A simultaneous-equation problem describes two conditions that must be true at the same time. Solving the system means finding the values that satisfy both conditions simultaneously. This is why graphical solutions appear as intersections and why substitution and elimination are logically equivalent routes to the same pair of values.
This Secondary 3 Mathematics Learning Guide develops substitution, elimination, graphical interpretation and modelling with simultaneous linear equations in two variables. It also explains when a system has one solution, no solution or infinitely many solutions as an extension of the core idea.
The current 2027 SEC G3 Mathematics syllabus listing from SEAB identifies Mathematics as K310. Its equations content includes simultaneous linear equations in two variables solved by substitution, elimination and graphical methods, together with formulation of equations to solve problems.
One Equation Is Usually Not Enough for Two Unknowns
The equation x+y=10 has many solutions: (1,9), (2,8), (3,7) and infinitely many other real pairs. A second independent condition narrows the possibilities.
If 2x+y=14 is also required, subtracting the first equation from the second gives x=4, then y=6. The pair (4,6) is the unique point satisfying both.
Substitution Replaces One Variable With an Equivalent Expression
If one equation already isolates a variable, substitution is often efficient. From y=11−2x, place that expression wherever y appears in the second equation.
The logic is valid because y and 11−2x represent the same quantity under the first condition.
Worked Example 1: Substitution
Solve y=11−2x and 3x+y=15.
Substitute: 3x+(11−2x)=15. Hence x+11=15, so x=4. Then y=11−8=3.
The solution is (x,y)=(4,3). Check both original equations before finishing.
Elimination Creates an Equation in One Variable
Elimination is useful when the coefficients of one variable can be made equal or opposite. Adding or subtracting the equations then removes that variable.
The operation preserves the common solution because equal quantities are being added to or subtracted from equal quantities.
Worked Example 2: Elimination
Solve 2x+3y=17 and 4x−3y=7.
Add the equations: 6x=24, so x=4. Substitute into the first equation: 8+3y=17, so y=3.
Again, (4,3) satisfies both conditions.
Choose the Route That Simplifies the Structure
Substitution is often efficient when one variable is already isolated or has coefficient 1. Elimination is often efficient when coefficients already match or can be matched with small multipliers.
There is no requirement to use the same route every time. Good mathematical control means selecting the route that reduces complexity without changing meaning.
Worked Example 3: Create Matching Coefficients
Solve 3x+2y=16 and 5x−y=19.
Multiply the second equation by 2: 10x−2y=38. Add to the first equation to obtain 13x=54, so x=54/13.
Then y=5x−19 = 270/13−247/13 = 23/13. A fractional solution is perfectly valid unless the context requires integers.
Graphical Solution Means Intersection
Each linear equation represents a straight line. A pair (x,y) satisfying both equations lies on both lines, so it must be their intersection.
A graph may give an approximate answer depending on scale and reading precision. Algebraic methods can provide exact values when the coefficients allow it.
Worked Example 4: Algebra and Graph Agree
Find the intersection of y=2x+1 and y=−x+7. Set the two expressions equal:
2x+1=−x+7, so 3x=6 and x=2. Then y=5. The intersection is (2,5).
This connects directly to the earlier Coordinate Geometry and Transformations guide.
Modelling Starts Before the Equations
Word problems require three stages: define variables, translate relationships, then solve. If the variables are unclear, the equations can be correct algebraically but represent the wrong quantities.
Always write what x and y mean, including units when relevant.
Worked Example 5: Tickets
A concert sells adult tickets for $18 and student tickets for $11. A total of 120 tickets produces $1710. Find the number of each type.
Let a be adult tickets and s student tickets. Then a+s=120 and 18a+11s=1710.
Multiply the first equation by 11: 11a+11s=1320. Subtract from the revenue equation: 7a=390, so a=390/7, not an integer.
This reveals that the given numerical data are inconsistent with whole ticket counts. A modelling check is therefore essential: not every pair of stated totals leads to a physically valid integer solution.
If the revenue were $1740 instead, subtraction would give 7a=420, so a=60 and s=60. This corrected version has a valid whole-number solution.
Worked Example 6: Mixture of Quantities
A shop sells 3 kg fruit boxes and 5 kg fruit boxes. It sells 42 boxes containing 162 kg of fruit. Find how many of each type were sold.
Let x be 3 kg boxes and y be 5 kg boxes. Then x+y=42 and 3x+5y=162.
Multiply the first equation by 3: 3x+3y=126. Subtract: 2y=36, so y=18 and x=24.
Check: 24+18=42 and 72+90=162.
Worked Example 7: Two-Digit Number
A two-digit number has digit sum 11. Reversing the digits decreases the number by 27. Find the number.
Let tens digit be x and units digit y. Then x+y=11. Original number=10x+y; reversed=10y+x.
(10x+y)−(10y+x)=27 gives 9x−9y=27, so x−y=3.
Solve x+y=11 and x−y=3: 2x=14, x=7 and y=4. The number is 74.
One Solution, No Solution, Infinitely Many
As an extension, the geometry of lines explains possible system behaviour. Two nonparallel lines intersect once, giving one solution. Distinct parallel lines never meet, giving no solution. Two equations representing the same line have infinitely many common points.
For example, x+y=5 and 2x+2y=10 are equivalent equations and therefore describe the same line.
Common Errors
Eliminating only one term but not the full equation: repair by multiplying every term on both sides.
Substituting into a modified equation incorrectly: repair by keeping the original equations visible.
Accepting a contextual fraction when counts must be whole: repair by returning to the variable definitions.
Stopping after finding x: a two-variable system usually requires both variables unless the question asks for only one.
Independent Practice
1. Solve x+y=13, x−y=5.
2. Solve 2x+3y=19, 2x−y=7.
3. Solve y=3x−4, x+y=16.
4. Find the intersection of y=4x−3 and y=−2x+9.
5. Solve 3x+2y=8, 6x+4y=16 and interpret the result.
6. Two numbers total 42 and differ by 8. Find them.
7. Pens cost $2 and notebooks $5. A customer buys 18 items for $57. Find each quantity.
8. A farm has chickens and goats. There are 32 animals and 94 legs. Find each number.
9. A two-digit number has digit sum 9 and original number exceeds reversed number by 45. Find it.
10. Explain why parallel distinct lines correspond to no simultaneous solution.
Explained Answers
1. Add equations: 2x=18, x=9, y=4.
2. Subtract second from first: 4y=12, y=3; then x=5.
3. x+3x−4=16, so x=5, y=11.
4. 4x−3=−2x+9 gives x=2, y=5.
5. The second equation is twice the first, so infinitely many solutions along the same line.
6. x+y=42, x−y=8 gives x=25, y=17.
7. p+n=18, 2p+5n=57 gives n=7, p=11.
8. c+g=32, 2c+4g=94 gives g=15, c=17.
9. x+y=9, x−y=5 gives x=7, y=2, so 72.
10. The two equations require a point lying on both lines, but distinct parallel lines share no point.
Continue the Secondary 3 Learning Route
Continue with Number Structure, Prime Factorisation, HCF and LCM, Algebraic Patterns, nth-Term Rules and Identities, and Compound Interest, Repeated Growth and Financial Reasoning.
Simultaneous equations become reliable when both equations remain meaningful descriptions of the same situation all the way to the final answer. Return to the Secondary Mathematics Hub.
