MathsICSE Class 9 · Chapter 5

Simultaneous Linear Equations ICSE Class 9: Elimination, Substitution and Cross-Multiplication

Updated 15 October 2026 · Fact-checked

Simultaneous linear equations are two equations in x and y that must both be true at once. ICSE Class 9 solves them by elimination (make one coefficient equal and add or subtract), substitution (express one variable from one equation and put it in the other) and cross-multiplication, and uses them to solve word problems.

What Simultaneous Linear Equations is about

A linear equation in two variables has the form $a x + b y + c = 0$ , where $a$ and $b$ are not both zero. On its own it has infinitely many solutions. Two such equations taken together are simultaneous equations: the solution is the pair $(x, y)$ that satisfies both.

The CISCE 2028 Class IX syllabus asks you to solve simultaneous linear equations in two variables, with numerical coefficients only, algebraically by elimination, substitution and cross-multiplication, and to solve simple problems by framing equations. Solving the same equations by drawing graphs is covered in Coordinate Geometry.

In the school exam expect 3-mark 'solve' questions (sometimes 'by cross-multiplication' is specified), equations with fractions or decimals to clear first, and 4 to 5-mark word problems on numbers, ages, fractions, digits, costs and speeds.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Simultaneous Linear Equations is in the 3. Algebra (iii) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

What a solution means

A solution of the pair is an ordered pair $(x, y)$ that makes both equations true. Check any answer by substituting into both original equations, not into a rearranged one.

Before solving, tidy each equation into the form $a x + b y = c$ : clear fractions by multiplying by the LCM, clear decimals by multiplying by 10 or 100, and collect like terms.

Worked example

Is $(3, - 1)$ a solution of $2 x + 3 y = 3$ and $x - y = 4$ ?

Try it first, then show the solutionSolution

First: $2 (3) + 3 (- 1) = 6 - 3 = 3$ ✓.
Second: $3 - (- 1) = 4$ ✓.
Both are true, so yes.

Answer: Yes, (3, −1) is the solution.

Elimination method

1. Multiply one or both equations by suitable numbers so that the coefficients of one variable become equal in size.
2. Add the equations if those coefficients have opposite signs; subtract if they have the same sign.
3. Solve the resulting one-variable equation.
4. Substitute back to find the other variable.

Choose the variable that needs the smaller multipliers.

Worked example

Solve $3 x + 2 y = 11$ and $2 x - 3 y = 3$ by elimination.

Try it first, then show the solutionSolution

Multiply the first by 3 and the second by 2: $9 x + 6 y = 33$ , $4 x - 6 y = 6$ .
Add: $13 x = 39$ , so $x = 3$ .
Substitute: $9 + 2 y = 11$ , so $y = 1$ .

Answer: x = 3, y = 1

Substitution method

1. From one equation, express one variable in terms of the other (choose a variable with coefficient 1 or $- 1$ if possible).
2. Substitute this expression into the other equation.
3. Solve for the remaining variable, then find the first.

Substitution is quickest when a coefficient is already 1.

Worked example

Solve $x + 2 y = 7$ and $3 x - y = 7$ by substitution.

Try it first, then show the solutionSolution

From the first: $x = 7 - 2 y$ .
Substitute: $3 (7 - 2 y) - y = 7$ , so $21 - 7 y = 7$ and $y = 2$ .
$x = 7 - 4 = 3$ .

Answer: x = 3, y = 2

Cross-multiplication method

Write both equations as $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ (everything on the left, 0 on the right). Then

$\frac{x}{b _{1} c _{2} - b _{2} c _{1}} = \frac{y}{c _{1} a _{2} - c _{2} a _{1}} = \frac{1}{a _{1} b _{2} - a _{2} b _{1}}$ .

A handy layout: write the coefficients in the order $b, c, a, b$ for each equation and cross-multiply down-right minus up-right. If $a_{1} b_{2} - a_{2} b_{1} = 0$ , the pair has no unique solution (the lines are parallel or the same).

Worked example

Solve by cross-multiplication: $2 x + 3 y - 8 = 0$ , $4 x - 5 y + 6 = 0$ .

Try it first, then show the solutionSolution

$a_{1} = 2, b_{1} = 3, c_{1} = - 8$ ; $a_{2} = 4, b_{2} = - 5, c_{2} = 6$ .
$b_{1} c_{2} - b_{2} c_{1} = 18 - 40 = - 22$ ; $c_{1} a_{2} - c_{2} a_{1} = - 32 - 12 = - 44$ ; $a_{1} b_{2} - a_{2} b_{1} = - 10 - 12 = - 22$ .
$x = \frac{- 22}{- 22} = 1$ , $y = \frac{- 44}{- 22} = 2$ .

Answer: x = 1, y = 2

Special patterns: swapped coefficients and reciprocals

Swapped coefficients: when the equations look like $a x + b y = c$ and $b x + a y = d$ , add them and subtract them. You get $x + y$ and $x - y$ directly.

Reciprocals: equations such as $\frac{2}{x} + \frac{3}{y} = 13$ are not linear in $x$ and $y$ , but they are linear in $p = \frac{1}{x}$ and $q = \frac{1}{y}$ . Solve for $p$ and $q$ , then invert.

Worked example

Solve $41 x + 53 y = 135$ and $53 x + 41 y = 147$ .

Try it first, then show the solutionSolution

Add: $94 x + 94 y = 282$ , so $x + y = 3$ .
Subtract (second minus first): $12 x - 12 y = 12$ , so $x - y = 1$ .
Add these: $2 x = 4$ , $x = 2$ ; then $y = 1$ .

Answer: x = 2, y = 1

Framing equations from word problems

1. Choose letters for the two unknowns and say what they stand for, with units.
2. Translate each condition into an equation.
3. Solve by any method.
4. Check the answer against the words, and write it as a sentence.

Standard set-ups:

Two-digit number with tens digit $t$ and units digit $u$ : number $= 10 t + u$ ; reversed $= 10 u + t$ .
Fraction $\frac{x}{y}$ : changes are applied to numerator and denominator separately.
Boat with still-water speed $b$ and stream speed $s$ : downstream $b + s$ , upstream $b - s$ .
Ages: 'after $n$ years' adds $n$ to each age; 'ago' subtracts.

Worked example

A father is three times as old as his son. After 12 years he will be twice as old as his son. Find their present ages.

Try it first, then show the solutionSolution

Let the son be $s$ years and the father $f$ years. $f = 3 s$ .
$f + 12 = 2 (s + 12)$ , so $3 s + 12 = 2 s + 24$ .
$s = 12$ , $f = 36$ .

Answer: Son 12 years, father 36 years.

Maths in real life: Simultaneous Linear Equations

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.

Two prices from two bills

Suppose one customer pays ₹110 for 3 pens and 2 notebooks, and another pays ₹115 for 2 pens and 3 notebooks at the same shop.

Find the price of a pen and of a notebook.

Try it first, then show the solutionSolution

$3 p + 2 n = 110$ and $2 p + 3 n = 115$ .
Add: $5 (p + n) = 225$ , so $p + n = 45$ . Subtract: $p - n = - 5$ .
$p = 20$ , $n = 25$ .

Answer: Pen ₹20, notebook ₹25.

Memory hook: Swapped coefficients: add and subtract.

Boat and stream

Imagine a boat on a river covers 30 km downstream in 2 hours and 18 km upstream in 2 hours.

Find the speed of the boat in still water and the speed of the stream.

Try it first, then show the solutionSolution

$b + s = 15$ km/h and $b - s = 9$ km/h.
$b = 12$ , $s = 3$ .

Answer: Boat 12 km/h, stream 3 km/h.

Memory hook: The stream helps one way and hinders the other.

Two cars, two meeting times

Suppose two towns are 100 km apart and two cars start from them at the same time. If they drive in the same direction, the faster catches the slower in 5 hours; if they drive towards each other, they meet in 1 hour.

Find the speeds of the cars.

Try it first, then show the solutionSolution

Towards each other: $u + v = 100$ .
Same direction: the gap closes at $u - v$ , so $5 (u - v) = 100$ , $u - v = 20$ .
$u = 60$ , $v = 40$ .

Answer: 60 km/h and 40 km/h.

Memory hook: Towards: speeds add. Chasing: speeds subtract.

Topics in this chapter

Simultaneous Linear Equations formula sheet

Standard form: $a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$; Needed for cross-multiplication.
Cross-multiplication: $\frac{x}{b _{1} c _{2} - b _{2} c _{1}} = \frac{y}{c _{1} a _{2} - c _{2} a _{1}} = \frac{1}{a _{1} b _{2} - a _{2} b _{1}}$; Requires a₁b₂ − a₂b₁ ≠ 0.
Two-digit number: $N = 10 t + u; N_{r e v} = 10 u + t$; N − N_rev = 9(t − u).
Boat and stream: $v_{d o w n} = b + s; v_{u p} = b - s$; Time = distance ÷ speed.
Swapped coefficients: $(a + b) (x + y) = c + d$; From ax + by = c and bx + ay = d, added.

How the school exam asks Simultaneous Linear Equations

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Solve by a named method	Solve by substitution, by elimination, or by cross-multiplication.	Use the method asked for; marks are given for the method as well as the answer.
Equations with fractions or decimals	x/2 + y/3 = 4 type, or 0.4x + 0.3y = 1.7.	Multiply by the LCM or by 10 to get whole-number coefficients first.
Swapped or large coefficients	41x + 53y = 135 and 53x + 41y = 147.	Add and subtract to get x + y and x − y.
Reciprocal equations	2/x + 3/y = 13 and 5/x − 4/y = −2.	Put p = 1/x, q = 1/y; solve; invert.
Word problems	Numbers, ages, fractions, two-digit numbers, costs, speeds, boats.	Define variables, form two equations, solve, check against the words.

Tricks to remember Simultaneous Linear Equations

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

Same signs subtract, different signs add
SSS, DSA: Same Sign Subtract, Different Sign Add.
Use it when: Elimination step.
b c a b
For cross-multiplication write b₁ c₁ a₁ b₁ over b₂ c₂ a₂ b₂ and cross-multiply each neighbouring pair.
Use it when: Remembering the three denominators in order.
Equal to zero first
Cross-multiplication only works with '= 0' on the right. Move c across first.
Use it when: Avoiding sign errors in c₁ and c₂.
Check in both
One equation proves nothing; check your pair in both originals.
Use it when: Finding errors before the examiner does.

Simultaneous Linear Equations: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Checking a solution · 1 mark · MCQExam level

The solution of $x + y = 7$ and $x - y = 1$ is:
(a) $(3, 4)$
(b) $(4, 3)$
(c) $(5, 2)$
(d) $(6, 1)$

Try it first, then show the solutionSolution

Add: $2 x = 8$ , $x = 4$ ; then $y = 3$ .

Answer: (b)

Example 2: Fractions first · 3 marks · Short answerExam level

Solve $\frac{x}{2} + \frac{y}{3} = 4$ and $\frac{x}{3} - \frac{y}{2} = \frac{1}{2}$ .

Try it first, then show the solutionSolution

Multiply by 6: $3 x + 2 y = 24$ and $2 x - 3 y = 3$ .
Multiply by 3 and 2: $9 x + 6 y = 72$ , $4 x - 6 y = 6$ . Add: $13 x = 78$ , $x = 6$ .
$18 + 2 y = 24$ , so $y = 3$ .

Answer: x = 6, y = 3

Example 3: Decimals by cross-multiplication · 3 marks · Short answerExam level

Solve by cross-multiplication: $0.4 x + 0.3 y = 1.7$ and $0.7 x - 0.2 y = 0.8$ .

Try it first, then show the solutionSolution

Multiply by 10: $4 x + 3 y - 17 = 0$ , $7 x - 2 y - 8 = 0$ .
$b_{1} c_{2} - b_{2} c_{1} = - 24 - 34 = - 58$ ; $c_{1} a_{2} - c_{2} a_{1} = - 119 + 32 = - 87$ ; $a_{1} b_{2} - a_{2} b_{1} = - 8 - 21 = - 29$ .
$x = \frac{- 58}{- 29} = 2$ , $y = \frac{- 87}{- 29} = 3$ .

Answer: x = 2, y = 3

Example 4: Reciprocal equations · 3 marks · Short answerHarder exam level

Solve $\frac{2}{x} + \frac{3}{y} = 13$ and $\frac{5}{x} - \frac{4}{y} = - 2$ .

Try it first, then show the solutionSolution

Let $p = \frac{1}{x}$ , $q = \frac{1}{y}$ : $2 p + 3 q = 13$ , $5 p - 4 q = - 2$ .
Multiply by 4 and 3: $8 p + 12 q = 52$ , $15 p - 12 q = - 6$ . Add: $23 p = 46$ , $p = 2$ ; then $q = 3$ .
$x = \frac{1}{2}$ , $y = \frac{1}{3}$ .

Answer: x = 1/2, y = 1/3

Example 5: Fraction problem · 4 marks · Long answerHarder exam level

If 1 is added to both the numerator and denominator of a fraction, it becomes $\frac{4}{5}$ . If 5 is subtracted from both, it becomes $\frac{1}{2}$ . Find the fraction.

Try it first, then show the solutionSolution

Let the fraction be $\frac{x}{y}$ . $\frac{x + 1}{y + 1} = \frac{4}{5}$ gives $5 x - 4 y = - 1$ .
$\frac{x - 5}{y - 5} = \frac{1}{2}$ gives $2 x - y = 5$ , so $y = 2 x - 5$ .
$5 x - 4 (2 x - 5) = - 1$ : $- 3 x + 20 = - 1$ , $x = 7$ ; $y = 9$ .
Check: $\frac{8}{10} = \frac{4}{5}$ , $\frac{2}{4} = \frac{1}{2}$ ✓.

Answer: 7/9

Example 6: Two-digit number · 4 marks · Long answerHarder exam level

The sum of the digits of a two-digit number is 9. The number exceeds the number formed by reversing its digits by 27. Find the number.

Try it first, then show the solutionSolution

Tens digit $t$ , units digit $u$ : $t + u = 9$ .
$(10 t + u) - (10 u + t) = 27$ , so $9 (t - u) = 27$ and $t - u = 3$ .
$t = 6$ , $u = 3$ . Number $= 63$ . Check: $63 - 36 = 27$ ✓.

Answer: 63

Example 7: Two possible numbers · 4 marks · Long answerHOTS / competency

The sum of a two-digit number and the number formed by reversing its digits is 110, and the digits differ by 4. Find all such numbers.

Try it first, then show the solutionSolution

$(10 t + u) + (10 u + t) = 110$ , so $11 (t + u) = 110$ and $t + u = 10$ .
'Differ by 4' gives $t - u = 4$ or $u - t = 4$ .
Case 1: $t = 7$ , $u = 3$ , number 73. Case 2: $t = 3$ , $u = 7$ , number 37.
Both check: $73 + 37 = 110$ .

Answer: 73 or 37

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

The school canteen

A school canteen sells sandwiches and juice. On Monday, 40 sandwiches and 30 juices brought in ₹2,900. On Tuesday, 25 sandwiches and 50 juices brought in ₹3,125. On Wednesday a combo of 1 sandwich and 1 juice is offered for ₹5 less than the separate prices.

(i) · 3 marks
Form two equations for the price $s$ of a sandwich and $j$ of a juice, in their simplest form.
Answer
Answer: $4 s + 3 j = 290$ ; $s + 2 j = 125$
Monday: $40 s + 30 j = 2900 \Rightarrow 4 s + 3 j = 290$ . Tuesday: $25 s + 50 j = 3125 \Rightarrow s + 2 j = 125$ .
(ii) · 3 marks
Solve them by substitution.
Answer
Answer: s = ₹41, j = ₹42
$s = 125 - 2 j$ ; $4 (125 - 2 j) + 3 j = 290 \Rightarrow 500 - 5 j = 290 \Rightarrow j = 42$ . Then $s = 125 - 84 = 41$ . Check: $4 (41) + 3 (42) = 164 + 126 = 290$ ✓; $41 + 84 = 125$ ✓. So $s = ₹41$ , $j = ₹42$ .
(iii) · 4 marks
On Wednesday 60 combos were sold. Find the takings and the discount given in total.
Answer
Answer: ₹4,680; ₹300
Combo price $= 41 + 42 - 5 = 78$ . Takings $= 60 \times 78 = ₹4, 680$ . Discount $= 60 \times 5 = ₹300$ .

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Solve by substitution: $2 x - y = 4$ and $3 x + 2 y = 13$ . (3 marks)

Step	Marks
$y = 2 x - 4$	1
$3 x + 2 (2 x - 4) = 13 \Rightarrow 7 x = 21 \Rightarrow x = 3$	1
$y = 2$	1

Solve by cross-multiplication: $3 x - 4 y = 10$ and $4 x + 3 y = 5$ . (4 marks)

Step	Marks
Standard form: $3 x - 4 y - 10 = 0$ , $4 x + 3 y - 5 = 0$	1
$b_{1} c_{2} - b_{2} c_{1} = 20 + 30 = 50$ ; $c_{1} a_{2} - c_{2} a_{1} = - 40 + 15 = - 25$	1
$a_{1} b_{2} - a_{2} b_{1} = 9 + 16 = 25$	1
$x = 2$ , $y = - 1$	1

Five years ago, a woman was seven times as old as her daughter. Five years from now, she will be three times as old as her daughter. Find their present ages. (5 marks)

Step	Marks
Let the present ages be $w$ and $d$ years	1
$w - 5 = 7 (d - 5) \Rightarrow w - 7 d = - 30$	1
$w + 5 = 3 (d + 5) \Rightarrow w - 3 d = 10$	1
Subtract: $4 d = 40$ , so $d = 10$	1
$w = 40$ : woman 40 years, daughter 10 years	1

Where students lose marks in Simultaneous Linear Equations

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

Mistake: Sign errors in cross-multiplication
Students use $c$ from $a x + b y = c$ without moving it to the left, so every $c$ has the wrong sign.
Wrong working
$3 x - 4 y = 10$ : $c_{1} = 10$
Right working
$3 x - 4 y - 10 = 0$ : $c_{1} = - 10$
Do this: Rewrite both equations as '... = 0' before reading off a, b, c.
Usually costs: The whole question.
Mistake: Multiplying only some terms
When scaling an equation to eliminate a variable, the constant on the right is often left unchanged.
Do this: Put the whole equation in brackets and multiply every term, including the right side.
Usually costs: Every mark after that line.
Mistake: Adding when you should subtract
Coefficients with the same sign must be subtracted to cancel.
Do this: Look at the signs: same sign subtract, different sign add.
Usually costs: 1 to 2 marks.
Mistake: Not defining the variables in word problems
Without 'Let the son's age be s years', the equations have no meaning and method marks can be lost.
Do this: Start every word problem with a 'Let …' line stating units.
Usually costs: 1 mark.
Mistake: Answering with x and y instead of the question's quantities
The question asks for 'the fraction' or 'the number', not $x = 7$ , $y = 9$ .
Do this: Finish with a sentence: 'The fraction is 7/9.'
Usually costs: 1 mark.

Practice questions on Simultaneous Linear Equations

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 · MCQExam level
If $2 x + y = 7$ and $x + 2 y = 8$ , then $x + y$ equals:
(a) 3
(b) 5
(c) 15
(d) 1
Hint
Add the equations.
Answer
(b): $3 (x + y) = 15$ .
Q2 · 2 marks · Short answerExam level
Solve by elimination: $5 x + 3 y = 21$ and $2 x - 3 y = 0$ .
Hint
Add.
Answer
$7 x = 21$ , $x = 3$ ; $y = 2$ .
Q3 · 2 marks · Short answerExam level
The sum of two numbers is 50 and their difference is 14. Find the numbers.
Hint
x + y = 50, x − y = 14.
Answer
32 and 18.
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): The pair $x + 2 y = 4$ and $2 x + 4 y = 9$ has no solution.
Reason (R): The second equation is twice the first.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Twice the first would be 2x + 4y = 8.
Answer
(a). Doubling the first gives $2 x + 4 y = 8$ , which contradicts $= 9$ , so there is no solution; but the second equation is not twice the first (9 ≠ 8).
Q5 · 3 marks · Short answerExam level
Solve by cross-multiplication: $x + y = 5$ and $3 x - 2 y = 5$ .
Hint
x + y − 5 = 0, 3x − 2y − 5 = 0.
Answer
Denominators: $(1) (- 5) - (- 2) (- 5) = - 15$ , $(- 5) (3) - (- 5) (1) = - 10$ , $(1) (- 2) - (3) (1) = - 5$ . $x = 3$ , $y = 2$ .
Q6 · 3 marks · Short answerHarder exam level
Solve $\frac{x + 1}{2} + \frac{y - 1}{3} = 8$ and $\frac{x - 1}{3} + \frac{y + 1}{2} = 9$ .
Hint
Multiply each by 6.
Answer
$3 x + 2 y = 47$ and $2 x + 3 y = 53$ . Add: $x + y = 20$ ; subtract: $x - y = - 6$ . $x = 7$ , $y = 13$ .
Q7 · 3 marks · Short answerHarder exam level
3 chairs and 2 tables cost ₹7,000, while 5 chairs and 3 tables cost ₹10,900. Find the cost of 2 chairs and 1 table.
Hint
Eliminate tables.
Answer
Multiply by 3 and 2: $9 c + 6 t = 21000$ , $10 c + 6 t = 21800$ ; $c = 800$ , $t = 2300$ . $2 c + t = ₹3, 900$ .
Q8 · 4 marks · Long answerHarder exam level
Five years ago, A was thrice as old as B. Ten years later, A will be twice as old as B. Find their present ages.
Hint
a − 5 = 3(b − 5); a + 10 = 2(b + 10).
Answer
$a - 3 b = - 10$ ; $a - 2 b = 10$ . $b = 20$ , $a = 50$ .
Q9 · 4 marks · Long answerHarder exam level
A man rows 40 km downstream and 24 km upstream, taking 4 hours each way. Find his speed in still water and the speed of the stream.
Hint
Speeds are 10 and 6 km/h.
Answer
$b + s = 10$ , $b - s = 6$ : $b = 8$ km/h, $s = 2$ km/h.
Q10 · 3 marks · Short answerHarder exam level
Solve $\frac{3}{x} - \frac{1}{y} = - 9$ and $\frac{2}{x} + \frac{3}{y} = 5$ .
Hint
p = 1/x, q = 1/y.
Answer
$3 p - q = - 9$ , $2 p + 3 q = 5$ . $q = 3 p + 9$ ; $2 p + 9 p + 27 = 5$ , $p = - 2$ , $q = 3$ . $x = - \frac{1}{2}$ , $y = \frac{1}{3}$ .
Q11 · 4 marks · Long answerHOTS / competency
In a cricket match, a batter hit only fours and sixes. He hit 13 boundaries in all and scored 64 runs from them. Some spectators claim he hit 6 sixes. Test the claim by solving the equations.
Hint
f + s = 13 and 4f + 6s = 64.
Answer
$4 f + 6 s = 64$ and $4 f + 4 s = 52$ ; subtract: $2 s = 12$ , $s = 6$ , $f = 7$ . The claim is correct: 7 fours and 6 sixes ( $28 + 36 = 64$ ).
Q12 · 4 marks · Long answerHOTS / competency
Priya says she thought of a two-digit number whose digits add up to 8 and whose reversed number is 30 more than it. Show by forming equations that she must have made a mistake. Then find the number if the difference had been 36.
Hint
Reversed minus original is 9(u − t).
Answer
$t + u = 8$ and $9 (u - t) = 30$ give $u - t = \frac{10}{3}$ , which is not a whole number, so no such digits exist. With 36: $u - t = 4$ , so $u = 6$ , $t = 2$ ; the number is 26 (check: $62 - 26 = 36$ ).

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 simultaneous equations?Show answer
An ordered pair (x, y) that satisfies both equations.
2.Write the cross-multiplication formula.Show answer
x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁).
3.In elimination, when do you add the equations?Show answer
When the coefficients to be eliminated have opposite signs.
4.How do you write a two-digit number with tens digit t and units digit u?Show answer
10t + u.
5.Downstream speed of a boat?Show answer
Speed in still water + speed of stream.
6.What substitution turns 2/x + 3/y = 13 into a linear equation?Show answer
p = 1/x, q = 1/y.
7.What happens if a₁b₂ − a₂b₁ = 0?Show answer
There is no unique solution.
8.What is the first step with x/2 + y/3 = 4?Show answer
Multiply by 6 to clear fractions: 3x + 2y = 24.

Best order to study it

Practise checking a given pair in both equations.
Master elimination and substitution with whole-number coefficients.
Learn cross-multiplication with the b c a b layout.
Practise fractions, decimals, swapped coefficients and reciprocal forms.
Finish with word problems of every type.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you choose the quicker of elimination and substitution?
Do you rewrite equations as '= 0' before cross-multiplying?
Do you clear fractions and decimals first?
Can you add and subtract to solve swapped-coefficient pairs?
Can you set up digit, fraction, age and boat problems?
Do you check your answer in both original equations?
Do you finish word problems with a sentence answer?

Simultaneous Linear Equations: frequently asked questions

Which method is best?

Any method gives the same answer. Use substitution when a coefficient is 1, elimination otherwise, and cross-multiplication when the question asks for it.

Are equations with letters as coefficients in the syllabus?

No. The Class IX syllabus says numerical coefficients only.

Where is the graphical method?

Solving simultaneous equations graphically is part of the Coordinate Geometry chapter in the Class IX syllabus.

Can a pair of linear equations have no solution?

Yes, if the lines are parallel, for example x + y = 2 and x + y = 5. Then a₁b₂ − a₂b₁ = 0.

Do I need to check my answer?

It is not always demanded, but a quick check in both equations catches most errors and takes seconds.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028