Simultaneous Equations

Section 1 — What Are Simultaneous Equations?

A simultaneous equation problem gives you two equations containing two unknown variables (usually x and y). The goal is to find the one pair of values (x, y) that satisfies both equations at the same time.

Why two equations? One equation with two unknowns has infinitely many solutions. For example, x + y = 5 is satisfied by (1, 4), (2, 3), (0, 5), and so on. A second equation narrows it down to exactly one pair (in most cases).

What does “satisfies both” mean? When you substitute your answer into both equations, both sides must balance.

Example to illustrate

Given:

x + y = 5 … (1) x - y = 1 … (2)

If x = 3, y = 2:

Equation (1): 3 + 2 = 5 ✓
Equation (2): 3 - 2 = 1 ✓

So x = 3, y = 2 is the simultaneous solution.

If x = 4, y = 1:

Equation (1): 4 + 1 = 5 ✓
Equation (2): 4 - 1 = 3 ✗

So (4, 1) satisfies equation (1) but not equation (2) — it is not a solution to the simultaneous equations.

Section 2 — Method 1: Substitution

How It Works

You rearrange one equation to express one variable in terms of the other, then substitute that expression into the second equation. This turns two equations with two unknowns into one equation with one unknown, which you can solve directly.

Step-by-Step Procedure

Label the equations (1) and (2).
Choose one equation and rearrange it to make x (or y) the subject.
Substitute that expression into the other equation.
Solve the resulting equation for the remaining variable.
Substitute that value back to find the other variable.
State the answer clearly and verify in both original equations.

Worked Example 1 (Straightforward)

Solve:

2x + y = 7 … (1) x - y = 2 … (2)

Step 1 — Rearrange equation (2) to make x the subject:

From (2): x = 2 + y … (3)

Step 2 — Substitute (3) into (1):

2(2 + y) + y = 7

4 + 2y + y = 7

4 + 3y = 7

3y = 3

y = 1

Step 3 — Substitute y = 1 into (3):

x = 2 + 1 = 3

Answer: x = 3, y = 1

Verification:

(1): 2(3) + 1 = 6 + 1 = 7 ✓
(2): 3 - 1 = 2 ✓

Worked Example 2 (Substituting into the y-variable)

Solve:

3x + 2y = 16 … (1) y = x + 1 … (2)

Step 1 — Equation (2) already gives y in terms of x. Substitute into (1):

3x + 2(x + 1) = 16

3x + 2x + 2 = 16

5x + 2 = 16

5x = 14

x = 14/5 = 2.8

Step 2 — Substitute x = 2.8 into (2):

y = 2.8 + 1 = 3.8

Answer: x = 2.8, y = 3.8

Verification:

(1): 3(2.8) + 2(3.8) = 8.4 + 7.6 = 16 ✓
(2): y = 2.8 + 1 = 3.8 ✓

Worked Example 3 (Negative Coefficients)

Solve:

5x - 2y = 11 … (1) x + 3y = -1 … (2)

Step 1 — Rearrange (2) to make x the subject:

x = -1 - 3y … (3)

Step 2 — Substitute (3) into (1):

5(-1 - 3y) - 2y = 11

-5 - 15y - 2y = 11

-5 - 17y = 11

-17y = 11 + 5

-17y = 16

y = 16 / (-17)

y = -16/17

Step 3 — Substitute y = -16/17 into (3):

x = -1 - 3(-16/17)

x = -1 + 48/17

x = -17/17 + 48/17

x = 31/17

Answer: x = 31/17, y = -16/17

Verification:

(1): 5(31/17) - 2(-16/17) = 155/17 + 32/17 = 187/17 = 11 ✓
(2): 31/17 + 3(-16/17) = 31/17 - 48/17 = -17/17 = -1 ✓

Note: Fractional answers are perfectly valid — do not round unless the question asks for it.

Section 3 — Method 2: Elimination

How It Works

You add or subtract the two equations to eliminate one variable. For this to work, the variable you want to eliminate must have the same coefficient in both equations (or coefficients that are negatives of each other). If they don’t match, you multiply one or both equations by a suitable constant first.

Step-by-Step Procedure

Label the equations (1) and (2).
Decide which variable to eliminate.
If needed, multiply one or both equations by constants so that the chosen variable has the same coefficient in both.
Add the equations (if the matching coefficients have opposite signs) or subtract one from the other (if the matching coefficients have the same sign).
Solve for the remaining variable.
Substitute back to find the other variable.
State the answer and verify.

Worked Example 4 (Direct elimination by subtraction)

Solve:

3x + 2y = 13 … (1) x + 2y = 7 … (2)

Step 1 — Both equations have 2y. Subtract (2) from (1) to eliminate y:

(3x + 2y) - (x + 2y) = 13 - 7

3x + 2y - x - 2y = 6

2x = 6

x = 3

Step 2 — Substitute x = 3 into (2):

3 + 2y = 7

2y = 4

y = 2

Answer: x = 3, y = 2

Verification:

(1): 3(3) + 2(2) = 9 + 4 = 13 ✓
(2): 3 + 2(2) = 3 + 4 = 7 ✓

Worked Example 5 (Direct elimination by addition)

Solve:

4x + 3y = 10 … (1) 2x - 3y = 2 … (2)

Step 1 — The y-terms are +3y and -3y. Add the equations to eliminate y:

(4x + 3y) + (2x - 3y) = 10 + 2

6x = 12

x = 2

Step 2 — Substitute x = 2 into (1):

4(2) + 3y = 10

8 + 3y = 10

3y = 2

y = 2/3

Answer: x = 2, y = 2/3

Verification:

(1): 4(2) + 3(2/3) = 8 + 2 = 10 ✓
(2): 2(2) - 3(2/3) = 4 - 2 = 2 ✓

Worked Example 6 (Multiplication needed to align coefficients)

Solve:

2x + 3y = 12 … (1) 5x - 2y = 11 … (2)

Step 1 — Neither variable has matching coefficients. Plan:

To eliminate y: multiply (1) by 2 and (2) by 3 so both have a y-coefficient of 6 (with opposite signs, so add).
Alternatively, to eliminate x: multiply (1) by 5 and (2) by 2 so both have an x-coefficient of 10 (with same sign, so subtract).

Using the y-route:

(1) × 2: 4x + 6y = 24 … (3)

(2) × 3: 15x - 6y = 33 … (4)

Step 2 — Add (3) and (4) to eliminate y:

4x + 6y + 15x - 6y = 24 + 33

19x = 57

x = 3

Step 3 — Substitute x = 3 into (1):

2(3) + 3y = 12

6 + 3y = 12

3y = 6

y = 2

Answer: x = 3, y = 2

Verification:

(1): 2(3) + 3(2) = 6 + 6 = 12 ✓
(2): 5(3) - 2(2) = 15 - 4 = 11 ✓

Tip: When multiplying to align coefficients, use the LCM of the two coefficients for the variable you want to eliminate.

Section 4 — Graphical Method

Concept

Each linear equation in two unknowns represents a straight line when drawn on a coordinate plane. The simultaneous solution is the point of intersection of the two lines.

How to Use the Graphical Method

Rearrange each equation into the form y = mx + c.
Choose at least two x-values, calculate the corresponding y-values, and plot the points for each line.
Draw both straight lines accurately on the same axes.
Read off the coordinates of the point where the lines cross — that is the solution.

Example

Solve graphically:

y = 2x - 1 … (1) y = -x + 5 … (2)

For line (1), y = 2x - 1:

x	0	1	3
y	-1	1	5

For line (2), y = -x + 5:

x	0	2	5
y	5	3	0

Plot both lines. They intersect at (2, 3).

So x = 2, y = 3.

Verify:

(1): 3 = 2(2) - 1 = 3 ✓
(2): 3 = -2 + 5 = 3 ✓

When to Use the Graphical Method

When the question specifically asks for a graphical solution.
When you want a visual check of an algebraic answer.
When the coefficients are awkward and you only need an approximate answer.

Limitation: Reading coordinates off a graph introduces error. For exact answers, always use substitution or elimination. In exams, if a graph is given, use it; otherwise use algebra.

Section 5 — Verification

Always substitute your answer back into BOTH original equations.

This is not optional — it is part of the solution. Exam mark schemes often include a method mark for checking.

Why Both Equations?

It is possible to make an error that still satisfies one equation but not the other. Checking only one equation gives you a false sense of security.

Verification Routine

After finding x = a, y = b:

Write “Check in equation (1):” and substitute x = a, y = b. State whether LHS = RHS.
Write “Check in equation (2):” and substitute x = a, y = b. State whether LHS = RHS.
If both check out, the answer is correct.

If one equation does not balance, go back and find the arithmetic error — do not just adjust the answer to make it fit.

Section 6 — Word Problems

Strategy for Setting Up Simultaneous Equations

Read carefully — identify what you are being asked to find.
Assign variables — let x = (first unknown) and y = (second unknown). State this clearly.
Translate each piece of information into an equation.
Solve using substitution or elimination.
Answer the question in a sentence, with units if applicable.
Verify in the context of the problem, not just the algebra.

Word Problem 1 — Age Problem

Question: The sum of Ali’s age and Beng’s age is 35. Ali is 7 years older than Beng. Find their ages.

Let x = Ali’s age and y = Beng’s age.

Equation (1): x + y = 35

Equation (2): x - y = 7 (Ali is 7 years older than Beng)

Solve by elimination — add (1) and (2):

2x = 42

x = 21

Substitute into (1):

21 + y = 35

y = 14

Answer: Ali is 21 years old and Beng is 14 years old.

Check in context: 21 + 14 = 35 ✓ and 21 - 14 = 7 ✓

Word Problem 2 — Cost Problem

Question: 3 pens and 4 erasers cost $5.00. 2 pens and 1 eraser cost $3.00. Find the cost of one pen and one eraser.

Let x = cost of one pen (cents) and y = cost of one eraser (cents).

Equation (1): 3x + 4y = 500

Equation (2): 2x + y = 300

Rearrange (2): y = 300 - 2x … (3)

Substitute (3) into (1):

3x + 4(300 - 2x) = 500

3x + 1200 - 8x = 500

-5x = -700

x = 140

y = 300 - 2(140) = 20

Answer: One pen costs $1.40 and one eraser costs $0.20.

Check in context:

(1): 3(140) + 4(20) = 500 ✓
(2): 2(140) + 20 = 300 ✓

Word Problem 3 — Coin Problem

Question: A jar contains 20c and 50c coins. There are 18 coins altogether worth $6.00 total. How many of each coin are there?

Let x = number of 20c coins, y = number of 50c coins.

Equation (1): x + y = 18 (total number of coins)

Equation (2): 20x + 50y = 600 (total value in cents)

Simplify (2) by dividing by 10:

2x + 5y = 60 … (2’)

From (1): x = 18 - y … (3)

Substitute (3) into (2’):

2(18 - y) + 5y = 60

36 - 2y + 5y = 60

36 + 3y = 60

3y = 24

y = 8

x = 18 - 8 = 10

Answer: There are 10 twenty-cent coins and 8 fifty-cent coins.

Check in context:

Total coins: 10 + 8 = 18 ✓
Total value: 10(20) + 8(50) = 200 + 400 = 600 cents = $6.00 ✓

Word Problem 4 — Distance / Speed Problem

Question: Two cyclists set off from the same place in the same direction. Cyclist A travels at a constant speed of x km/h and Cyclist B at y km/h. After 3 hours, A is 12 km ahead of B. If they had both started from the same place and gone in opposite directions, after 2 hours they would be 50 km apart. Find the speed of each cyclist.

Setting up the equations:

After 3 hours travelling in the same direction:

Distance A travels - Distance B travels = 12

3x - 3y = 12

Divide by 3: x - y = 4 … (1)

After 2 hours travelling in opposite directions:

Distance A + Distance B = 50

2x + 2y = 50

Divide by 2: x + y = 25 … (2)

Add (1) and (2):

2x = 29

x = 14.5

Substitute into (2):

14.5 + y = 25

y = 10.5

Answer: Cyclist A travels at 14.5 km/h and Cyclist B travels at 10.5 km/h.

Check:

(1): 14.5 - 10.5 = 4 ✓
(2): 14.5 + 10.5 = 25 ✓

Word Problem 5 — Number Problem

Question: The sum of two numbers is 84. Three times the larger number minus twice the smaller number equals 152. Find both numbers.

Let x = the larger number, y = the smaller number.

Equation (1): x + y = 84

Equation (2): 3x - 2y = 152

From (1): x = 84 - y … (3)

Substitute (3) into (2):

3(84 - y) - 2y = 152

252 - 3y - 2y = 152

252 - 5y = 152

-5y = -100

y = 20

x = 84 - 20 = 64

Answer: The two numbers are 64 and 20.

Check:

(1): 64 + 20 = 84 ✓
(2): 3(64) - 2(20) = 192 - 40 = 152 ✓

Section 7 — Choosing Between Substitution and Elimination

Both methods always give the same answer. The choice is about efficiency.

Situation	Recommended Method
One equation already has a lone variable (e.g. y = 3x - 1 or x = 5)	Substitution — no rearranging needed
One variable in one equation has coefficient 1 or -1	Substitution — easy to rearrange
Both equations are in ax + by = c form with “nice” coefficients	Elimination
The coefficients of one variable are already equal (or negatives)	Elimination — no multiplication needed
You need to multiply both equations by large numbers to align coefficients	Consider substitution instead

Quick Decision Rule

If you can isolate x or y in one step with no fractions, use substitution.
If adding or subtracting the equations directly removes a variable, use elimination.
If neither is obvious, use elimination with multiplication — it is generally safer.

Section 8 — Special Cases

Case 1 — No Solution (Parallel Lines)

If the two equations represent parallel lines, they never intersect. There is no pair (x, y) that satisfies both equations.

How to recognise algebraically: When you attempt to eliminate a variable, both variables disappear and you are left with a false statement such as 0 = 5.

Example:

2x + 4y = 8 … (1) x + 2y = 7 … (2)

Multiply (2) by 2: 2x + 4y = 14 … (3)

Subtract (1) from (3): 0 = 6

This is false — no solution exists. The lines are parallel (notice that 2x + 4y = 8 simplifies to x + 2y = 4, while equation (2) is x + 2y = 7; same left-hand side, different right-hand side).

Graphically: The two lines have the same gradient but different y-intercepts. They never meet.

Case 2 — Infinitely Many Solutions (Same Line)

If the two equations are multiples of each other, they represent the same line. Every point on the line satisfies both equations.

How to recognise algebraically: When you eliminate a variable, both variables disappear and you are left with a true statement such as 0 = 0.

Example:

3x + 6y = 12 … (1) x + 2y = 4 … (2)

Multiply (2) by 3: 3x + 6y = 12 … (3)

Subtract (1) from (3): 0 = 0

This is always true — infinitely many solutions. Equation (1) is simply 3 × equation (2); they are the same line.

Graphically: The two lines lie exactly on top of each other.

Summary of Cases

Algebraic outcome	Geometric meaning	Number of solutions
Unique values of x and y	Lines intersect at one point	Exactly one
False statement (e.g. 0 = 5)	Lines are parallel	None
True identity (e.g. 0 = 0)	Lines are the same	Infinitely many

Section 9 — Common Errors

Error 1 — Losing a Negative Sign When Substituting

Wrong: From x = 3 - 2y, substituting into 4x + y = 5:

4(3 - 2y) + y = 5

12 - 2y + y = 5 ← forgot to multiply 4 × (-2y) correctly

Correct: 4(3 - 2y) + y = 5

12 - 8y + y = 5

12 - 7y = 5

Rule: When substituting an expression with subtraction, distribute the coefficient carefully to every term inside the bracket.

Error 2 — Subtracting Equations Incorrectly

Wrong: (3x + 2y) - (x + 2y) = 10 - 4

3x + 2y - x + 2y = 6 ← sign error: -x + 2y should be -x - 2y

Correct: 3x + 2y - x - 2y = 6

2x = 6

Rule: When subtracting an equation, change the sign of every term in the equation being subtracted.

Error 3 — Forgetting to Substitute Back

After finding one variable, some students stop there. You must substitute back to find the other variable and give a complete answer.

Wrong answer: “x = 4”

Correct answer: “x = 4, y = -1”

Error 4 — Substituting Back into a Derived Equation Instead of an Original

If you substitute back to find the second variable, use one of the original equations — not an equation you derived (e.g., after multiplying through), as arithmetic errors during manipulation can make a derived equation wrong.

Error 5 — Not Checking in BOTH Original Equations

Even if your answer looks reasonable, always verify in both equations. A consistent error across both equations will not be caught if you only check one.

Error 6 — Incorrect Handling of Fractions After Division

When dividing to isolate a variable, keep fractions exact rather than rounding prematurely.

Wrong: y = 7/3 ≈ 2.3, then use 2.3 in subsequent steps (rounding error compounds).

Correct: Keep y = 7/3 and substitute the exact fraction.

Section 10 — Practice Questions

Pure Solving (Questions 1–6)

Question 1 Solve the simultaneous equations:

3x + y = 11 x - y = 1

Question 2 Solve the simultaneous equations:

2x + 5y = 19 3x - 5y = 6

Question 3 Solve the simultaneous equations:

y = 3x - 4 2x + y = 11

Question 4 Solve the simultaneous equations:

4x + 3y = 10 2x - y = 4

Question 5 Solve the simultaneous equations:

3x + 4y = 5 2x + 3y = 4

Question 6 Solve the simultaneous equations:

5x - 2y = 7 3x + 4y = -1

Word Problems (Questions 7–10)

Question 7 The perimeter of a rectangle is 54 cm. The length is 9 cm more than the width. Find the length and the width.

Question 8 Two numbers differ by 13. Twice the larger number added to three times the smaller number equals 121. Find the two numbers.

Question 9 A school tuck shop sells chicken rice at $x each and mee goreng at $y each. On Monday, 25 plates of chicken rice and 15 plates of mee goreng were sold for $97.50. On Tuesday, 10 plates of chicken rice and 30 plates of mee goreng were sold for $87.00. Find the price of each dish.

Question 10 The sum of the digits of a two-digit number is 9. When the digits are reversed, the new number is 27 more than the original number. Find the original two-digit number.

Full Answers

Answer 1

3x + y = 11 … (1) x - y = 1 … (2)

Add (1) and (2):

4x = 12

x = 3

Substitute into (2): 3 - y = 1, so y = 2

x = 3, y = 2

Check: (1) 3(3) + 2 = 11 ✓ | (2) 3 - 2 = 1 ✓

Answer 2

2x + 5y = 19 … (1) 3x - 5y = 6 … (2)

Add (1) and (2) (the y-terms cancel):

5x = 25

x = 5

Substitute into (1): 2(5) + 5y = 19 → 10 + 5y = 19 → 5y = 9 → y = 9/5

x = 5, y = 9/5

Check: (1) 10 + 5(9/5) = 10 + 9 = 19 ✓ | (2) 15 - 5(9/5) = 15 - 9 = 6 ✓

Answer 3

y = 3x - 4 … (1) 2x + y = 11 … (2)

Substitute (1) into (2):

2x + (3x - 4) = 11

5x - 4 = 11

5x = 15

x = 3

y = 3(3) - 4 = 9 - 4 = 5

x = 3, y = 5

Check: (1) 5 = 3(3) - 4 = 5 ✓ | (2) 2(3) + 5 = 11 ✓

Answer 4

4x + 3y = 10 … (1) 2x - y = 4 … (2)

From (2): y = 2x - 4 … (3)

Substitute (3) into (1):

4x + 3(2x - 4) = 10

4x + 6x - 12 = 10

10x = 22

x = 2.2

y = 2(2.2) - 4 = 4.4 - 4 = 0.4

x = 2.2, y = 0.4

Check: (1) 4(2.2) + 3(0.4) = 8.8 + 1.2 = 10 ✓ | (2) 2(2.2) - 0.4 = 4.4 - 0.4 = 4 ✓

Answer 5

3x + 4y = 5 … (1) 2x + 3y = 4 … (2)

Eliminate x: multiply (1) by 2 and (2) by 3:

6x + 8y = 10 … (3) 6x + 9y = 12 … (4)

Subtract (3) from (4):

y = 2

Substitute into (2): 2x + 3(2) = 4 → 2x = -2 → x = -1

x = -1, y = 2

Check: (1) 3(-1) + 4(2) = -3 + 8 = 5 ✓ | (2) 2(-1) + 3(2) = -2 + 6 = 4 ✓

Answer 6

5x - 2y = 7 … (1) 3x + 4y = -1 … (2)

Multiply (1) by 2: 10x - 4y = 14 … (3)

Add (3) and (2) to eliminate y:

13x = 13

x = 1

Substitute into (1): 5(1) - 2y = 7 → -2y = 2 → y = -1

x = 1, y = -1

Check: (1) 5(1) - 2(-1) = 5 + 2 = 7 ✓ | (2) 3(1) + 4(-1) = 3 - 4 = -1 ✓

Answer 7

Let x = length (cm) and y = width (cm).

Perimeter: 2x + 2y = 54 → x + y = 27 … (1)

Length is 9 cm more than width: x - y = 9 … (2)

Add (1) and (2): 2x = 36 → x = 18

y = 27 - 18 = 9

Length = 18 cm, Width = 9 cm

Check: 2(18) + 2(9) = 36 + 18 = 54 ✓ | 18 - 9 = 9 ✓

Answer 8

Let x = the larger number, y = the smaller number.

x - y = 13 … (1)

2x + 3y = 121 … (2)

From (1): x = y + 13 … (3)

Substitute (3) into (2):

2(y + 13) + 3y = 121

2y + 26 + 3y = 121

5y = 95

y = 19

x = 19 + 13 = 32

The two numbers are 32 and 19.

Check: 32 - 19 = 13 ✓ | 2(32) + 3(19) = 64 + 57 = 121 ✓

Answer 9

Let x = price of chicken rice ($) and y = price of mee goreng ($).

25x + 15y = 97.50 … (1)

10x + 30y = 87.00 … (2)

Simplify (1) by dividing by 5: 5x + 3y = 19.50 … (1’)

Simplify (2) by dividing by 10: x + 3y = 8.70 … (2’)

Subtract (2’) from (1’): 4x = 10.80 → x = 2.70

Substitute into (2’): 2.70 + 3y = 8.70 → 3y = 6.00 → y = 2.00

Chicken rice costs $2.70 and mee goreng costs $2.00.

Check: (1) 25(2.70) + 15(2.00) = 67.50 + 30.00 = 97.50 ✓ | (2) 10(2.70) + 30(2.00) = 27.00 + 60.00 = 87.00 ✓

Answer 10

Let the tens digit = x and the units digit = y. Original number = 10x + y. Reversed number = 10y + x.

Sum of digits: x + y = 9 … (1)

Reversed number is 27 more: (10y + x) - (10x + y) = 27

9y - 9x = 27

y - x = 3 … (2)

Add (1) and (2): 2y = 12 → y = 6

x = 9 - 6 = 3

Original number = 10(3) + 6 = 36.

The original two-digit number is 36.

Check: 3 + 6 = 9 ✓ | Reversed: 63. 63 - 36 = 27 ✓

Quick Reference Summary

Topic	Key Point
What they are	Two equations, two unknowns — find the pair satisfying BOTH
Substitution	Isolate one variable, substitute into the other equation
Elimination	Match coefficients (multiply if needed), add or subtract to remove one variable
Graphical	Point of intersection of two straight lines
Verify	Check answer in BOTH original equations
No solution	Elimination gives a false statement; lines are parallel
Infinite solutions	Elimination gives 0 = 0; equations are the same line
Best choice	Substitution if one variable is already isolated; elimination if coefficients are easy to match