Simultaneous equations are not two unrelated algebra questions placed side by side. They describe two conditions that must be true at the same time. The solution is the value or pair of values that satisfies both relationships together.
This eighteenth Secondary 4 Mathematics Learning Guide develops simultaneous linear equations through substitution, elimination, graph intersections and real-world modelling. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
Current syllabus connection: solving simultaneous linear equations and interpreting intersections form part of the Number and Algebra and graph systems used in the 2026 O-Level Mathematics 4052 and 2027 SEC G3 Mathematics K310 syllabuses. The examples below are original teaching material.
What makes the equations simultaneous?
Consider:
x + y = 10
2x + y = 14
The first equation describes every pair whose sum is 10. The second describes every pair whose weighted sum is 14. The simultaneous solution is the pair that belongs to both sets of solutions.
Graphically, each equation is a line. The solution is their intersection. Algebraically, elimination or substitution finds the same point without drawing the full graph.
Elimination: remove one variable by combining equations
Elimination works when the coefficients of one variable are equal or can be made equal. Adding or subtracting the equations then removes that variable.
Worked Example 1 | Direct elimination
Solve:
x + y = 10
2x + y = 14
Subtract the first equation from the second:
x = 4.
Substitute into x + y = 10:
4 + y = 10, so y = 6.
The solution is (4, 6). Verify in both equations: 4+6=10 and 2(4)+6=14.
Substitution: replace a variable with an equivalent expression
Substitution is especially efficient when one equation already gives a variable in terms of the other, such as y = 3x − 2.
The replacement must preserve brackets. If y = 3x − 2 and another equation contains 2y, substitute 2(3x − 2), not 6x − 2.
Worked Example 2 | Substitution
Solve:
y = 3x − 2
2x + y = 18
Substitute 3x − 2 for y:
2x + 3x − 2 = 18
5x = 20
x = 4.
Then y = 3(4) − 2 = 10.
The solution is (4, 10). Check in the second equation: 2(4)+10=18.
Choose the method that creates the shortest stable route
| Equation structure | Useful method |
|---|---|
| One variable already isolated | Substitution |
| Matching or easily matched coefficients | Elimination |
| Graph already supplied | Read intersection if graphical solution is requested |
| Context produces equations naturally in totals | Either method after modelling |
The “best” method is not the one with the most steps. It is the one that preserves the relationships clearly and can be verified easily.
Worked Example 3 | Elimination after scaling
Solve:
2x + 3y = 17
5x − 2y = 4
To eliminate y, multiply the first equation by 2 and the second by 3:
4x + 6y = 34
15x − 6y = 12.
Add:
19x = 46, so x = 46/19.
Substitute into the first equation:
2(46/19)+3y=17.
92/19+3y=323/19.
3y=231/19.
y=77/19.
Exact fractions are valid. A system does not need integer answers to be correct.
Model first: define what the variables mean
In word problems, solving is usually not the first difficult step. The hard part is building two faithful equations.
Write variable definitions with units. For example, let a be the number of adult tickets and c the number of child tickets. Then a+c=total tickets is meaningful because both variables count tickets. A money equation such as 12a+7c=total revenue carries dollars through the coefficients.
Worked Example 4 | Ticket model
A theatre sells 120 tickets. Adult tickets cost $15 and student tickets cost $9. Total revenue is $1512. How many of each ticket were sold?
Let a be adult tickets and s be student tickets.
a + s = 120
15a + 9s = 1512
Multiply the first equation by 9:
9a + 9s = 1080.
Subtract from the revenue equation:
6a = 432, so a = 72.
Then s = 48. Therefore 72 adult tickets and 48 student tickets were sold.
Check revenue: 72×15 + 48×9 = 1080 + 432 = 1512.
Worked Example 5 | Mixture model
A shop blends two types of coffee costing $12/kg and $18/kg to make 10 kg of a blend costing $15/kg. Find the mass of each type.
Let x kg be the $12 coffee and y kg be the $18 coffee.
x + y = 10
12x + 18y = 150
Multiply the first equation by 12:
12x + 12y = 120.
Subtract:
6y = 30, so y = 5 and x = 5.
The blend uses 5 kg of each coffee. Check that the weighted average lies exactly halfway between $12 and $18 because the masses are equal.
Graphs: the same solution appears as an intersection
Rewrite two linear equations in graphable form. For example:
y = 10 − x
y = 14 − 2x
Their intersection is (4,6), the same solution found algebraically in Worked Example 1.
A graphical answer may be approximate depending on scale and reading precision. Algebra can provide an exact pair when the equations permit it.
Worked Example 6 | Interpret an intersection
Plan A costs $10 plus $0.40 per unit. Plan B costs $22 plus $0.20 per unit. At what usage do the plans cost the same?
Let x be usage units and y be total cost.
y = 10 + 0.40x
y = 22 + 0.20x
Set the expressions equal:
10 + 0.40x = 22 + 0.20x
0.20x = 12
x = 60.
At 60 units, both plans cost $34. The point (60,34) is the intersection of their cost graphs.
The break-even point is not the end of interpretation. Below 60 units, Plan A is cheaper; above 60, Plan B is cheaper. A graph makes that changing comparison visible.
No solution and infinitely many solutions
Two distinct parallel lines never intersect, so their simultaneous equations have no solution.
If two equations describe exactly the same line, every point on that line satisfies both, so there are infinitely many solutions.
These cases can be detected algebraically when elimination removes both variables and produces either a false statement such as 0=5 or a true identity such as 0=0.
Worked Example 7 | No solution
Solve:
2x + y = 5
4x + 2y = 13
Doubling the first equation gives 4x+2y=10, but the second says 4x+2y=13. The same left-hand expression cannot equal both 10 and 13. Therefore there is no solution.
Graphically, the lines have the same gradient but different intercepts.
Worked Example 8 | Same line
Consider:
3x + 6y = 12
x + 2y = 4
The first equation is exactly three times the second. They describe the same line, so there are infinitely many common solutions.
A problem asking for one unique pair would therefore be underdetermined unless another independent condition is supplied.
Units help verify models
In the ticket model, a+s counts tickets, while 15a+9s counts dollars. Adding 15a and s directly would mix dollars and tickets and signal a modelling error.
Dimensional consistency is therefore a useful check before the algebra becomes complicated.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Eliminates by adding when subtraction is needed | Signs not inspected | Write the target coefficients directly above one another |
| Substitutes y=3x−2 into 2y as 6x−2 | Whole expression not bracketed | Write 2(3x−2) first |
| Finds x but forgets y | Intermediate answer mistaken for full solution | Return to one original equation |
| Forms two versions of the same condition | Model lacks independent information | Identify what each equation represents |
| Accepts impossible negative counts | Context not checked | Return the algebraic pair to variable meaning |
| Reads a graph intersection too precisely | Display precision mistaken for exactness | Match answer accuracy to graph scale |
Verification routine
- Substitute the final pair into both original equations.
- Check units in each model equation.
- Confirm the values satisfy practical restrictions such as non-negative counts.
- If using a graph, verify that the intersection is visually consistent with the algebra.
- For break-even models, test one value on each side of the intersection.
Independent practice
- Solve x+y=11 and 2x−y=4.
- Solve y=2x+1 and 3x+y=21.
- Solve 3x+2y=18 and 5x−2y=14.
- Thirty-five tickets are sold. Adult tickets cost $12, child tickets cost $7 and total revenue is $330. Find the number of each.
- Plan A costs 8+0.5x and Plan B costs 20+0.3x. Find the break-even usage and cost.
- State whether the system 2x+4y=8 and x+2y=4 has a unique solution, no solution or infinitely many solutions.
Explained answers
1. Add the equations: 3x=15, so x=5 and y=6.
2. Substitute: 3x+2x+1=21, so 5x=20, x=4, y=9.
3. Add the equations: 8x=32, x=4. Then 12+2y=18, so y=3.
4. Let a adults and c children. a+c=35 and 12a+7c=330. Multiply the first by 7 and subtract: 5a=85, so a=17 and c=18.
5. 8+0.5x=20+0.3x gives 0.2x=12, so x=60. Cost = 8+30=$38.
6. The first equation is twice the second, so they describe the same line. There are infinitely many solutions.
Teaching sequence: condition first, method second
Begin with pairs of equations and ask learners what each one represents before solving. Then compare elimination and substitution on the same system so method choice becomes visible rather than habitual.
Next move to ticket, mixture and break-even models. Require variable definitions and units before equations are accepted. Finish with graph intersections and the special cases of parallel or identical lines.
Connect this guide to Algebra, Functions and Graphs Under Mixed-Topic Conditions and Coordinate Geometry and Transformations as Representation.
Final thought
A simultaneous solution is where two conditions agree. Whether you eliminate, substitute or inspect an intersection, the purpose is the same: find the value that survives every condition at once.
Do not solve two equations separately. Solve the relationship they share.
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