Linear simultaneous equations

Overview

Simultaneous equations are equations that are true at the same time. A pair of linear simultaneous equations usually contains two unknowns, such as $x$ and $y$, and the goal is to find the one pair of values that satisfies both equations.

This topic matters because many real problems involve two linked quantities: two prices, two ages, two numbers, or two mixtures. Instead of guessing, simultaneous equations give a clear way to translate the information and solve both unknowns together.

The central idea is "two facts, two unknowns". One equation usually gives only a relationship, not enough information for one exact answer. A second independent equation can narrow the possibilities to a single pair of values.

Prerequisites

• Algebraic expressions and equations - Learners should solve linear equations in one unknown before solving two unknowns.
• Rational, irrational, and real numbers - Elimination often involves negative coefficients, fractions, and decimal answers.
• Repeating decimals and fractions - Some solutions are fractional or recurring decimal values.
• Coordinate geometry: gradient and straight-line equations - Each linear equation can also be viewed as a straight line.
• Graphical solution of simultaneous equations - The graph topic gives a visual method for the same idea.

Learning Scope

This chapter covers the meaning of simultaneous equations, checking ordered-pair solutions, solving by substitution, solving by elimination, forming simultaneous equations from word problems, and recognizing simple special cases.

This page does not fully teach matrix methods, determinants, graphical solution, quadratic simultaneous equations, or linear programming. Those topics build on the same idea but require additional methods.

Subtopics

Meaning Of A Simultaneous Solution

A solution to simultaneous equations must satisfy every equation in the system. For example, consider:

$$ \begin{cases} x + y = 7 \\ x - y = 1 \end{cases} $$

The pair $x = 4$, $y = 3$ works because:

$$ 4 + 3 = 7 $$

and:

$$ 4 - 3 = 1 $$

Key insight: A value that works in only one equation is not a simultaneous solution.

Think of each equation as a test. A correct ordered pair must pass both tests. For the first equation above, $x=6$, $y=1$ also passes because $6+1=7$. But it fails the second equation because $6-1=5$, not $1$. That is why both equations must be used.

Checking A Pair Of Values

Before solving, learners should know how to test a proposed answer. Substitute the values into both equations.

For the system:

$$ \begin{cases} 2x + y = 11 \\ x + 3y = 18 \end{cases} $$

test $x = 3$, $y = 5$:

$$ 2(3) + 5 = 11 $$

but:

$$ 3 + 3(5) = 18 $$

Both statements are true, so the pair is a solution.

Key insight: Checking is not a separate skill for after the exam. It catches arithmetic errors while solving.

Write checks as two short substitutions:

$$ \begin{aligned} 2(3)+5 &= 11 \quad \text{true} \\ 3+3(5) &= 18 \quad \text{true} \end{aligned} $$

If one line is false, the pair is not the solution, even if the other line is true.

Solving By Substitution

Substitution is useful when one equation already gives one unknown in terms of the other.

For example:

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

Replace $y$ in the second equation with $x + 2$:

$$ \begin{aligned} 2x + (x + 2) &= 11 \\ 3x + 2 &= 11 \\ 3x &= 9 \\ x &= 3 \end{aligned} $$

Then:

$$ y = x + 2 = 3 + 2 = 5 $$

So the solution is $x = 3$, $y = 5$.

Key insight: Substitution turns two equations in two unknowns into one equation in one unknown.

Substitution is usually easiest when one equation already says $x=...$ or $y=...$. If not, first rearrange a simple equation. For example, from $x+y=10$, it is easy to write $y=10-x$ or $x=10-y$.

Misconception note: When substituting $y=x+2$ into $2x+y=11$, replace the whole $y$ with the whole expression $x+2$. Use brackets so the structure is clear: $2x+(x+2)=11$.

Solving By Elimination

Elimination is useful when adding or subtracting equations can remove one unknown.

For example:

$$ \begin{cases} 2x + y = 13 \\ 3x - y = 12 \end{cases} $$

Add the equations:

$$ \begin{aligned} (2x + y) + (3x - y) &= 13 + 12 \\ 5x &= 25 \\ x &= 5 \end{aligned} $$

Substitute $x = 5$ into $2x + y = 13$:

$$ \begin{aligned} 2(5) + y &= 13 \\ 10 + y &= 13 \\ y &= 3 \end{aligned} $$

So the solution is $x = 5$, $y = 3$.

Key insight: Choose addition or subtraction so that one variable disappears.

Elimination works because adding equal quantities to equal quantities gives equal results. If

$$ 2x+y=13 $$

and

$$ 3x-y=12 $$

then their left sides can be added and their right sides can be added. The $+y$ and $-y$ cancel because their sum is $0$.

Misconception note: Cancelling does not mean deleting terms at random. The coefficients must be equal and opposite for addition, or equal and the same sign for subtraction.

Making Coefficients Equal

Sometimes the coefficients do not cancel immediately. Multiply one or both equations first.

For example:

$$ \begin{cases} 2x + 3y = 18 \\ x + y = 7 \end{cases} $$

Multiply the second equation by $2$:

$$ 2x + 2y = 14 $$

Now subtract:

$$ \begin{aligned} (2x + 3y) - (2x + 2y) &= 18 - 14 \\ y &= 4 \end{aligned} $$

Then:

$$ x + 4 = 7 $$

so $x = 3$.

Key insight: Multiplying an entire equation by the same number keeps its solutions unchanged.

When multiplying to match coefficients, multiply every term in the equation, including the right side. If $x+y=7$ is multiplied by $2$, the result is:

$$ 2x+2y=14 $$

not $2x+2y=7$.

Forming Simultaneous Equations From Words

A word problem often gives two facts about two unknown quantities. Use one equation for each fact.

If $2$ exercise books and $1$ pen cost $1,700$ shillings, while $3$ exercise books and $2$ pens cost $2,900$ shillings, let:

$$ b = \text{cost of one exercise book} $$

and:

$$ p = \text{cost of one pen} $$

Then:

$$ \begin{cases} 2b + p = 1,700 \\ 3b + 2p = 2,900 \end{cases} $$

Key insight: Define the variables with units before writing the equations.

A reliable word-problem routine is:

1. Choose letters and write what each one means.
2. Turn the first fact into an equation.
3. Turn the second fact into another equation.
4. Solve by substitution or elimination.
5. Interpret both values with units.

For cost questions, keep item names separate. If $b$ is the cost of a book and $p$ is the cost of a pen, then $2b+p$ means "two books and one pen", not "two prices plus one object".

Special Cases

Some systems have no solution or many solutions. At Form I level, the main expectation is usually to solve systems with one solution, but learners should notice when the equations do not behave normally.

For example:

$$ \begin{cases} x + y = 5 \\ 2x + 2y = 10 \end{cases} $$

The second equation is just twice the first, so it does not add new information. Many pairs satisfy both equations.

Key insight: If both equations describe the same relationship, one exact pair cannot be found.

Another special case is no solution:

$$ \begin{cases} x+y=5 \\ x+y=8 \end{cases} $$

The same expression cannot equal $5$ and $8$ at the same time, so no pair of values can satisfy both equations.

Key Terms

• Simultaneous equations: Two or more equations that must be true at the same time.
• Linear equation: An equation in which the unknowns have power $1$ only.
• Unknown: A quantity represented by a letter whose value is to be found.
• Ordered pair: A pair such as $(x, y)$ showing values for two unknowns in order.
• Substitution: Replacing one variable with an equivalent expression or value.
• Elimination: Removing one variable by adding or subtracting equations.
• Coefficient: The number multiplying a variable.
• Solution: The value or pair of values that satisfies the equation or system.
• Consistent system: A system with at least one solution.
• Dependent equations: Equations that represent the same relationship.

Worked Examples

Example 1: Solve By Substitution

Solve:

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

Substitute $y = 2x - 1$ into $x + y = 8$:

$$ \begin{aligned} x + (2x - 1) &= 8 \\ 3x - 1 &= 8 \\ 3x &= 9 \\ x &= 3 \end{aligned} $$

Find $y$:

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

So $x = 3$ and $y = 5$.

Example 2: Solve By Elimination

Solve:

$$ \begin{cases} 4x + y = 19 \\ 2x - y = 5 \end{cases} $$

Add the equations:

$$ \begin{aligned} 6x &= 24 \\ x &= 4 \end{aligned} $$

Substitute into $2x - y = 5$:

$$ \begin{aligned} 2(4) - y &= 5 \\ 8 - y &= 5 \\ -y &= -3 \\ y &= 3 \end{aligned} $$

So $x = 4$ and $y = 3$.

Example 3: Make Coefficients Match

Solve:

$$ \begin{cases} 3x + 2y = 16 \\ x + 2y = 8 \end{cases} $$

Subtract the second equation from the first:

$$ \begin{aligned} (3x + 2y) - (x + 2y) &= 16 - 8 \\ 2x &= 8 \\ x &= 4 \end{aligned} $$

Substitute into $x + 2y = 8$:

$$ \begin{aligned} 4 + 2y &= 8 \\ 2y &= 4 \\ y &= 2 \end{aligned} $$

So $x = 4$ and $y = 2$.

Example 4: Word Problem

Two mangoes and one orange cost $1,100$ shillings. One mango and three oranges cost $1,300$ shillings. Find the cost of one mango and one orange.

Let $m$ be the cost of one mango and $o$ be the cost of one orange.

$$ \begin{cases} 2m + o = 1,100 \\ m + 3o = 1,300 \end{cases} $$

From the first equation:

$$ o = 1,100 - 2m $$

Substitute into the second equation:

$$ \begin{aligned} m + 3(1,100 - 2m) &= 1,300 \\ m + 3,300 - 6m &= 1,300 \\ -5m &= -2,000 \\ m &= 400 \end{aligned} $$

Then:

$$ o = 1,100 - 2(400) = 300 $$

One mango costs $400$ shillings and one orange costs $300$ shillings.

Example 5: Elimination After Multiplying Both Equations

Solve:

$$ \begin{cases} 2x + 3y = 23 \\ 3x + 2y = 22 \end{cases} $$

The coefficients do not cancel immediately. Make the $x$ coefficients equal. Multiply the first equation by $3$ and the second by $2$:

$$ \begin{aligned} 6x + 9y &= 69 \\ 6x + 4y &= 44 \end{aligned} $$

Subtract the second new equation from the first:

$$ \begin{aligned} (6x+9y)-(6x+4y) &= 69-44 \\ 5y &= 25 \\ y &= 5 \end{aligned} $$

Substitute $y=5$ into $2x+3y=23$:

$$ \begin{aligned} 2x+3(5) &= 23 \\ 2x+15 &= 23 \\ 2x &= 8 \\ x &= 4 \end{aligned} $$

The solution is $x=4$, $y=5$.

Example 6: Age Context

A father is $4$ times as old as his son. In $6$ years, their ages will add to $72$. Find their present ages.

Let $f$ be the father's present age and $s$ be the son's present age.

The first fact gives:

$$ f=4s $$

In $6$ years, their ages will be $f+6$ and $s+6$, so:

$$ (f+6)+(s+6)=72 $$

Simplify:

$$ f+s+12=72 $$

so:

$$ f+s=60 $$

Now solve:

$$ \begin{cases} f=4s \\ f+s=60 \end{cases} $$

Substitute $f=4s$:

$$ \begin{aligned} 4s+s &= 60 \\ 5s &= 60 \\ s &= 12 \end{aligned} $$

Then:

$$ f=4(12)=48 $$

The son is $12$ years old and the father is $48$ years old. In $6$ years they will be $18$ and $54$, and $18+54=72$.

Common Mistakes

• Solving only one equation: A simultaneous solution must satisfy both equations.
• Adding equations when subtraction is needed: Check whether coefficients have the same sign or opposite signs.
• Multiplying only part of an equation: If an equation is multiplied by $3$, every term on both sides must be multiplied.
• Substituting into the wrong expression: Use an equation that is easy, but keep the signs exactly as written.
• Forgetting to find the second unknown: After finding $x$, substitute to find $y$.
• Writing word-problem equations without defining variables: This makes it easy to mix up costs, quantities, or units.
• Treating $(x, y)$ as unordered: In coordinate or graph contexts, $(3, 5)$ is not the same as $(5, 3)$.

Practice Tasks

Foundation

1. Check whether $x = 2$, $y = 4$ satisfies $x + y = 6$ and $2x + y = 8$.
2. Check whether $(3,5)$ satisfies $2x+y=11$ and $x+3y=18$.
3. Explain why a pair that satisfies only one equation is not a simultaneous solution.

Substitution Practice

1. Solve by substitution: $y = x + 3$ and $x + y = 11$.
2. Solve by substitution: $x = 2y$ and $x + y = 15$.
3. Solve by substitution: $y = 10 - x$ and $3x + y = 18$.

Elimination Practice

1. Solve by elimination: $x + y = 9$ and $x - y = 1$.
2. Solve by elimination: $3x + y = 14$ and $2x - y = 6$.
3. Solve: $2x + 3y = 19$ and $2x + y = 9$.
4. Solve: $4x + 2y = 22$ and $x + 2y = 10$.
5. Solve: $2x+5y=31$ and $3x+2y=24$.

Applications And Reasoning

1. Form simultaneous equations for: two pencils and three pens cost $2,400$ shillings; four pencils and one pen cost $2,200$ shillings.
2. Solve the equations formed in Task 12.
3. Find two numbers whose sum is $21$ and whose difference is $5$.
4. A mother is three times as old as her child. In $8$ years, their ages will add to $64$. Form and solve simultaneous equations.
5. Create your own price problem that leads to two simultaneous equations, then solve it.
6. Give an example of two equations that have many solutions, and explain why.
7. Give an example of two equations that have no solution, and explain why.

Generated Question Layer

• Direct understanding questions: Ask learners to identify whether a pair satisfies both equations.
• Skill questions: Generate systems suitable for substitution, immediate elimination, and elimination after multiplying one equation.
• Word-problem questions: Use prices, ages, numbers, transport tickets, and mixture contexts.
• Progressive sets: Start with equations such as $y = x + a$, then move to matched coefficients and finally to coefficients requiring multiplication.
• Edge cases: Include systems with repeated equations, sign traps, fractional answers, and variables that cancel unexpectedly.
• LLM tutor: Prompt the tutor to ask, "Which variable is easiest to eliminate?" before showing a full method.

Learner Aid Opportunities

• diagram: Two-balance illustration showing two facts about the same unknown quantities.
• chart: Method-choice chart comparing substitution, immediate elimination, and elimination after multiplication, with a sample trigger for each.
• graph: Coordinate-plane view showing the solution as the intersection of two straight lines, linked back to the same algebraic pair.
• animation: Elimination animation where $+y$ and $-y$ cancel to $0$ as equations are added.
• interactive: System solver where learners choose substitution or elimination and receive step-level feedback after each algebra line.
• interactive: Word-problem builder where learners assign variables to two quantities before writing the two equations.
• video: Short worked example translating a market-price problem into two equations, including a check in shillings.
• LLM tutor: Guided questioning that asks learners to define variables, choose the easier method, write two equations, and justify each operation.

Exam-Derived Signals

The first automatic 2021-2025 Paper 1 mapping lists Linear simultaneous equations in the low-or-no-coverage group for direct primary records. This is an unreviewed extraction signal, not evidence that the topic is unimportant.

The 2022 examination format crosswalk maps the format group Algebra/Quadratic equations to this topic, together with algebraic expressions, quadratic equations, and inequalities. The crosswalk record is marked official, but it gives a broad group-level signal rather than a separate weight for simultaneous equations.

Unreviewed secondary extracted signals include:

| Year | Question ID | Signal | | ---: | --- | --- | | 2021 | csee_041_2021_p1_q13_a | Solving simultaneous equations by matrix method, mapped primarily to Form IV matrices. | | 2025 | csee_041_2025_p1_q13_b | Solving simultaneous equations by matrix method, mapped primarily to Form IV matrices and marked with a missing-marks review flag. |

These records show that simultaneous-equation language appears later inside matrix-method assessment. They should not be treated as reviewed Form I past-question references until checked manually against the original papers.

Source And Review Notes

• Official syllabus status: The topic identity, form placement, competence, source topic ID, and hub come from the 2023 CSEE Mathematics syllabus through data/curriculum_map.json.
• Official source reference: The cited syllabus file is raw/syllabuses/csee/2023/csee_mathematics_syllabus_2023.pdf.
• Exam signal status: The 2021-2025 signals come from data/topic_frequency_2021_2025.json and data/question_map_2021_2025.jsonl; they are unreviewed and should not be treated as audited past-question references.
• Exam format status: The 2022 format crosswalk maps Algebra/Quadratic equations to this page as one member of a broader algebra group.
• Content authorship status: Explanations, worked examples, and practice tasks are original learner-facing prose written from the syllabus topic and assessment signals, not copied from textbooks, Wikipedia, or extracted solutions.
• Renderer QA: This page uses $...$ and $$...$$ math notation for compatibility with Obsidian, KaTeX, and MathJax. Some plain Markdown viewers may show the raw delimiters.