Linear Equations in One Variable Class 8: Chapter PDF

The official NCERT PDF for Linear Equations in One Variable class 8 is right below. This is Chapter 2 of the NCERT Class 8 Mathematics textbook — six printed pages in the official file — and the download is the exact edition NCERT publishes on ncert.nic.in. Take the link for the chapter itself; keep reading for what the chapter teaches.

Download the NCERT Class 8 Linear Equations in One Variable PDF

The official version, exactly as NCERT released it, is the file your school’s copy is printed from. Open it to read the chapter in full, follow the worked examples line by line, or work the exercises on paper.

NCERT Class 8 Maths Chapter 2 Linear Equations in One Variable (official PDF) — this is the exact file NCERT publishes on ncert.nic.in, so the text matches the copy your school uses.

What is inside the Linear Equations in One Variable chapter

The chapter moves in one direction. Section 2.1 refreshes what expressions and equations are, and sets up the balancing idea used to solve them. Section 2.2 solves equations with variable terms on both sides, and Exercise 2.1 practises that skill.

Section 2.3 then shows how to simplify messy equations — fractions over several denominators, and brackets — before solving, with Exercise 2.2 for practice. The chapter closes with “What Have We Discussed?”, its own six-point summary. The table below maps the chapter part by part.

What this chapter covers: from a number on one side to variables on both sides

In earlier classes, the equations you solved usually had a number on one side — in \(2x – 3 = 7\) the right side is just 7. This chapter removes that assumption. An equation like \(2x – 3 = x + 2\) carries expressions with variables on both sides.

So the chapter’s argument runs: first, learn to move variable terms between sides (Section 2.2). Then learn the two simplification tools — multiplying both sides by the LCM of the denominators, and opening brackets (Section 2.3). The two exercises build the same skill twice. The comparison below shows what each set practises.

Notice what the chapter does not contain: no separate word-problem section. Everything here is the solving method itself. The closing summary simply lists the applications the method serves — problems on numbers, ages, perimeters and combination of currency notes — so the skill you build in these pages is the engine those word problems run on later.

Key concepts: expressions, equations, LHS and RHS

Solving this chapter’s equations is one job in two steps. First, name the parts correctly — expression or equation, LHS or RHS. Second, move terms until the variable stands alone. The three ideas below carry the whole chapter.

Expression or equation: the equality sign test

An expression is a combination of numbers, variables and operations with no equality sign: \(5x\), \(2x – 3\), \(3x + y\), \(x^2 + 1\) are all expressions. The moment you add an equality sign you have an equation, because two sides now claim to be equal. That is the chapter’s test: equations use the equality sign; expressions do not.

In \(2x – 3 = 7\), the LHS is \(2x – 3\) and the RHS is 7. The two sides have equal values only for certain values of \(x\), and those values are the solutions. For \(x = 5\), LHS \(= 2 \times 5 – 3 = 7\), which equals the RHS, so \(x = 5\) is a solution. For \(x = 10\), LHS \(= 2 \times 10 – 3 = 17\), which does not equal 7, so \(x = 10\) is not a solution.

What linear in one variable means

“Linear” names the highest power of the variable; “in one variable” names how many variables there are. A linear expression in one variable is one in which the highest power of the variable is 1: \(2x\), \(2x + 1\), \(3y – 7\) and \(12 – 5z\) all qualify.

That is why \(x^2 + 1\) is not linear even though it has only one variable: its highest power is 2, and the same failure hits \(y + y^2\) and \(1 + z + z^2 + z^3\). A second variable also disqualifies an expression — \(2xy + 5\) is not part of this chapter because there are two variables in it.

So a linear equation in one variable is an equation formed from linear expressions in exactly one variable: \(2x – 3 = x + 2\) is one.

The balance method: keeping LHS and RHS equal

The chapter assumes the two sides of an equation are balanced, like a weighing scale. Whatever operation you do to one side, you must do to the other — otherwise the balance breaks and the equality stops being true.

Written as transposing, the same steps become \(7x – 4 = 2x + 11 \Rightarrow 7x – 2x = 11 + 4 \Rightarrow 5x = 15 \Rightarrow x = 3\). Notice the signs: \(2x\) crossed to the left and became \(-2x\); \(-4\) crossed to the right and became \(+4\). The chapter makes this exact point when it solves \(2x – 3 = x + 2\) — subtracting \(x\) from both sides amounts to transposing \(x\) to the other side.

Keep one memory aid: whatever you do to one side, do to the other, and a term that crosses the equals sign flips its sign.

The balance figure: why we do the same to both sides

Section 2.1 opens with a diagram of the idea behind every step of this method.

The diagram shows a balance with equal weight on both pans. If you remove the same weight from both pans, the balance stays level; remove weight from one pan only, and it tips. Solving an equation copies this exactly.

Take the chapter’s own equation \(2x – 3 = x + 2\). Subtracting \(x\) from both sides removes the same weight from both pans: the \(x\)-term vanishes from the right side and from the left side equally, and the equality survives. That is why the answer found this way is correct — nothing was done to one side that was not done to the other.

Definitions you need from this chapter

Homework rarely stalls on the algebra in this chapter — it stalls on the vocabulary. These are the terms Section 2.1 and the chapter summary use, in plain words.

Common mistakes when solving linear equations

The errors in this chapter are rarely about algebra facts. They are about breaking the balance, losing a sign, or skipping the check — and every one is caught by substituting the answer back in.

The mistake that costs most marks is clearing denominators unevenly. Watch the full method on a fresh example, then see the wrong version. Solve \(\frac{2x+3}{5} = \frac{x-1}{2} + 1\).

Now the wrong version: multiplying only the fraction terms by 10 and leaving the \(+1\) alone.

\[ 2(2x+3) = 5(x-1) + 1 \]

\[ 4x + 6 = 5x – 5 + 1 \quad \Rightarrow \quad 4x + 6 = 5x – 4 \quad \Rightarrow \quad x = 10 \]

The check rejects it: for \(x = 10\), LHS \(= \frac{23}{5}\) but RHS \(= \frac{11}{2}\), and the two are not equal. The chapter’s Example 16 multiplies every term by 6 — including the lone 1 — for exactly this reason.

Exam notes: solve cleanly and check your answer

Exam guidance here is deliberately modest: no guarantees, no marks promises. The chapter demands one thing — a solving method — so revise the method, not a list of answers.

• Write every step in order. One line per operation makes sign slips visible and keeps the working honest.
• Check by substitution. The chapter’s own worked examples always verify LHS = RHS, and Exercise 2.1 asks for the check on every question. Make it part of every answer.
• Test the vocabulary before solving. Is it an expression or an equation? What is the highest power of the variable? These two questions stop most careless errors.
• Know where the skill leads. The chapter summary lists the problems linear equations serve — numbers, ages, perimeters, combination of currency notes — so the word problems of later classes run on this same method.
• Stuck on a variable on both sides? Go back to Section 2.2. Stuck on denominators or brackets? Section 2.3. Both sections are short and each has two worked examples.
• Textbook contents and the examinable syllabus are not always identical — check the current official syllabus before deciding what to revise.

What to remember from this chapter, in six points

The chapter closes with its own six-point recap, in the section “What Have We Discussed?”. The same six points, in our words:

1. An equation is an equality involving variables. It states that the value of the expression on one side of the equality sign equals the value on the other side.
2. The equations of Classes VI–VIII are linear in one variable. One variable only, and its highest power is 1.
3. Both sides may carry variable expressions. The equations of earlier classes had just a number on one side; this chapter does not assume that.
4. Variables transpose just like numbers — moving a term across the equals sign reverses its sign.
5. Sometimes simplify first. Multiply both sides by a suitable expression (for instance the LCM of the denominators) or open brackets before solving. Some equations only become linear after this step.
6. The utility is in applications. Problems on numbers, ages, perimeters and combinations of currency notes are solved with linear equations.

Related chapters and resources

This listing is maintained for the 2026-27 academic session using the NCERT textbook information available to us. NCERT remains the authority for confirming the latest edition.

Move around the Class 8 Mathematics book, or jump to the notes that explain each chapter’s method.

• Class 8 Mathematics book page — the full list of chapters in this textbook.
• Rational Numbers (Chapter 1) — the chapter before this one.
• Understanding Quadrilaterals (Chapter 3) — the chapter that follows.
• Class 8 book hub — every NCERT Class 8 subject.
• CBSE notes for Class 8 Mathematics — revision pages for the same chapters.
• Rational Numbers notes — the previous chapter’s method in note form.
• A Story of Numbers notes — related Class 8 Mathematics material.
• Class 8 revision hub — notes for every subject.
• CBSE notes library — every class in note form.

Need the chapter again? Reopen the official NCERT Class 8 Maths Chapter 2 PDF — it is the same edition as the download above.

Sources and data verification

• This page describes Chapter 2 of the NCERT Class 8 Mathematics textbook, in the official edition published on ncert.nic.in. The download at the top of this page is that official file.
• This chapter page covers Chapter 2 only — not the whole book, and not other classes. Neighbouring chapters have their own pages.
• It is maintained for the current academic session.
• NCERT settles textbooks, editions and PDFs (ncert.nic.in); CBSE settles curriculum, syllabus and examinations. For the examinable scope of Class 8 Mathematics, consult the current official CBSE syllabus.

Reference: NCERT Class 8 Mathematics textbook, chapter 2, official edition on ncert.nic.in.

FAQs on Linear Equations in One Variable Class 8

Why can we subtract a variable like x from both sides of an equation?

Because a variable stands for a number. Subtracting the same number from both sides keeps the equality true, and since \(x\) is the same on both sides, it cancels on one side while the rest of the equation stays balanced. Section 2.2 makes the point that variables are also numbers, so they obey the same balance rule as constants.

Why do we multiply both sides by the LCM of the denominators?

To clear fractions in one step. The LCM is the smallest number that every denominator divides, so each fraction’s denominator cancels and the equation becomes one without fractions. The chapter’s Example 16 uses 6 because 6 is the LCM of 3 and 6.

What is the difference between an algebraic expression and an equation?

The equality sign. An expression — like \(5x\) or \(2x – 3\) — has no equals sign. An equation, like \(2x – 3 = 7\), uses the equality sign, so it has a left hand side and a right hand side whose values are claimed to be equal.

What does linear mean in a linear equation in one variable?

It means the highest power of the variable is 1. So \(x^2 + 1\) is not linear, even though it has only one variable, because the variable appears to the power 2. The chapter also requires exactly one variable — \(2xy + 5\) has two, so it stays out.

How do we check whether a value is the solution of an equation?

Substitute the value into both sides separately. If LHS = RHS, it is a solution; if not, it is not. For example, \(x = 5\) solves \(2x – 3 = 7\) because \(2 \times 5 – 3 = 7\), while \(x = 10\) fails because \(2 \times 10 – 3 = 17\), which is not 7.

Which kinds of problems can linear equations in one variable solve?

The chapter’s closing summary names the families: problems on numbers, ages, perimeters, and combinations of currency notes. This chapter teaches the solving method only; those applications build on it in later work.