MathsICSE Class 9 · Simultaneous Linear Equations

Simultaneous Linear Equations: Elimination and Substitution Methods (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

A solution of two linear equations in x and y is a pair (x, y) that satisfies both. By elimination, make the coefficients of one variable equal in size, add or subtract to remove it, and solve; by substitution, express one variable from one equation and substitute into the other. Always check in both equations.

The solution of two equations is the single point where their lines cross; substitution finds that same point, (2, 1), by algebra.

Illustration drawn for QuestionBuddy.

Understand Solving Simultaneous Linear Equations by Elimination and Substitution

What a solution is. A pair $(x, y)$ satisfying both $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ . Graphically it is the point where the lines cross.

Elimination. (1) Multiply one or both equations so that the coefficients of one variable are equal in size. (2) Add if the signs differ, subtract if they are the same. (3) Solve the single equation. (4) Substitute back to find the other variable. (5) Check in both equations.

*Example.* $7 x - 3 y = 31$ and $9 x - 5 y = 41$ . Multiply the first by 5 and the second by 3: $35 x - 15 y = 155$ , $27 x - 15 y = 123$ . Subtract: $8 x = 32$ , $x = 4$ ; then $28 - 3 y = 31$ , $y = - 1$ . Check: $36 + 5 = 41$ . ✓

Substitution. Make one variable the subject of the simpler equation. For $y = 2 x - 3$ and $3 x + 2 y = 8$ : $3 x + 2 (2 x - 3) = 8$ , $7 x = 14$ , $x = 2$ , $y = 1$ .

Clearing fractions. Multiply each equation by the LCM of its denominators first. $\frac{x}{2} + \frac{y}{3} = 4$ becomes $3 x + 2 y = 24$ ; $\frac{x}{4} - \frac{y}{6} = 0$ becomes $3 x - 2 y = 0$ . Adding: $6 x = 24$ , $x = 4$ , $y = 6$ .

Unknown constants. If $(3, 2)$ satisfies $a x + b y = 12$ and $b x - a y = 5$ , substitute to get a pair of equations in $a, b$ : $3 a + 2 b = 12$ and $- 2 a + 3 b = 5$ . Solving gives $a = 2$ , $b = 3$ .

Which method? Elimination suits equations in standard form; substitution suits an equation already giving $y$ in terms of $x$ (or with coefficient $\pm 1$ ).

Level. Class IX is a school-exam class. Fraction clearing and unknown-constant questions are at the hardest school-exam level.

Solving Simultaneous Linear Equations by Elimination and Substitution: formulas to remember

System: $a_{1} x + b_{1} y = c_{1}, a_{2} x + b_{2} y = c_{2}$; Two unknowns.
Elimination: $equal coefficients, then add or subtract$; Same sign: subtract; opposite: add.
Substitution: $y = f (x) \Rightarrow replace y in the other$; One equation in x.
Unique solution: $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$; Lines intersect.

How to solve Solving Simultaneous Linear Equations by Elimination and Substitution questions

Clear fractions, choose the method, solve, back-substitute, check.

1Write both equations in the form $a x + b y = c$ (clear fractions first).
2Multiply to equalise one variable's coefficients, or isolate a variable.
3Add or subtract (or substitute) to get one equation in one variable.
4Solve it, then substitute back for the second variable.
5Check in both original equations.
6State the answer as $x = \dots, y = \dots$ .

Solving Simultaneous Linear Equations by Elimination and Substitution in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Stationery prices

Suppose 2 pencils and 3 erasers cost ₹19 and 3 pencils and 2 erasers cost ₹21.

Find the price of each and the cost of 4 pencils and 5 erasers.

Try it first, then show the solutionSolution

$2 p + 3 e = 19$ and $3 p + 2 e = 21$ .
Multiply by 3 and 2: $6 p + 9 e = 57$ , $6 p + 4 e = 42$ . Subtract: $5 e = 15$ , $e = 3$ , $p = 5$ .
$4 (5) + 5 (3) = 35$ .

Answer: Pencil ₹5, eraser ₹3; ₹35.

Memory hook: Two purchases give two equations.

Splitting a total

Imagine two numbers whose sum is 10 and whose difference is 4.

Find them.

Try it first, then show the solutionSolution

$x + y = 10$ , $x - y = 4$ . Add: $2 x = 14$ , $x = 7$ , $y = 3$ .

Answer: 7 and 3.

Memory hook: Add when the signs differ.

Tricks to remember Solving Simultaneous Linear Equations by Elimination and Substitution

Short memory aids. Say each one aloud once, then use it on a question the same day.

Same sign subtract
If the coefficients have the same sign, subtract; opposite signs, add.
Use it when: Eliminating cleanly.
Fractions first
Multiply through by the LCM before solving.
Use it when: Messy equations.
Check twice
Substitute into both equations.
Use it when: Catching slips.

Solving Simultaneous Linear Equations by Elimination and Substitution: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A quick solve · 1 mark · MCQBoard level

The solution of $2 x + y = 7$ and $x - y = 2$ is:
(a) $x = 2, y = 3$
(b) $x = 3, y = 1$
(c) $x = 1, y = 5$
(d) $x = 4, y = - 1$

Try it first, then show the solutionSolution

Add: $3 x = 9$ , $x = 3$ ; then $y = 7 - 6 = 1$ .

Answer: (b)

Example 2: Elimination · 3 marks · Short answerBoard level

Solve $3 x + 2 y = 16$ and $2 x - y = 6$ .

Try it first, then show the solutionSolution

Multiply the second by 2: $4 x - 2 y = 12$ .
Add to the first: $7 x = 28$ , $x = 4$ .
$2 (4) - y = 6$ , so $y = 2$ . Check: $12 + 4 = 16$ . ✓

Answer: x = 4, y = 2

Example 3: Clearing fractions · 4 marks · Short answerHarder board level

Solve $\frac{x}{2} + \frac{y}{3} = 4$ and $\frac{x}{4} - \frac{y}{6} = 0$ .

Try it first, then show the solutionSolution

First: multiply by 6: $3 x + 2 y = 24$ .
Second: multiply by 12: $3 x - 2 y = 0$ .
Add: $6 x = 24$ , $x = 4$ ; subtract: $4 y = 24$ , $y = 6$ .
Check: $2 + 2 = 4$ and $1 - 1 = 0$ . ✓

Answer: x = 4, y = 6

Example 4: Unequal coefficients · 4 marks · Short answerHarder board level

Solve $7 x - 3 y = 31$ and $9 x - 5 y = 41$ .

Try it first, then show the solutionSolution

Multiply the first by 5: $35 x - 15 y = 155$ ; the second by 3: $27 x - 15 y = 123$ .
Subtract: $8 x = 32$ , $x = 4$ .
$28 - 3 y = 31$ , so $y = - 1$ . Check: $36 + 5 = 41$ . ✓

Answer: x = 4, y = −1

Example 5: Finding unknown constants · 5 marks · Long answerHOTS / competency

The pair $x = 3$ , $y = 2$ satisfies both $a x + b y = 12$ and $b x - a y = 5$ . Find $a$ and $b$ .

Try it first, then show the solutionSolution

Substitute: $3 a + 2 b = 12$ ... (1) and $3 b - 2 a = 5$ ... (2).
Multiply (1) by 3 and (2) by 2: $9 a + 6 b = 36$ and $- 4 a + 6 b = 10$ .
Subtract: $13 a = 26$ , $a = 2$ . From (1): $6 + 2 b = 12$ , $b = 3$ .
Check (2): $9 - 4 = 5$ . ✓

Answer: a = 2, b = 3

Where students lose marks in Solving Simultaneous Linear Equations by Elimination and Substitution

The slips examiners see most often, and what to write instead.

Mistake: Adding when the signs are the same
Same-sign terms need subtraction to cancel.
Wrong working
$(3 y) + (3 y)$ to eliminate $y$ .
Right working
Subtract the equations.
Do this: Look at the signs.
Usually costs: Whole answer.
Mistake: Not multiplying every term
Multiply both sides entirely.
Wrong working
$5 (7 x - 3 y = 31)$ becomes $35 x - 15 y = 31$ .
Right working
$35 x - 15 y = 155$ .
Do this: Multiply the right side too.
Usually costs: Whole answer.
Mistake: Not checking
Slips go unnoticed.
Wrong working
Stop at x = 4.
Right working
Find y and test in both equations.
Do this: Check twice.
Usually costs: Accuracy marks.

Exam tips

Class IX is examined in school papers; choose the method you can do cleanly.
Number your equations (1) and (2) and label the multiplication.
For fractions, state the LCM you use.
Write the final answer as a pair.

Practice questions on Solving Simultaneous Linear Equations by Elimination and Substitution

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $x + y = 10$ and $x - y = 4$ , then $x$ equals:
(a) 3
(b) 6
(c) 7
(d) 14
Hint
Add the equations.
Answer
(c) 7
Q2 · 3 marks · Short answerHarder board level
Solve $2 x + 3 y = 13$ and $5 x - 2 y = 4$ .
Hint
Multiply by 2 and 3 to match y.
Answer
$4 x + 6 y = 26$ and $15 x - 6 y = 12$ ; add: $19 x = 38$ , $x = 2$ , $y = 3$ .
Q3 · 3 marks · Short answerHarder board level
Solve by substitution: $y = 2 x - 3$ and $3 x + 2 y = 8$ .
Hint
Replace y.
Answer
$3 x + 4 x - 6 = 8$ , $x = 2$ , $y = 1$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The pair $x + y = 5$ and $2 x + 2 y = 10$ has infinitely many solutions.
Reason (R): The two equations represent the same line.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
The second is twice the first.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: 2 pencils and 3 erasers cost ₹19; 3 pencils and 2 erasers cost ₹21.
(a) Form the equations.
(b) Find the price of a pencil and of an eraser.
(c) Find the cost of 4 pencils and 5 erasers.
Hint
Multiply by 3 and 2.
Answer
(a) 2p + 3e = 19, 3p + 2e = 21 (b) pencil ₹5, eraser ₹3 (c) ₹35

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is a solution of a pair of linear equations?Show answer
A pair (x, y) satisfying both.
2.When do you add to eliminate?Show answer
When the coefficients have opposite signs.
3.What do you do first with fractions?Show answer
Multiply through by the LCM.
4.How many solutions if the lines intersect?Show answer
Exactly one.

Solving Simultaneous Linear Equations by Elimination and Substitution: frequently asked questions

Which method is expected in the exam?

Any correct method is accepted; show the working.

What if y has coefficient 1?

Substitution is quick.

Do I check in the original equations?

Yes, always.