One equation can describe many possible pairs. Two independent equations can narrow the possibilities to one shared pair. That is the central idea behind simultaneous linear equations. The algebraic methods are useful because they preserve both conditions until the common solution becomes visible.
This Secondary 2 Mathematics Learning Guide develops substitution, elimination, graphical solving and problem formulation as different routes to the same mathematical object: a pair of values satisfying two linear conditions at once.
Secondary Mathematics Hub: S1–S4 Capability Map · Secondary 2 Learning Guide, Batch 2, Guide 3. Companion guides cover quadratic functions, linear inequalities, and quadratic equations.
Course boundary. The current MOE G2/G3 Mathematics syllabus includes Secondary Two simultaneous linear equations in two variables solved by substitution, elimination and graphical methods, together with formulating a pair of linear equations to solve problems. This guide follows that scope and emphasises modelling and checking rather than introducing nonlinear systems.
Navigate: Meaning · Substitution · Elimination · Graphical solution · Formulating models · Special cases · Practice and answers · Teaching and transfer.
1. A simultaneous solution must satisfy both equations
The equation x + y = 10 has many real solutions: (0, 10), (4, 6), (7.5, 2.5) and many more. Add the second equation x − y = 2, and most of those pairs disappear. The pair that satisfies both is x = 6, y = 4.
Checking one equation is not enough. The pair (7, 3) satisfies x + y = 10 but fails x − y = 2. Simultaneous means both conditions hold at the same time.
Worked example 1: verify a proposed pair
Does (x, y) = (3, 5) solve x + y = 8 and 2x + y = 11? The first gives 3 + 5 = 8, true. The second gives 6 + 5 = 11, true. Therefore the pair is a simultaneous solution.
Now test (4, 4). It satisfies x + y = 8, but 2x + y = 12 rather than 11. The pair fails the system. This simple substitution check should remain part of every later method.
The graph interpretation
Each linear equation in two variables represents a straight line. A simultaneous solution is therefore a point lying on both lines: their intersection. Algebraic solving and graphical solving are not separate topics; they are different ways of locating the same shared pair.
This interpretation makes the number of possible solutions visible. Distinct non-parallel lines meet once. Parallel distinct lines never meet. Coincident lines share every point on the line. These cases will return later in the guide.
2. Substitution replaces one variable with an equivalent expression
If y = 12 − x, then every occurrence of y in the other equation may be replaced by 12 − x. This does not introduce a new assumption. It uses the first equation to express the same quantity in another form.
Worked example 2: direct substitution
Solve y = 12 − x and 2x + y = 16. Substitute y = 12 − x into the second equation: 2x + 12 − x = 16. Therefore x + 12 = 16, so x = 4.
Then y = 12 − 4 = 8. Check both originals: 4 + 8 = 12 and 2(4) + 8 = 16. The solution is (4, 8).
Worked example 3: rearrange first
Solve 2x + y = 13 and 3x − 2y = 4. Rearrange the first equation to y = 13 − 2x. Substitute into the second: 3x − 2(13 − 2x) = 4.
Expand carefully: 3x − 26 + 4x = 4. Therefore 7x = 30, so x = 30/7. Then y = 13 − 60/7 = 31/7. Fractions are valid solutions; do not force integers merely because previous examples used them.
A decimal approximation may be useful for interpretation, but the exact fraction pair preserves the exact algebra. Substitute into both original equations if an exact check is required.
Choose substitution when one variable is already isolated
Substitution is often efficient when an equation already has the form x = expression or y = expression, or when one coefficient is 1 or −1. It may be less efficient when isolating a variable creates awkward fractions before the other equation is used.
Method choice should respond to the structure. The question may require a particular method for practice, but outside that instruction, choose a route that keeps the algebra controlled.
3. Elimination combines equations to remove one variable
Elimination works because adding or subtracting equal quantities preserves equality. If two equations have matching coefficients for one variable, adding or subtracting them can remove that variable and leave an equation in only one unknown.
Worked example 4: subtract matching coefficients
Solve 3x + y = 17 and 2x + y = 13. Subtract the second equation from the first. The y-terms cancel: x = 4. Substitute into 2x + y = 13 to obtain y = 5.
The solution is (4, 5). Check: 3(4) + 5 = 17 and 2(4) + 5 = 13.
Worked example 5: create matching coefficients first
Solve 2x + 3y = 19 and 3x − 2y = 4. The coefficients do not match. Multiply the first equation by 2: 4x + 6y = 38. Multiply the second by 3: 9x − 6y = 12.
Add the new equations: 13x = 50, so x = 50/13. Substitute into either original equation to obtain y = 49/13. Again, fractional answers are mathematically acceptable unless the context later imposes a whole-number requirement.
Multiplying an entire equation preserves its solution set
When 2x + 3y = 19 is multiplied by 2, every term must be multiplied: 4x + 6y = 38. Multiplying only one coefficient would create a different line and therefore a different system.
It is useful to think geometrically. Multiplying a complete equation by a non-zero constant changes its written form but not its graph. The new equation represents the same line.
Worked example 6: add rather than subtract
Solve 4x + 3y = 25 and 2x − 3y = 5. Add the equations because +3y and −3y cancel. This gives 6x = 30, so x = 5.
Substituting into 2x − 3y = 5 gives 10 − 3y = 5, so y = 5/3. The pair is (5, 5/3). A sign-aware choice between addition and subtraction keeps the elimination short.
4. Graphical solving makes the shared condition visible
To solve two equations graphically, draw both lines on the same coordinate axes and read the intersection. The result is often approximate because of plotting and scale. If an exact algebraic method is also available, it can check the graphical estimate.
The graph must represent the equations faithfully. A wrong scale, swapped coordinates or inaccurate line can produce an incorrect intersection even when the concept is understood.
Worked example 7: solve from two graph equations
Solve y = 8 − x and y = 2x − 1. Their intersection satisfies both, so set the expressions equal: 8 − x = 2x − 1. This gives 9 = 3x, so x = 3 and y = 5.
On a correctly drawn graph, the lines meet at (3, 5). The algebra and graph agree because they locate the same shared point.
Graphical accuracy depends on scale
If the true intersection is (2.4, 5.7), a coarse graph may only support a reading such as x ≈ 2.4 and y ≈ 5.7. Reporting six decimal places from such a drawing would suggest precision the graph does not contain.
When the question asks for a graphical solution, show the graph route. When it asks for an exact solution, algebra is usually preferable. Match the representation to the required output.
5. Formulating the equations is often the real problem
In a word problem, the equations do not arrive ready-made. The learner must decide what the variables represent and translate each independent condition. A correct algebraic solution to incorrectly formulated equations still answers the wrong problem.
Worked example 8: two ticket types
A fictional event sells 42 tickets. Adult tickets cost 9 dollars and student tickets cost 5 dollars. Total revenue is 282 dollars. Find the number of each type. Let a be adult tickets and s be student tickets.
The count condition gives a + s = 42. The money condition gives 9a + 5s = 282. Multiply the first equation by 5: 5a + 5s = 210. Subtract to get 4a = 72, so a = 18 and s = 24.
Check both meanings: 18 + 24 = 42 tickets and 9(18) + 5(24) = 282 dollars. The units help distinguish the equations: one counts tickets; the other counts dollars.
Worked example 9: two quantities with sum and difference
Two numbers have sum 56 and difference 14. Find them. Let x be the larger number and y the smaller. Then x + y = 56 and x − y = 14.
Add the equations: 2x = 70, so x = 35. Then y = 21. The pair satisfies both conditions. If the variable roles were reversed, the difference equation would need to change accordingly.
Worked example 10: a mixture-style model
A fictional box contains 30 items of two types. Type A weighs 2 kg and Type B weighs 5 kg. The total weight is 96 kg. Find the counts. Let a and b be the respective counts. Then a + b = 30 and 2a + 5b = 96.
Multiply the first equation by 2: 2a + 2b = 60. Subtract from the weight equation: 3b = 36, so b = 12 and a = 18. Check: 18 + 12 = 30 and 36 + 60 = 96 kg.
Choose variables that make the story easier to state
Variables should represent quantities the question actually needs or quantities that make the conditions easy to express. For a count problem, let x and y be the counts rather than the total prices unless a different choice is clearly more efficient.
Write a short definition such as let x be the number of adult tickets. This prevents later confusion when the same symbols appear in several equations.
6. Not every pair of linear equations has one unique solution
Consider y = 2x + 3 and y = 2x − 1. The lines have the same gradient but different intercepts, so they are parallel and never meet. The system has no solution.
Now consider y = 2x + 3 and 2y = 4x + 6. Dividing the second equation by 2 gives the first. Both equations represent the same line, so every point on that line satisfies both. The system has infinitely many real solutions.
Elimination reveals these cases algebraically
If elimination produces a false statement such as 0 = 4, the original conditions are inconsistent and there is no simultaneous solution. If it produces an identity such as 0 = 0, the equations may be equivalent, giving infinitely many solutions unless another independent condition exists.
Do not treat 0 = 0 as x = 0 or 0 = 4 as a calculator error automatically. Interpret what the eliminated system is telling you about the relationship between the two lines.
Context can reject an algebraically valid pair
Suppose a count model gives x = 6.5 and y = 8.5 people. The algebra may solve the stated equations exactly, but the result cannot represent counts of whole people. This may indicate that the data in the model do not permit a whole-number solution, or that the equations were formulated incorrectly.
Always return the pair to the original meaning. Variables representing length may need positivity; counts may need whole numbers; time may have a stated domain. Algebra produces candidates, context decides whether they are admissible.
7. Mixed practice: choose a route, then verify both conditions
Use substitution when one variable is already isolated or easily isolated. Use elimination when coefficients can be matched cleanly. Use graphs when the question asks for a graphical answer or when comparison ranges matter. Whatever the method, check the final pair in both original equations.
Questions 1–6. 1. Solve x + y = 11 and x − y = 3. 2. Solve y = 9 − x and 2x + y = 13. 3. Solve 3x + y = 14 and x + y = 8. 4. Solve 2x + 3y = 16 and 2x − y = 4. 5. Solve y = 2x + 1 and y = 10 − x. 6. Verify whether (2, 5) solves x + y = 7 and 3x + y = 11.
Questions 7–12. 7. A total of 25 items consists of two types. Type A costs 4 dollars and Type B costs 7 dollars. The total cost is 130 dollars. Find the counts. 8. Two numbers sum to 48 and differ by 8. Find them. 9. A box contains 20 objects weighing either 2 kg or 3 kg. Total weight is 54 kg. Find the counts. 10. Explain why y = 3x + 2 and y = 3x − 5 have no simultaneous solution. 11. Explain why x + y = 6 and 2x + 2y = 12 have infinitely many solutions. 12. A solution gives 7.5 adults and 10.5 children. Explain why the model must be reconsidered.
Questions 13–18. 13. Form equations: two ticket types total 60 tickets, priced 8 and 5 dollars, with total revenue 390 dollars. 14. Solve the system from Question 13. 15. Form equations: two numbers sum to 75; the larger is 21 more than the smaller. 16. Solve. 17. A graph of two lines intersects at (4, −2). What does this mean about the simultaneous equations represented? 18. If two accurately drawn lines never meet and have equal gradient, what does that imply?
Explained answers: questions 1–6
1. Add to get 2x = 14, so x = 7, y = 4. 2. Substitute to obtain x = 4, y = 5. 3. Subtract the second from the first: 2x = 6, so x = 3, y = 5.
4. Subtract the second equation from the first: 4y = 12, so y = 3, x = 3.5. 5. 2x + 1 = 10 − x gives x = 3 and y = 7. 6. Yes: 2 + 5 = 7 and 3(2) + 5 = 11.
Explained answers: questions 7–12
7. a + b = 25 and 4a + 7b = 130. Solving gives b = 10 and a = 15. 8. 28 and 20. 9. a + b = 20 and 2a + 3b = 54, giving b = 14 and a = 6.
10. Same gradient, different intercepts: parallel distinct lines. 11. The second equation is exactly twice the first, so both represent the same line. 12. Adult and child counts must be whole numbers, so either the data or formulation does not fit the intended context.
Explained answers: questions 13–18
13. Let a and s be the ticket counts: a + s = 60 and 8a + 5s = 390. 14. Multiply the first by 5 and subtract: 3a = 90, so a = 30 and s = 30.
15. Let x be larger and y smaller: x + y = 75 and x − y = 21. 16. Add to get 2x = 96, so x = 48 and y = 27. 17. x = 4 and y = −2 satisfy both equations. 18. The system has no solution.
8. Teach the shared condition before the algorithms
A useful introduction gives one equation and asks for three valid pairs. Then add a second equation and ask which pair survives. Plot both lines and locate the intersection. Only after the meaning is visible should substitution and elimination be treated as efficient algebraic procedures.
For learners repairing algebra, choose coefficients that keep the focus on the system rather than arithmetic complexity. For learners keeping up, mix substitution, elimination and formulation. For learners moving ahead, ask them to compare routes, construct systems with no solution or infinitely many solutions, and explain the graph meaning.
Diagnose by the first broken link
If the equations are formulated incorrectly, more elimination practice will not repair the model. If the equations are correct but coefficients are not multiplied across the whole line, repair equivalence. If the pair is solved correctly but only one equation is checked, repair the meaning of simultaneous.
If graph and algebra disagree, compare the plotted equations, axis scales and arithmetic. The disagreement is useful evidence. Do not simply choose whichever answer looks more convenient.
Questions parents can ask
What does x represent? What does y represent? Which condition produced the first equation? Which produced the second? Why did you choose substitution or elimination? Does the final pair satisfy both original equations? These questions expose reasoning without supplying the algebra.
9. The transfer test: same mathematics, changed surface
A fictional study centre has 28 chairs arranged in two room types. Small rooms contain 2 chairs and large rooms contain 5 chairs. There are 8 rooms altogether. Let s be small rooms and l be large rooms. Then s + l = 8 and 2s + 5l = 28.
Multiply the first equation by 2: 2s + 2l = 16. Subtract from the chair equation: 3l = 12, so l = 4 and s = 4. Check: 4 + 4 = 8 rooms and 2(4) + 5(4) = 28 chairs.
Now change the surface from rooms and chairs to containers and objects, while preserving the two equations. The algebra is unchanged because the quantitative structure is unchanged. This is the transfer target: recognise the system beneath the story.
Use Linear Graphs, Coordinates and Relationships when the intersection meaning needs strengthening. Use Algebraic Factorisation and Structural Control when manipulation is the weak link. Continue to Quadratic Equations when one unknown appears in a second-degree relationship rather than two independent linear ones.
Define the variables. Preserve both conditions. Choose a route. Verify the pair in both originals. Return to the context.
