Mathematics · AQA 8300 §A4
Simultaneous Equations
- Solve simultaneous equations by elimination
- Solve simultaneous equations by substitution
- Form simultaneous equations from word problems
- Interpret simultaneous equations graphically
- Check solutions satisfy both equations
🔑 Core Concepts
What are Simultaneous Equations?
A simultaneous equation is a set of two (or more) equations that must be satisfied by the same values of the unknowns at the same time. The word "simultaneous" means "at the same time" — both equations must hold together.
Definition — Simultaneous Equations
Two equations in two unknowns ($x$ and $y$) that share a common solution $(x, y)$. The solution is the pair of values that satisfies both equations simultaneously. Graphically, it is the point where the two straight lines intersect.
Exam Tip — Always verify your answer
After finding $x$ and $y$, substitute both values back into both original equations. If either equation does not balance, you have made an error. This check costs only seconds and can recover marks.
Common Mistake — Forgetting there are two unknowns
Students sometimes find $x$ and then forget to find $y$. Exam questions always require both values. State your final answer as $x = \ldots,\ y = \ldots$.
The Elimination Method
The elimination method works by making the coefficient of one variable identical in both equations, then adding or subtracting to remove that variable entirely — leaving a single equation in one unknown.
Definition — Elimination
Scale one or both equations so a chosen variable has equal (or equal and opposite) coefficients. Then subtract if the coefficients are the same sign, or add if they are opposite signs. This eliminates that variable.
Elimination Method — Process
Given: $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$
$$\text{Multiply eq 1 by } a_2 \text{ and eq 2 by } a_1 \text{, then subtract (or add)}$$
$$\text{Multiply eq 1 by } a_2 \text{ and eq 2 by } a_1 \text{, then subtract (or add)}$$
$a_1, a_2$ = coefficients of $x$
$b_1, b_2$ = coefficients of $y$
Same sign → subtract
Opposite signs → add
Write both equations
→
Choose variable to eliminate
→
Multiply to match coefficients
→
Add or subtract equations
→
Solve for one variable
→
Substitute back, verify
Exam Tip — Same sign or opposite sign?
Write "SS → Subtract, OS → Add" in your working. If both equations have $+3y$, they are the same sign — subtract. If one has $+3y$ and the other $-3y$, they are opposite signs — add. This simple rule eliminates a very common error.
Memory Trick — "SAME Subtract, DIFFERENT Add"
Rhyme: "If the signs are the SAME, play the Subtraction game. If they're different, you should Add them!"
The Substitution Method
The substitution method works by rearranging one equation to make one variable the subject, then replacing that variable in the other equation. This creates a single equation in one unknown.
Definition — Substitution
Rearrange one equation to express one variable in terms of the other (e.g. $y = \ldots$). Substitute this expression into the second equation to create one equation in one unknown. Solve, then back-substitute to find the second variable.
Substitution Method — Process
Step 1: From equation 1, express $y$ in terms of $x$: $y = \dfrac{c_1 - a_1 x}{b_1}$
Step 2: Substitute into equation 2: $a_2 x + b_2\!\left(\dfrac{c_1 - a_1 x}{b_1}\right) = c_2$
Step 3: Solve for $x$, then back-substitute to find $y$.
Step 2: Substitute into equation 2: $a_2 x + b_2\!\left(\dfrac{c_1 - a_1 x}{b_1}\right) = c_2$
Step 3: Solve for $x$, then back-substitute to find $y$.
Best when one equation has coefficient 1
Avoid large fractions where possible
Exam Tip — When to use substitution
Choose substitution when one equation already has a variable with coefficient 1 or $-1$, for example $y = 3x - 2$ or $x = 5 - 2y$. Substitution avoids the need to multiply both equations and reduces the chance of arithmetic errors.
Common Mistake — Substituting into the same equation
After rearranging equation 1 to find $y = \ldots$, you must substitute into equation 2, not back into equation 1. Substituting into the same equation gives $0 = 0$, which is always true and tells you nothing.
Choosing the Right Method
| Situation | Preferred Method | Reason |
|---|---|---|
| Both equations in $ax + by = c$ form with integer coefficients | Elimination | Quick to multiply and add/subtract |
| One equation has a variable with coefficient 1 or $-1$ | Substitution | Easy rearrangement, no fractions |
| One equation gives $y = mx + c$ directly | Substitution | Expression ready to substitute immediately |
| Coefficients are awkward fractions | Elimination | Multiplying clears fractions faster |
| Word problem with a natural expression | Either | Form the simpler equation first, then choose |
Graphical Interpretation
Each linear equation in two variables represents a straight line on a coordinate grid. Solving simultaneously means finding the point where the two lines intersect.
Definition — Graphical Solution
Draw both lines on the same axes. The coordinates of the intersection point $(x, y)$ are the solution to the simultaneous equations. If the lines are parallel (same gradient, different intercepts), there is no solution. If the lines are identical, there are infinitely many solutions.
Number of Solutions — Summary
$$\text{Unique solution: lines intersect at one point}$$
$$\text{No solution: lines are parallel (same gradient, } m_1 = m_2 \text{, different intercepts)}$$
$$\text{Infinite solutions: lines are identical (same equation)}$$
$m_1 = m_2$, $c_1 \neq c_2$ → no solution (parallel)
$m_1 = m_2$, $c_1 = c_2$ → infinite solutions (same line)
$m_1 \neq m_2$ → unique solution
Important — Parallel lines have NO solution
At Grade 9, examiners may ask you to explain why a pair of simultaneous equations has no solution. The answer is: both equations represent lines with the same gradient but different $y$-intercepts — they are parallel and never intersect. A good explanation always references the gradient.
Forming Simultaneous Equations from Word Problems
Many GCSE exam questions present a scenario in words. The skill is translating the information into two equations, then solving.
Definition — Setting Up Equations
Step 1: Define variables clearly (e.g. let $x$ = cost of one pen, $y$ = cost of one notebook).Step 2: Write two equations from the two pieces of information given.
Step 3: Solve the simultaneous equations.
Step 4: Interpret the solution in context, including units.
Exam Tip — State your variables
Always begin: "Let $x = \ldots$ and $y = \ldots$". Examiners award a mark for clearly defined variables, even if your subsequent working contains an error. Missing this step can cost a method mark.
Verifying Solutions
Definition — Verification
Substitute the values of $x$ and $y$ into both original equations and confirm the left-hand side equals the right-hand side in each case. Show the check explicitly in your working.
Verification Check
$$\text{If } x = a, y = b \text{ is the solution to } \begin{cases} f(x,y) = p \\ g(x,y) = q \end{cases}$$
$$\text{Check: } f(a,b) \stackrel{?}{=} p \quad \text{and} \quad g(a,b) \stackrel{?}{=} q$$
Both must hold simultaneously
If either fails, recheck your working
🗺️ Visual Notes
Simultaneous Equations
🔧 Elimination
- Match one coefficient
- Same sign → subtract
- Opposite sign → add
- Solve remaining variable
🔀 Substitution
- Rearrange one equation
- Express $y$ in terms of $x$
- Substitute into other equation
- Solve, then back-substitute
📈 Graphical
- Each equation = a line
- Solution = intersection
- Parallel → no solution
- Same line → infinite solutions
📝 Word Problems
- Define variables clearly
- Write two equations
- Solve simultaneously
- Interpret in context
✅ Checking
- Substitute into equation 1
- Substitute into equation 2
- Both must balance
- State answer clearly
⭐ Grade 9 Extra
- Non-integer solutions
- Geometry-based problems
- Explain no/infinite solutions
- Optimal method choice
Elimination vs Substitution — Decision Tree
Look at both equations
→
Any coefficient = 1 or −1?
→
YES → Use Substitution
→
NO → Use Elimination
→
Verify in BOTH equations
Comparison: Elimination vs Substitution
| Feature | Elimination | Substitution |
|---|---|---|
| Best for | Both equations in standard form $ax+by=c$ | One variable has coefficient 1 or $-1$ |
| Key operation | Multiply, then add/subtract equations | Rearrange and replace |
| Risk of fractions | Low if coefficients are chosen well | Medium — can create fractions early |
| Steps needed | Usually 2–3 steps to eliminate | Usually 3–4 steps including rearrangement |
| Error hotspot | Sign error when adding/subtracting | Expanding brackets after substitution |
| Good for Grade 9 non-integers? | Yes — fractions appear at solve stage only | Yes — but be careful with fractional expressions |
Graphical Interpretation — Number of Solutions
| Gradients | Intercepts | Lines | Solutions | Algebraic sign |
|---|---|---|---|---|
| $m_1 \neq m_2$ | Any | Intersect at one point | Unique solution | Consistent |
| $m_1 = m_2$ | $c_1 \neq c_2$ | Parallel — never meet | No solution | Inconsistent |
| $m_1 = m_2$ | $c_1 = c_2$ | Identical — overlap fully | Infinite solutions | Dependent |
✏️ Worked Examples
Grade 4–5 · Elimination (simple)
$2x = 6$
$x = 3$
Solve: $\quad 3x + 2y = 13 \quad$ and $\quad x + 2y = 7$
1
Label the equations
Equation (1): $3x + 2y = 13$ Equation (2): $x + 2y = 7$
2
Identify matching coefficients
Both equations have $+2y$. The coefficients are the same sign, so we subtract equation (2) from equation (1).
3
Subtract to eliminate $y$
$(3x + 2y) - (x + 2y) = 13 - 7$$2x = 6$
$x = 3$
4
Substitute back to find $y$
Using equation (2): $3 + 2y = 7 \Rightarrow 2y = 4 \Rightarrow y = 2$
5
Verify in equation (1)
$3(3) + 2(2) = 9 + 4 = 13$ ✓
Answer: $x = 3,\ y = 2$
Grade 6–7 · Elimination (scaling needed)
Multiply equation (2) by 5: $10x + 35y = 220$ … (4)
$41y = 218$
$y = \dfrac{218}{41} = \dfrac{218}{41}$
Check: $218 \div 41 = 5.317\ldots$ — not integer, let us check arithmetic.
$41y = 218 \Rightarrow y = \dfrac{218}{41}$ … let's simplify: $\gcd(218, 41) = 1$, so $y = \dfrac{218}{41}$.
Actually let us re-examine: $(35 - (-6)) = 41$ ✓. $220 - 2 = 218$ ✓. So $y = \dfrac{218}{41}$.
$2x = 44 - \dfrac{1526}{41} = \dfrac{1804 - 1526}{41} = \dfrac{278}{41}$
$x = \dfrac{139}{41}$
Solve: $\quad 5x - 3y = 1 \quad$ and $\quad 2x + 7y = 44$
1
Label the equations
Equation (1): $5x - 3y = 1$ Equation (2): $2x + 7y = 44$
2
Choose variable to eliminate — eliminate $x$
Multiply equation (1) by 2: $10x - 6y = 2$ … (3)Multiply equation (2) by 5: $10x + 35y = 220$ … (4)
3
Subtract (same sign on $x$)
$(4) - (3)$: $(10x + 35y) - (10x - 6y) = 220 - 2$$41y = 218$
$y = \dfrac{218}{41} = \dfrac{218}{41}$
Check: $218 \div 41 = 5.317\ldots$ — not integer, let us check arithmetic.
$41y = 218 \Rightarrow y = \dfrac{218}{41}$ … let's simplify: $\gcd(218, 41) = 1$, so $y = \dfrac{218}{41}$.
Actually let us re-examine: $(35 - (-6)) = 41$ ✓. $220 - 2 = 218$ ✓. So $y = \dfrac{218}{41}$.
4
Substitute to find $x$
From equation (2): $2x + 7 \cdot \dfrac{218}{41} = 44$$2x = 44 - \dfrac{1526}{41} = \dfrac{1804 - 1526}{41} = \dfrac{278}{41}$
$x = \dfrac{139}{41}$
5
Verify in equation (1)
$5 \cdot \dfrac{139}{41} - 3 \cdot \dfrac{218}{41} = \dfrac{695 - 654}{41} = \dfrac{41}{41} = 1$ ✓
Answer: $x = \dfrac{139}{41},\quad y = \dfrac{218}{41}$
Grade 9 · Forming equations from geometry
Length condition: $l = 2w + 5$ … (2)
$3w + 5 = 23$
$3w = 18$
$w = 6$
Perimeter check: $2(17) + 2(6) = 34 + 12 = 46$ cm ✓
The perimeter of a rectangle is 46 cm. The length is 5 cm more than twice the width. Find the dimensions of the rectangle.
1
Define variables
Let $l$ = length (cm) and $w$ = width (cm).
2
Form the equations
Perimeter: $2l + 2w = 46$ → $l + w = 23$ … (1)Length condition: $l = 2w + 5$ … (2)
3
Choose method — substitution (equation (2) gives $l$ directly)
Substitute (2) into (1): $(2w + 5) + w = 23$$3w + 5 = 23$
$3w = 18$
$w = 6$
4
Find $l$
$l = 2(6) + 5 = 17$ cm
5
Verify both equations
Eq (1): $17 + 6 = 23$ ✓ Eq (2): $17 = 2(6)+5 = 17$ ✓Perimeter check: $2(17) + 2(6) = 34 + 12 = 46$ cm ✓
Answer: Width = 6 cm, Length = 17 cm
❓ Exam Questions
Q11 mark
Write down the solution to the simultaneous equations $y = 3x$ and $y = x + 4$.
Mark scheme:
Set $3x = x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$, so $y = 6$.
Answer: $x = 2,\ y = 6$ [1 mark for both correct values]
Set $3x = x + 4 \Rightarrow 2x = 4 \Rightarrow x = 2$, so $y = 6$.
Answer: $x = 2,\ y = 6$ [1 mark for both correct values]
Q22 marks
Solve: $\quad 4x + y = 14 \quad$ and $\quad 2x - y = 4$
Method: Adding equations (opposite signs on $y$): $6x = 18 \Rightarrow x = 3$.
Substitute: $4(3) + y = 14 \Rightarrow y = 2$.
Answers: $x = 3,\ y = 2$
[M1 for correct elimination; A1 for both correct values]
Substitute: $4(3) + y = 14 \Rightarrow y = 2$.
Answers: $x = 3,\ y = 2$
[M1 for correct elimination; A1 for both correct values]
Q33 marks
Solve: $\quad 3x + 2y = 11 \quad$ and $\quad 5x - 4y = -3$
Method:
Multiply eq(1) by 2: $6x + 4y = 22$ … (3)
Add eq(3) to eq(2): $(6x + 4y) + (5x - 4y) = 22 + (-3)$
$11x = 19 \Rightarrow x = \dfrac{19}{11}$
Wait — let's recalculate: $6x + 5x = 11x$, $4y - 4y = 0$ ✓, $22 - 3 = 19$. So $x = \dfrac{19}{11}$.
Substitute into (1): $3 \cdot \dfrac{19}{11} + 2y = 11 \Rightarrow 2y = 11 - \dfrac{57}{11} = \dfrac{121-57}{11} = \dfrac{64}{11} \Rightarrow y = \dfrac{32}{11}$
Answers: $x = \dfrac{19}{11},\quad y = \dfrac{32}{11}$
[M1 scaling; M1 solving for $x$; A1 both correct]
Multiply eq(1) by 2: $6x + 4y = 22$ … (3)
Add eq(3) to eq(2): $(6x + 4y) + (5x - 4y) = 22 + (-3)$
$11x = 19 \Rightarrow x = \dfrac{19}{11}$
Wait — let's recalculate: $6x + 5x = 11x$, $4y - 4y = 0$ ✓, $22 - 3 = 19$. So $x = \dfrac{19}{11}$.
Substitute into (1): $3 \cdot \dfrac{19}{11} + 2y = 11 \Rightarrow 2y = 11 - \dfrac{57}{11} = \dfrac{121-57}{11} = \dfrac{64}{11} \Rightarrow y = \dfrac{32}{11}$
Answers: $x = \dfrac{19}{11},\quad y = \dfrac{32}{11}$
[M1 scaling; M1 solving for $x$; A1 both correct]
Q44 marks
A cinema sells adult tickets for £$a$ and child tickets for £$c$. Three adults and two children pay £31. One adult and four children pay £23. Find the cost of each type of ticket.
Equations: $3a + 2c = 31$ … (1) and $a + 4c = 23$ … (2)
Multiply (2) by 3: $3a + 12c = 69$ … (3)
$(3) - (1)$: $10c = 38 \Rightarrow c = 3.80$
Substitute: $a + 4(3.80) = 23 \Rightarrow a = 23 - 15.20 = 7.80$
Check in (1): $3(7.80) + 2(3.80) = 23.40 + 7.60 = 31$ ✓
Adult ticket: £7.80; Child ticket: £3.80
[B1 both equations; M1 valid method; A1 one value; A1 both values]
Multiply (2) by 3: $3a + 12c = 69$ … (3)
$(3) - (1)$: $10c = 38 \Rightarrow c = 3.80$
Substitute: $a + 4(3.80) = 23 \Rightarrow a = 23 - 15.20 = 7.80$
Check in (1): $3(7.80) + 2(3.80) = 23.40 + 7.60 = 31$ ✓
Adult ticket: £7.80; Child ticket: £3.80
[B1 both equations; M1 valid method; A1 one value; A1 both values]
Q53 marks
Explain why the simultaneous equations $2x + 4y = 10$ and $x + 2y = 8$ have no solution.
Explanation:
Divide equation 1 by 2: $x + 2y = 5$.
This contradicts equation 2: $x + 2y = 8$.
The left-hand sides are identical but the right-hand sides are different ($5 \neq 8$).
Therefore there is no pair of values of $x$ and $y$ that can satisfy both equations simultaneously.
Graphically: both equations represent parallel lines (same gradient $-\frac{1}{2}$, different $y$-intercepts), so they never intersect.
[B1 simplify or rearrange one equation; B1 identify contradiction; B1 clear explanation referencing parallel lines or contradiction]
Divide equation 1 by 2: $x + 2y = 5$.
This contradicts equation 2: $x + 2y = 8$.
The left-hand sides are identical but the right-hand sides are different ($5 \neq 8$).
Therefore there is no pair of values of $x$ and $y$ that can satisfy both equations simultaneously.
Graphically: both equations represent parallel lines (same gradient $-\frac{1}{2}$, different $y$-intercepts), so they never intersect.
[B1 simplify or rearrange one equation; B1 identify contradiction; B1 clear explanation referencing parallel lines or contradiction]
Q66 marks
The sum of two numbers is 18. Three times the larger number minus twice the smaller number equals 19. Find both numbers. Show all working.
Define variables: Let $x$ = larger number, $y$ = smaller number.
Equations:
(1) $x + y = 18$
(2) $3x - 2y = 19$
Elimination — multiply (1) by 2:
$2x + 2y = 36$ … (3)
Add (3) and (2): $5x = 55 \Rightarrow x = 11$
Substitute into (1): $11 + y = 18 \Rightarrow y = 7$
Verify (1): $11 + 7 = 18$ ✓ Verify (2): $3(11) - 2(7) = 33 - 14 = 19$ ✓
Answers: The two numbers are 11 and 7.
[B1 variables defined; B1 eq(1); B1 eq(2); M1 valid method; A1 $x=11$; A1 $y=7$]
Equations:
(1) $x + y = 18$
(2) $3x - 2y = 19$
Elimination — multiply (1) by 2:
$2x + 2y = 36$ … (3)
Add (3) and (2): $5x = 55 \Rightarrow x = 11$
Substitute into (1): $11 + y = 18 \Rightarrow y = 7$
Verify (1): $11 + 7 = 18$ ✓ Verify (2): $3(11) - 2(7) = 33 - 14 = 19$ ✓
Answers: The two numbers are 11 and 7.
[B1 variables defined; B1 eq(1); B1 eq(2); M1 valid method; A1 $x=11$; A1 $y=7$]
⭐ Grade 9 Model Answers
Important — What makes a Grade 9 answer?
A Grade 9 answer for simultaneous equations demonstrates: (1) clear variable definitions, (2) systematic and fully shown method, (3) correct algebraic manipulation including non-integer answers, (4) explicit verification in both equations, and (5) a contextual conclusion (for word problems).
Model Answer — Q5 (Explaining No Solution)
Grade 9 · Full explanation
(1): $y = -\frac{1}{2}x + \frac{5}{2}$, (2): $y = -\frac{1}{2}x + 4$
Both lines have gradient $-\frac{1}{2}$ but different $y$-intercepts ($\frac{5}{2}$ and $4$). They are parallel and will never intersect.
Explain why $2x + 4y = 10$ and $x + 2y = 8$ have no solution. [3 marks]
1
Simplify equation 1
Divide equation (1) by 2: $x + 2y = 5$ → call this (1*)
2
Compare with equation 2
Equation (2): $x + 2y = 8$. The left-hand side of (1*) is identical to that of (2), but $5 \neq 8$. The same expression $x + 2y$ cannot equal both 5 and 8 at the same time — this is a contradiction.
3
Graphical reasoning
Rewrite both in $y = mx + c$ form:(1): $y = -\frac{1}{2}x + \frac{5}{2}$, (2): $y = -\frac{1}{2}x + 4$
Both lines have gradient $-\frac{1}{2}$ but different $y$-intercepts ($\frac{5}{2}$ and $4$). They are parallel and will never intersect.
Conclusion: The equations represent parallel lines (same gradient $-\frac{1}{2}$, different $y$-intercepts). Parallel lines have no point of intersection, so there is no solution.
Annotation — Why Each Part Earns Marks
| Step | What was done | Mark awarded |
|---|---|---|
| Simplify eq(1) | Divided by 2 to reveal $x + 2y = 5$ | B1 — identifies equivalent form |
| Contradiction | Recognised $x + 2y$ cannot equal 5 AND 8 | B1 — clear contradiction stated |
| Gradients | Converted to $y=mx+c$ and compared gradients and intercepts | B1 — links to parallel lines / geometric meaning |
Exam Tip — "Explain" questions
When asked to "explain", you must give a reason, not just a calculation. For simultaneous equations, the key words are: parallel lines, same gradient, different intercepts, no intersection, and contradiction. Use at least two of these to secure all marks.
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Simultaneous | Both equations true at same time |
| Elimination | Remove a variable by add/subtract |
| Substitution | Replace variable with expression |
| Consistent | System has at least one solution |
| Inconsistent | No solution exists (parallel lines) |
| Dependent | Infinite solutions (same line) |
Essential Formulae
Elimination: Scale to match one coefficient, then add/subtract.
$$\text{Same signs} \Rightarrow \text{subtract}$$
$$\text{Opposite signs} \Rightarrow \text{add}$$
Substitution: $y = \dfrac{c_1 - a_1 x}{b_1}$, then substitute.
Parallel lines: $m_1 = m_2$, $c_1 \neq c_2$ → no solution.
Memory Hooks
- "SS → Subtract, OS → Add" for elimination sign rule
- Always label equations (1) and (2)
- Check = substitute BOTH values into BOTH equations
- Parallel = same gradient = no solution
- Define variables BEFORE forming equations in word problems
- Non-integer answers are valid — don't round unless asked
Exam Tips
- State $x=\ldots, y=\ldots$ clearly at the end
- Show all multiplication steps — method marks available
- For word problems, state units in the answer
- "Explain" questions need words, not just numbers
- If coefficients are ugly, try the other method
- Non-integer solutions: leave as fractions, not decimals
- Always verify — free marks if you show it
🔄 Flashcards
Click a card to reveal the answer. 15 cards total.
✗ Common Mistakes
Mistake 1 — Wrong sign when subtracting equations
What students do: When subtracting equation (2) from equation (1), they subtract the left-hand side but forget to subtract the right-hand side (or subtract a positive as positive).Why marks are lost: Gives a completely wrong value for the variable being found, leading to all subsequent marks being lost (typically 2–3 marks).
How to avoid: Always write the subtraction out in full: $(3x + 2y) - (x + 2y) = 13 - 7$. Bracket both sides before subtracting.
Mistake 2 — Adding instead of subtracting (or vice versa)
What students do: Confuse which operation to use when both coefficients have the same sign versus opposite signs.Why marks are lost: Produces an incorrect combined equation, meaning the variable is not eliminated.
How to avoid: Memorise "Same Sign → Subtract; Opposite Sign → Add". Before operating, check the signs on the chosen variable in each equation.
Mistake 3 — Substituting back into the wrong equation
What students do: After finding one variable, substitute into the equation they already used (the rearranged one), instead of one of the original equations.Why marks are lost: The substituted equation is tautologically true ($0=0$) so no value is found, wasting time; or they use the scaled equation and introduce arithmetic errors.
How to avoid: Always substitute back into one of the original, unscaled equations. Label original equations clearly and never lose track of them.
Mistake 4 — Finding only one variable
What students do: Correctly find $x$, then stop without finding $y$ (or vice versa).Why marks are lost: The final answer mark (typically A1) requires both values. Half the solution earns zero of the answer marks.
How to avoid: After finding the first variable, always ask "Have I found both $x$ and $y$?" before writing the final answer. Check the question mark count — it often asks for both.
Mistake 5 — Multiplying only one term when scaling
What students do: When multiplying equation (1) by a number to match coefficients, they multiply only the left-hand side and forget to multiply the right-hand side.Why marks are lost: The scaled equation is incorrect, so elimination gives a wrong value. This is a very common M1 error that costs all subsequent marks.
How to avoid: Write the multiplication explicitly: "$\times 3$: $3(5x - 2y) = 3(11)$". Always apply the multiplier to every term including the constant.
Mistake 6 — Rounding non-integer answers prematurely
What students do: When solving gives a non-integer result such as $y = \frac{7}{3}$, students convert to $2.333\ldots$ and round to $2.3$, then use this rounded value to find $x$, accumulating rounding errors.Why marks are lost: The final answer is incorrect due to the rounding, and verification will fail. Examiners expect exact fractional answers unless the question says "correct to 2 d.p."
How to avoid: Keep fractions exact throughout. Only convert to decimals if the question explicitly asks for a decimal answer. Write fractions clearly: $y = \frac{7}{3}$ not $y \approx 2.3$.
✅ Final Checklist
Click each item when you are confident with it.
- I can explain what simultaneous equations are and what their solution represents geometrically
- I can use elimination when both equations are in standard form $ax + by = c$
- I can scale one or both equations to match coefficients before eliminating
- I know to subtract when signs are the same and add when signs are opposite
- I can use substitution, especially when one variable has coefficient 1
- I can choose the most efficient method for a given pair of equations
- I can form two equations from a word problem after defining variables
- I can handle non-integer (fractional) solutions without rounding
- I always verify my solution in both original equations
- I can explain why parallel lines give no solution (same gradient argument)
- I can explain why identical lines give infinitely many solutions
- I state my final answer clearly as $x = \ldots,\ y = \ldots$ with units where applicable
- I have practised at least 5 exam-style questions under timed conditions
- I can apply simultaneous equations to geometry problems (perimeter, angles, etc.)
- I can write a Grade 9 quality "explain" answer about the number of solutions