Simultaneous Linear Equations Class 9 ICSE — Chapter 6

Written by Greya LakshmiPublished on 12 October 2026

Simultaneous linear equations class 9 ICSE teaches pairs of linear equations in two variables and the three algebraic methods used to solve them. This chapter from the Class 9 ICSE Mathematics book covers substitution, equating coefficients, cross-multiplication, equations reducible to linear form, and word problems. Each section below explains the method and shows a worked pattern with fresh numbers.

Download Simultaneous Linear Equations Including Problems Class 9 ICSE Mathematics Chapter 6 PDF

Download the Simultaneous Linear Equations Including Problems Class 9 ICSE Mathematics chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 18 pages, 7 sections, 35 questions.

What is inside Chapter 6

18 pages · 7 sections · 12 worked examples · 35 exercise questions

Section	Title
6.1	INTRODUCTION
6.2	METHODS OF SOLVING SIMULTANEOUS EQUATIONS
6.3	METHOD OF ELIMINATION BY SUBSTITUTION
6.4	METHOD OF ELIMINATION BY EQUATING COEFFICIENTS
6.5	METHOD OF CROSS-MULTIPLICATION
6.6	EQUATIONS REDUCIBLE TO LINEAR EQUATIONS
6.7	PROBLEMS BASED ON SIMULTANEOUS EQUATIONS

EXERCISE 6 (D): 10 questions
EXERCISE 6 (E): 15 questions
EXERCISE 6 (F): 10 questions

6.1 INTRODUCTION

An equation of the form $ax + by + c = 0$ is a linear equation. In it, $a, b, c$ are real constants and $x, y$ are variables of degree 1. When two such equations share the same two variables, they form a simultaneous pair.

The solution of the pair is the values of $x$ and $y$ that satisfy both equations together. To verify a proposed solution, substitute the pair into each equation. If both sides balance in both equations, the pair is the solution. This chapter continues the algebra you started in the rest of the Class 9 ICSE Maths book, and follows the syllabus published on the official CISCE website.

Consider checking $x = 2, y = 1$ against $2x + 3y = 7$ and $5x - y = 9$ . In the first equation, $2(2) + 3(1) = 4 + 3 = 7$ . The left side equals the right side. In the second equation, $5(2) - 1 = 9$ . Both hold, so $x = 2, y = 1$ is the solution of the pair.

6.2 METHODS OF SOLVING SIMULTANEOUS EQUATIONS

To solve two simultaneous equations means to find the values of both variables. This chapter teaches three algebraic methods for doing this.

Method of elimination by substitution.
Method of elimination by equating coefficients.
Method of cross-multiplication.

Each method suits a different shape of equation. The table below summarises when each one is the natural choice.

Method	Core idea	Easiest when
Substitution	Express one variable in terms of the other	A variable has coefficient 1
Equating coefficients	Equalise one coefficient, then add or subtract	Coefficients are close in size
Cross-multiplication	Apply the formula directly	Both equations are in $ax + by + c = 0$ form

6.3 METHOD OF ELIMINATION BY SUBSTITUTION

Section 6.3 gives the method of elimination by substitution in three steps. First, from either equation, find one variable in terms of the other. Second, substitute that expression into the other equation and solve it. Third, put the value you found back into the first expression to get the remaining variable.

Solve $2x + y = 11$ and $x - y = 1$ by substitution. Step 1: from $x - y = 1$ , get $x = y + 1$ . Step 2: substitute into $2x + y = 11$ , giving $2(y + 1) + y = 11$ .

This simplifies to $2y + 2 + y = 11$ , so $3y = 9$ and $y = 3$ .

Step 3: back-substitute to get $x = 3 + 1 = 4$ . Check: $2(4) + 3 = 11$ and $4 - 3 = 1$ . The solution is $x = 4, y = 3$ .

You can also express $x$ in terms of $y$ instead of the reverse; the result is the same. When the equations contain fractions, multiply each equation through by the LCM of its denominators before substituting, as the chapter’s second example shows. Forgetting to clear fractions is a common error that makes the later steps messy.

6.4 METHOD OF ELIMINATION BY EQUATING COEFFICIENTS

Section 6.4 presents the method of elimination by equating coefficients in four steps. First, multiply one or both equations so that the coefficients of $x$ or $y$ become numerically equal. Second, add or subtract the equations so that the equal terms cancel. Third, solve the resulting equation for one variable. Fourth, substitute that value into either original equation to find the other variable.

Solve $2x + 3y = 8$ and $3x + 2y = 7$ . Multiply the first equation by 3 and the second by 2: $6x + 9y = 24$ and $6x + 4y = 14$ . Subtract the second from the first: $5y = 10$ , so $y = 2$ .

Substitute $y = 2$ into $2x + 3y = 8$ to get $2x + 6 = 8$ , so $x = 1$ . The solution is $x = 1, y = 2$ .

The chapter also treats the special case where coefficients are swapped, such as $65x - 33y = 97$ and $33x - 65y = 1$ . Here add the equations to get one simple equation, then subtract them to get another, and solve that smaller pair.

6.5 METHOD OF CROSS-MULTIPLICATION

Section 6.5 gives the cross-multiplication formula. Write the equations as $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ . Then

$x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \quad \text{and} \quad y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$

The book’s arrow layout connects the coefficients. The denominator $a_1b_2 - a_2b_1$ comes from the $a$ and $b$ columns, and the numerators come from the $b,c$ and $c,a$ columns, following the same diagonal pattern. The arrows tell you which products to subtract.

Solve $3x + 4y = 10$ and $2x + 5y = 9$ by cross-multiplication. First bring the constants to the left: $3x + 4y - 10 = 0$ and $2x + 5y - 9 = 0$ . Then $a_1 = 3, b_1 = 4, c_1 = -10$ and $a_2 = 2, b_2 = 5, c_2 = -9$ .

$x = \frac{4(-9) - 5(-10)}{3(5) - 2(4)} = \frac{-36 + 50}{15 - 8} = \frac{14}{7} = 2$

$y = \frac{(-10)(2) - (-9)(3)}{15 - 8} = \frac{-20 + 27}{7} = 1$

Check: $3(2) + 4(1) = 10$ and $2(2) + 5(1) = 9$ . The solution is $x = 2, y = 1$ .

Sign errors are the main trap. You must bring both constant terms to the left side before reading off $c_1$ and $c_2$ . If you leave a constant on the right, every numerator is wrong. Also keep the subtraction order of the arrow products fixed.

6.6 EQUATIONS REDUCIBLE TO LINEAR EQUATIONS

Section 6.6 covers equations reducible to linear equations. When the variables appear only in denominators, substitute $\frac{1}{x} = a$ and $\frac{1}{y} = b$ . The equations become linear in $a$ and $b$ . Solve that pair, then invert to get $x = \frac{1}{a}$ and $y = \frac{1}{b}$ . The chapter notes that you do not need the LCM of the denominators here.

Solve $\frac{5}{x} + \frac{3}{y} = 7$ and $\frac{2}{x} - \frac{1}{y} = 5$ . Substitute to get $5a + 3b = 7$ and $2a - b = 5$ . From the second, $b = 2a - 5$ . Substitute into the first: $5a + 3(2a - 5) = 7$ , which gives $11a - 15 = 7$ , so $a = 2$ . Then $b = -1$ .

Invert: $x = \frac{1}{2}$ and $y = -1$ .

Check: $\frac{5}{1/2} + \frac{3}{-1} = 10 - 3 = 7$ . The same idea works when the denominators are expressions such as $x - y$ and $x + y$ . Set $\frac{1}{x - y} = a$ and $\frac{1}{x + y} = b$ , solve for $a$ and $b$ , then solve $x - y = \frac{1}{a}$ and $x + y = \frac{1}{b}$ .

You may also find the index laws from Chapter 7 useful when simplifying such equations.

6.7 PROBLEMS BASED ON SIMULTANEOUS EQUATIONS

Section 6.7 applies simultaneous equations to word problems. The setup follows three steps: assign $x$ and $y$ to the two unknowns, translate the statement into two equations, then solve the pair.

The chapter organises problems by type: numbers, fractions, two-digit numbers, ages, cost price and selling price, time and work, and miscellaneous exchanges. In time and work problems, one day’s work of a person is $\frac{1}{x}$ if that person alone finishes in $x$ days. These problems also draw on the expansions you met in Chapter 4.

For a two-digit number with tens digit $x$ and unit digit $y$ , the number is $10x + y$ and the reversed number is $10y + x$ . Use this to form the second equation.

Example: the digits of a two-digit number sum to 9, and reversing the digits gives a number 27 more. Let the tens digit be $x$ and the unit digit be $y$ . Then $x + y = 9$ .

The reversed number $10y + x$ is larger, so $10y + x - (10x + y) = 27$ , which simplifies to $9y - 9x = 27$ , or $y - x = 3$ .

Solving $x + y = 9$ and $y - x = 3$ gives $y = 6, x = 3$ . The number is $10(3) + 6 = 36$ .

Here is the rule the chapter stresses. If reversing the digits increases the number, the unit digit is larger, so the digit difference is $y - x$ . If reversing decreases the number, the tens digit is larger, so the difference is $x - y$ . Check which case the statement describes before you write the equation.

What to practise from Simultaneous Linear Equations Class 9 ICSE

Exercise 6 (D) — 10 questions on equations reducible to linear form.
Exercise 6 (E) — 15 questions on numbers, fractions, and two-digit numbers.
Exercise 6 (F) — 10 questions on ages, cost price and selling price, time and work, and miscellaneous problems.

Simultaneous Linear Equations Class 9 ICSE frequently asked questions

How do I solve simultaneous linear equations in class 9 ICSE?

Use one of three methods: elimination by substitution, elimination by equating coefficients, or cross-multiplication. Choose the method that fits the coefficients of the pair, then verify your answer in both equations.

Which method should I use to solve simultaneous equations?

Use substitution when a variable has coefficient 1. Use equating coefficients when the coefficients are already close in size. Use cross-multiplication when both equations are easy to write as $ax + by + c = 0$ .

How do I solve equations where the variables are in the denominator?

Set $\frac{1}{x} = a$ and $\frac{1}{y} = b$ , solve the linear system for $a$ and $b$ , then invert to get $x$ and $y$ . You do not need the LCM of the denominators.

How do I set up equations from a two digit number word problem?

Let the tens digit be $x$ and the unit digit be $y$ . Then the number is $10x + y$ and its reverse is $10y + x$ . Form one equation from the digit sum and another from the relationship between the number and its reverse.

Where can I find the chapter PDF for simultaneous linear equations class 9 ICSE?

The chapter PDF is provided on this page. It contains the full text of Chapter 6, including exercises 6 (D), 6 (E) and 6 (F).

Reference: ICSE Class 9 Mathematics textbook material, chapter 6: Simultaneous Linear Equations Including Problems — CISCE-aligned.

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