How To Solve Simultaneous Equations (Singapore Secondary Level Tutorial)

Solving simultaneous equations is one of those topics that keeps coming back in Secondary school maths – from Sec 2 all the way to O-Level E-Maths and even in A-Maths word problems.

If you’re in Singapore, you already know: MOE loves testing this in many forms.

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The good news? Once you master a few clear methods, most questions follow the same patterns. In this tutorial, I’ll walk you through:

• Step-by-step methods (elimination & substitution)
• How to choose the fastest method in exams
• Typical O-Level style twists (fractions, parameters, word problems)
• Practice questions with harder variants
• Common mistakes that cause you to lose marks

And if you want instant, 24/7 practice that’s aligned to the Singapore syllabus, I’ll also show you how to use Tutorly.sg effectively. It’s a website (not an app) that thousands of students in Singapore already use, and it’s even been mentioned on Channel NewsAsia (CNA).

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Step-by-step tutorial

Let’s focus on what you actually see in your Sec 2 / O-Level E-Maths paper: two equations with two unknowns.

Typical forms:

• Linear–linear (both equations are straight lines)
• Linear–nonlinear (e.g. one is linear, the other is quadratic)

We’ll start with the basics, then move to the trickier types.

1. Method 1: Elimination (your main workhorse)

You use elimination when it’s easy to make one variable cancel out.

Example 1 (basic O-Level style)

Solve:

$$\begin{cases} 2𝑥 + 3𝑦 = 12 \\ 5𝑥 - 3𝑦 = 3 \end{cases}$$

Step 1: Line up the equations

Write them one under the other:

1. 2𝑥 + 3𝑦 = 12
2. 5𝑥 - 3𝑦 = 3

Notice the +3𝑦 and -3𝑦. Perfect for elimination.

Step 2: Add or subtract to eliminate one variable

Add equation (1) and (2):

$$(2𝑥 + 3𝑦) + (5𝑥 - 3𝑦) = 12 + 3$$

3𝑦 and -3𝑦 cancel:

$$7𝑥 = 15$$

So:

$$𝑥 = \frac{15}{7}$$

Step 3: Substitute back to find the other variable

Use equation (1): 2𝑥 + 3𝑦 = 12

Substitute $𝑥 = \frac{15}{7}$:

$$2\left(\frac{15}{7}\right) + 3𝑦 = 12$$ $$\frac{30}{7} + 3𝑦 = 12$$

Move $\frac{30}{7}$ to the right:

$$3𝑦 = 12 - \frac{30}{7} = \frac{84}{7} - \frac{30}{7} = \frac{54}{7}$$

So:

$$𝑦 = \frac{54}{7} \div 3 = \frac{54}{21} = \frac{18}{7}$$

Final answer: $𝑥 = \frac{15}{7}$, $𝑦 = \frac{18}{7}$

In exams, don’t panic if you see fractions. MOE doesn’t always give nice integers.

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2. Method 2: Substitution (great when a variable is already “alone”)

Use substitution when one equation already has 𝑥 or 𝑦 alone, or can be made alone easily.

Example 2 (Sec 2 / O-Level)

Solve:

$$\begin{cases} 𝑦 = 2𝑥 + 1 \\ 3𝑥 + 2𝑦 = 17 \end{cases}$$

Step 1: Identify the easy equation

You already have 𝑦 = 2𝑥 + 1. That’s very convenient.

Step 2: Substitute into the other equation

In 3𝑥 + 2𝑦 = 17, replace 𝑦 with (2𝑥 + 1):

$$3𝑥 + 2(2𝑥 + 1) = 17$$

Expand:

$$3𝑥 + 4𝑥 + 2 = 17$$

Combine like terms:

$$7𝑥 + 2 = 17$$

So:

$$7𝑥 = 15 \Rightarrow 𝑥 = \frac{15}{7}$$

Step 3: Find the other variable

Use 𝑦 = 2𝑥 + 1:

$$𝑦 = 2\left(\frac{15}{7}\right) + 1 = \frac{30}{7} + \frac{7}{7} = \frac{37}{7}$$

Final answer: $𝑥 = \frac{15}{7}$, $𝑦 = \frac{37}{7}$

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3. How to choose: Elimination vs Substitution

When you see a question in your Sec 3/4 test or O-Level paper, ask yourself:

• Is any variable already alone (e.g. 𝑦 = 3𝑥 - 2)?
  • Yes → Use substitution, usually faster.
• Can I easily make coefficients match (e.g. 2𝑥 + 3𝑦 and 4𝑥 - 3𝑦)?
  • Yes → Use elimination.
• Are there many fractions or decimals?
  • Often elimination is cleaner if you first clear denominators.

You don’t get extra marks for choosing a “fancier” method. Choose the one that gives fewer chances to make careless mistakes.

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4. Fraction and decimal equations (very common in O-Levels)

MOE loves to throw in fractions to test your algebra discipline.

Example 3 (with fractions)

Solve:

$$\begin{cases} \frac{𝑥}{2} + \frac{𝑦}{3} = 5 \\ \frac{𝑥}{4} - \frac{𝑦}{6} = 1 \end{cases}$$

Step 1: Clear denominators

For the first equation, LCM of 2 and 3 is 6. Multiply the whole equation by 6:

$$3𝑥 + 2𝑦 = 30 \quad (1)$$

For the second, LCM of 4 and 6 is 12. Multiply by 12:

$$3𝑥 - 2𝑦 = 12 \quad (2)$$

Step 2: Eliminate

Now add (1) and (2):

$$(3𝑥 + 2𝑦) + (3𝑥 - 2𝑦) = 30 + 12$$ $$6𝑥 = 42 \Rightarrow 𝑥 = 7$$

Substitute 𝑥 = 7 into (1):

$$3(7) + 2𝑦 = 30 \Rightarrow 21 + 2𝑦 = 30 \Rightarrow 2𝑦 = 9 \Rightarrow 𝑦 = \frac{9}{2}$$

Final answer: 𝑥 = 7, $𝑦 = \frac{9}{2}$

Key idea: Clear fractions first, then use normal elimination. This is exactly the kind of manipulation you need for O-Level E-Maths Paper 1.

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5. Linear–quadratic simultaneous equations (harder exam type)

These show up in Sec 3/4 and O-Levels, often in coordinate geometry or word problems.

Example 4 (linear + quadratic)

Solve:

$$\begin{cases} 𝑦 = 𝑥 + 1 \\ 𝑥^2 + 𝑦^2 = 25 \end{cases}$$

Step 1: Substitute the linear into the quadratic

From 𝑦 = 𝑥 + 1, substitute into $𝑥^2 + 𝑦^2 = 25$:

$$𝑥^2 + (𝑥 + 1)^2 = 25$$

Expand:

$$𝑥^2 + (𝑥^2 + 2𝑥 + 1) = 25$$ $$2𝑥^2 + 2𝑥 + 1 = 25$$

Bring everything to one side:

$$2𝑥^2 + 2𝑥 + 1 - 25 = 0$$ $$2𝑥^2 + 2𝑥 - 24 = 0$$

Divide by 2:

$$𝑥^2 + 𝑥 - 12 = 0$$

Step 2: Solve the quadratic

Factorise:

$$𝑥^2 + 𝑥 - 12 = (𝑥 + 4)(𝑥 - 3) = 0$$

So:

• 𝑥 = -4 or 𝑥 = 3

Step 3: Find the corresponding 𝑦 values

If 𝑥 = -4:

$$𝑦 = 𝑥 + 1 = -4 + 1 = -3$$

If 𝑥 = 3:

$$𝑦 = 𝑥 + 1 = 4$$

Final answers:
(𝑥, 𝑦) = (-4, -3) and (3, 4)

For linear–quadratic questions, substitution is almost always the way to go.

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Exam strategy guide

Knowing the methods is one thing. Scoring well under O-Level exam pressure is another. Here’s how to approach simultaneous equations strategically.

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1. Read the question type carefully

You’ll typically see simultaneous equations in:

• Pure algebra questions: “Solve the simultaneous equations …”
• Word problems: speed, number of pens & pencils, price discounts, etc.
• Coordinate geometry: intersection of lines / line and curve

Identify which type it is first. That tells you how much time to spend and what form your final answer should take (e.g. (𝑥, 𝑦) coordinates vs context-based answer like “the number of boys is 12”).

2. Time management (especially for O-Level E-Maths)

• For a straightforward “solve these equations” question (2–3 marks), aim for 3–4 minutes.
• For word problems (4–6 marks), you may need 6–8 minutes, including forming the equations.

If you’re stuck for more than 2 minutes just forming the equations, move on and come back later. Don’t let one part destroy your whole Paper 2.

3. Decide your method quickly

When you see the equations:

• If one is already $𝑦 = \dots$ or $𝑥 = \dots$ → Substitution.
• If coefficients look like they can match easily (e.g. 2𝑥+3𝑦 and 4𝑥-3𝑦) → Elimination.
• If there are many fractions → Clear denominators, then Elimination.

Train yourself to decide in under 10 seconds. This reduces hesitation and panic.

4. Always check your answers (fast but effective)

For O-Level, you don’t have time to fully re-do the question, but you can:

• Substitute your final 𝑥 and 𝑦 quickly into both original equations.
• If one doesn’t match, you know there’s an error somewhere.

This is where a tool like Tutorly.sg helps. You can key in a similar practice question, type your final answer, and let the AI tutor show you the step-by-step solution. Compare your working to see where you went wrong and fix your pattern of mistakes before the exam.

5. For word problems: define variables clearly

MOE loves to hide simultaneous equations inside stories. The biggest trap is unclear variables.

Example pattern:

A shop sells pens at $2 each and markers at $3 each. Ali buys 10 items and spends $26.
How many pens and markers did he buy?

Let:

• 𝑥 = number of pens
• 𝑦 = number of markers

Then:

• 𝑥 + 𝑦 = 10 (total items)
• 2𝑥 + 3𝑦 = 26 (total cost)

From here, it’s a standard elimination question.

Exam tip:
Write your variable definitions in words. It can earn you method marks even if your algebra later has mistakes.

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Worksheet practice

Use this section like a mini worksheet. Try each question on your own first, then check the outline of the solution.

If you want fully worked solutions (step-by-step), you can enter similar questions into Tutorly.sg and see how to solve them clearly.

A. Basic practice (Sec 2 / early Sec 3 level)

Question 1
Solve:

$$\begin{cases} 3𝑥 + 𝑦 = 10 \\ 2𝑥 - 𝑦 = 1 \end{cases}$$

Outline of solution:

1. Add equations to eliminate 𝑦: $$(3𝑥 + 𝑦) + (2𝑥 - 𝑦) = 10 + 1 \Rightarrow 5𝑥 = 11$$ So $𝑥 = \frac{11}{5}$.
2. Substitute 𝑥 back into either equation to find 𝑦.

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Question 2
Solve:

$$\begin{cases} 𝑦 = 4𝑥 - 3 \\ 2𝑥 + 𝑦 = 11 \end{cases}$$

Outline:

1. Substitute 𝑦 = 4𝑥 - 3 into 2𝑥 + 𝑦 = 11.
2. Solve for 𝑥, then find 𝑦.

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B. Intermediate practice (fractions & decimals)

Question 3
Solve:

$$\begin{cases} \frac{2𝑥}{3} + \frac{𝑦}{2} = 7 \\ \frac{𝑥}{6} - \frac{𝑦}{4} = 1 \end{cases}$$

Outline:

1. Clear denominators for each equation:
   • First: multiply by 6 → 4𝑥 + 3𝑦 = 42
   • Second: multiply by 12 → 2𝑥 - 3𝑦 = 12
2. Use elimination (add equations) to find 𝑥.
3. Substitute back to find 𝑦.

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Question 4
Solve:

$$\begin{cases} 1.5𝑥 + 2𝑦 = 13 \\ 0.5𝑥 - 𝑦 = 1 \end{cases}$$

Outline:

1. Multiply both equations by 2 to remove decimals:
   • 3𝑥 + 4𝑦 = 26
   • 𝑥 - 2𝑦 = 2
2. Use elimination or substitution to solve.

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C. Harder exam variants (O-Level style)

These are closer to what you’ll see in O-Level E-Maths Paper 2, including parameters and linear–quadratic systems.

Question 5 (with parameter 𝑘)

Solve for 𝑥 and 𝑦 in terms of 𝑘:

$$\begin{cases} 2𝑥 + ky = 8 \\ 𝑥 - 3𝑦 = 1 \end{cases}$$

Outline:

1. From second equation: 𝑥 = 1 + 3𝑦.
2. Substitute into first: 2(1 + 3𝑦) + ky = 8.
3. Simplify: $2 + 6𝑦 + ky = 8 \Rightarrow (6 + 𝑘)𝑦 = 6$.
4. So $𝑦 = \dfrac{6}{6 + 𝑘}$ (state restriction $𝑘 \neq -6$).
5. Then $𝑥 = 1 + 3𝑦 = 1 + \dfrac{18}{6 + 𝑘}$.

This kind of “in terms of 𝑘” question is common in Sec 4 and O-Levels to test algebra manipulation.

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Question 6 (linear–quadratic)

Solve:

$$\begin{cases} 𝑦 = 2𝑥 - 1 \\ 𝑥^2 + xy = 15 \end{cases}$$

Outline:

1. Substitute 𝑦 = 2𝑥 - 1 into $𝑥^2 + xy = 15$: $$𝑥^2 + 𝑥(2𝑥 - 1) = 15$$
2. Expand: $𝑥^2 + 2𝑥^2 - 𝑥 = 15 \Rightarrow 3𝑥^2 - 𝑥 - 15 = 0$.
3. Solve the quadratic (factorisation or formula).
4. For each 𝑥 value, find the corresponding 𝑦 using 𝑦 = 2𝑥 - 1.
5. State both pairs (𝑥, 𝑦).

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Question 7 (word problem – number of items)

A bookshop sells files at $3 each and notebooks at $2 each. A student buys 9 items and spends $22.

(a) Form two simultaneous equations.
(b) Hence, find how many files and notebooks the student bought.

Outline:

Let:

• 𝑥 = number of files
• 𝑦 = number of notebooks

(a)

• Total items: 𝑥 + 𝑦 = 9
• Total cost: 3𝑥 + 2𝑦 = 22

(b)
Use elimination:

• From 𝑥 + 𝑦 = 9, get 𝑦 = 9 - 𝑥.
• Substitute into 3𝑥 + 2𝑦 = 22: $$3𝑥 + 2(9 - 𝑥) = 22$$
• Solve for 𝑥 (files), then find 𝑦 (notebooks).

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Question 8 (harder word problem – speed)

Two cyclists, A and B, travel between Town P and Town Q, a distance of 60 km.

• Cyclist A travels from P to Q at 𝑥 km/h and takes 3 hours less than Cyclist B.
• Cyclist B travels at (𝑥 - 10) km/h.

(a) Write down an expression for the time taken by each cyclist.
(b) Form a pair of simultaneous equations in 𝑥 and the time taken by B.
(c) Hence, find the value of 𝑥.

Outline:

(a)

• Time taken by A: $\dfrac{60}{𝑥}$
• Time taken by B: $\dfrac{60}{𝑥 - 10}$

Given A takes 3 hours less than B:

$$\frac{60}{𝑥} = \frac{60}{𝑥 - 10} - 3$$

This is already one equation in 𝑥.

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You can also introduce a second variable 𝑡 (time taken by B), then relate it:

• $𝑡 = \dfrac{60}{𝑥 - 10}$
• $𝑡 - 3 = \dfrac{60}{𝑥}$

From there, you can form two equations in 𝑥 and 𝑡. Then eliminate 𝑡 to get a quadratic in 𝑥.

This is the kind of multi-step question where practice really matters. If you struggle with forming the equations, try similar word problems on Tutorly.sg and let the AI tutor walk you through the reasoning.

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D. How to use Tutorly.sg for extra practice

If you’re serious about improving before your Sec 3/4 exams or O-Levels, you need consistent practice, not just last-minute cramming.

On Tutorly.sg:

• You can ask it to generate new simultaneous equations questions at your level (Sec 2 / Sec 3 / O-Level).
• After you try the question on your own, type in your final answer.
• If it’s wrong, Tutorly will show you a clear, step-by-step worked solution so you can see exactly how to do it.
• You can then ask for similar questions to drill your weak spots (e.g. “more with fractions”, “more linear-quadratic ones”, “more word problems”).

Because it’s available 24/7 as a website, you can revise whenever you have a bit of free time – between CCA, tuition, or on weekends.

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Common mistakes

Even strong students lose marks on simultaneous equations because of small, repeated errors. Here are the big ones to avoid.

1. Sign errors when adding/subtracting equations

Example:

$$\begin{cases} 2𝑥 + 3𝑦 = 11 \\ 4𝑥 - 3𝑦 = 5 \end{cases}$$

If you add them:

$$(2𝑥 + 3𝑦) + (4𝑥 - 3𝑦) = 11 + 5 \Rightarrow 6𝑥 = 16$$

Some students accidentally write 6𝑥 + 6𝑦 = 16 because they forget that +3𝑦 and -3𝑦 cancel.

Fix:

• Write the equations neatly, line up 𝑥 and 𝑦 terms.
• Use a light pencil mark to show which terms are cancelling.

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2. Forgetting to substitute back to find the second variable

Sometimes you solve for 𝑥 and then rush to the next question, forgetting to find 𝑦.

Fix:
Train yourself: every time you get one variable, immediately write “Find 𝑦:” (or 𝑥) and do it. Make it a habit.

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3. Messy handling of fractions

When you have:

$$\frac{𝑥}{2} + \frac{𝑦}{3} = 4$$

Some students try to do elimination directly and get lost.

Fix:

1. Always clear denominators first:
   • LCM of 2 and 3 is 6 → multiply whole equation by 6: $$3𝑥 + 2𝑦 = 24$$
2. Then treat it like a normal integer-coefficient equation.

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4. Wrong substitution (mixing up expressions)

Example:

$$\begin{cases} 𝑦 = 3𝑥 - 2 \\ 2𝑥 + 𝑦 = 10 \end{cases}$$

Some students copy wrongly as 𝑦 = 3 - 2𝑥 or substitute 𝑦 = 3𝑥 + 2.

Fix:

• When substituting, circle the expression you’re replacing.
• Rewrite the second equation with brackets: 2𝑥 + (3𝑥 - 2) = 10.
• Expand step by step.

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5. Not checking for “no solution” or “infinite solutions”

Sometimes, after elimination, you may get:

• 0 = 5 → No solution (inconsistent equations, lines are parallel)
• 0 = 0 → Infinitely many solutions (same line)

MOE occasionally tests this concept.

Example:

$$\begin{cases} 2𝑥 + 4𝑦 = 8 \\ 𝑥 + 2𝑦 = 4 \end{cases}$$

If you multiply the second equation by 2, you get the first one. So they represent the same line → infinitely many solutions.

Fix:

• If variables disappear and you get a false statement → “No solution”.
• If variables disappear and you get a true statement → “Infinitely many solutions”.

State this clearly in your final answer.

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6. Careless algebra in linear–quadratic questions

When you substitute into a quadratic, one small expansion error can ruin everything.

Fix:

• Write $(x

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