing one price at a time is slow. Write both conditions as equations, make one unknown match, and subtract the two balanced statements. This page is an optional algebra extension for learners who are ready; ordinary PSLE problem sums can often use a model drawing instead.

Parents: let your child translate the two purchase statements before showing the elimination step, and remind them that this extension is enrichment rather than a required PSLE method.

By the end you will be able to name two unknowns, translate two linked conditions, scale every term equally, eliminate one unknown, and check the answer in the story.

When this extension helps

Use simultaneous equations when two unknown quantities obey two independent conditions and matching one term makes subtraction direct. The variable keeps the unknown price visible while you work. A bar model, ratio model or the units-and-parts page may be friendlier for other questions.

Name the unknowns and write both conditions

Here is the full question:

At a book fair, 2 notebooks and 3 pens cost $19. Four notebooks and 1 pen cost $23. What is the cost of one notebook and one pen?

Let n be the cost of one notebook and p be the cost of one pen, in dollars. Translate each purchase separately. Keep the two original equations visible; they are the conditions you must satisfy at the end.

Translate the story

$$2n + 3p = 19$$

$$4n + p = 23$$

$$n = \text{notebook price},\quad p = \text{pen price}$$

Make one unknown match

The second equation has 4n. Double the first equation so its 2n becomes 4n. An equation stays balanced only when every term on both sides is multiplied by the same number.

Scale the complete equation

$$2 \times (2n + 3p) = 2 \times 19$$

$$4n + 6p = 38$$

$$4n + 6p = 38$$

$$4n + p = 23$$

The two 4n groups now match. Subtract the second equation from the doubled first equation. The 4n terms disappear, leaving only pen terms and their difference in cost.

Eliminate, then substitute

Subtract the untouched second equation from the doubled first. Four notebook terms cancel. Five pen terms cost $15, so one pen costs $3. Substitute that price into an original equation to find the notebook price.

Eliminate and solve

$$(4n + 6p) - (4n + p) = 38 - 23$$

$$5p = 15$$

$$p = 15 \div 5 = 3$$

$$4n + 3 = 23$$

$$4n = 20,\quad n = 20 \div 4 = 5$$

One notebook costs $5 and one pen costs $3. Use the interactive price slider to choose a pen price. It compares the notebook price implied by each original equation; the correct pen price is the one that makes both implications agree.

🔬

Choose the pen price

🤔 Predict first: What should agree at the correct pen price?

The notebook price from both equationsThe two totals only

Check the original conditions

$$2 \times \$5 + 3 \times \$3 = \$19$$

$$4 \times \$5 + \$3 = \$23$$

$$\text{Both conditions are satisfied}$$

Catch the balance mistake

If you double only 2n and leave the other terms unchanged, you have changed the first purchase. The new line would not describe the same $19 purchase. It may still produce neat arithmetic, but it cannot be trusted.

Algebra is optional enrichment

The Singapore primary syllabus emphasises arithmetic, fractions, ratio and model reasoning. A learner can solve many two-unknown questions with a well-drawn model. Use this page to practise a clean algebra representation when you want an extension, and never treat solving simultaneous equations as a required shortcut for every PSLE question.

Watch out — easily mixed up

Quick recap

Graduated practice

1. Erasers and rulers

Three erasers and 2 rulers cost $16. One eraser and 2 rulers cost $10. Find each price.

Show solution

Subtract the second condition from the first. The ruler terms match and disappear, leaving the price of two erasers.

Practice 1

$$3e + 2r = 16$$

$$e + 2r = 10$$

$$2e = 6$$

$$e = 6 \div 2 = 3$$

$$3 + 2r = 10$$

$$r = 7 \div 2 = 3.50$$

One eraser costs $3 and one ruler costs $3.50. Check: 3 × 3 + 2 × 3.50 = 16 and 3 + 2 × 3.50 = 10.

2. Adult and child tickets

Two adult tickets and 3 child tickets cost $39. Four adult tickets and 1 child ticket cost $53. Find both prices.

Show solution

Double the first equation to make the adult terms match the second equation, then subtract.

Practice 2

$$2a + 3c = 39$$

$$4a + c = 53$$

$$4a + 6c = 78$$

$$5c = 25,\quad c = 5$$

$$4a + 5 = 53$$

$$a = 48 \div 4 = 12$$

An adult ticket costs $12 and a child ticket costs $5. Check: 2 × 12 + 3 × 5 = 39 and 4 × 12 + 5 = 53.

3. A non-money pair

Two notebooks and 5 stickers cost 19 units. Four notebooks and 1 sticker cost 17 units. Find the value of one notebook and one sticker.

Show solution

Let n be one notebook and s be one sticker. Double the first equation, then subtract the second to eliminate notebooks.

Practice 3

$$2n + 5s = 19$$

$$4n + s = 17$$

$$4n + 10s = 38$$

$$9s = 21,\quad s = \frac{7}{3}$$

$$4n + \frac{7}{3} = 17$$

$$n = \frac{11}{3}$$

The values are one sticker = $\frac{7}{3}$ units and one notebook = $\frac{11}{3}$ units. This fractional answer is mathematically valid; a school question would usually choose numbers that give convenient whole or decimal values. Substitution checks both original totals.

🎯 Mastery check

Answer all 7 — your progress is saved on this device.

1. What should each variable represent?

   Any number that changes meaningOne unknown quantity with one fixed meaningOnly the final total

2. What is the correct translation of 2 notebooks and 3 pens costing $19?

   2n + 3p = 192n + 3p = 52 + 3np = 19

3. When doubling 2n + 3p = 19, what must happen?

   Double only 2nDouble only 19Double every term

4. Subtracting 4n + p = 23 from 4n + 6p = 38 leaves…

   4n = 155p = 157p = 61

5. If 5p = 15, what is p?

   5153

6. After p = 3, which original equation gives 4n = 20?

   4n + p = 232n + 3p = 19Both are impossible

7. Why is this page described as an optional algebra extension?

   Algebra has no checksA model drawing can also solve many two-unknown PSLE problemsVariables cannot represent prices

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