CBSE Class 8 Maths — Linear Equations in One Variable

Chapter Overview & Weightage

Linear Equations in One Variable is a foundational chapter in CBSE Class 8 Maths. It builds directly on the basic algebra of Class 7 and sets up the logical framework students will use through Class 10 and beyond.

This chapter typically carries 6–10 marks in the Class 8 annual exam. Expect one or two word problems (3 marks each) and one straightforward solving question (2–3 marks). Word problems on ages, fractions, and perimeters are high-frequency.

Exam Year	Marks (approx.)	Focus Area
2023-24	8	Word problem on ages + fractions
2022-23	7	Perimeter problem + equation solving
2021-22	6	Solving equations with fractions

Key Concepts You Must Know

Linear equation in one variable: An equation of the form $ax + b = 0$ where $a \neq 0$ and $x$ is the variable
Solution: The value of the variable that satisfies the equation
Transposing: Moving a term from one side to the other, changing its sign
Equations reducible to linear form: Rational equations like $\frac{x+1}{x-2} = \frac{3}{4}$ that can be cross-multiplied and simplified

The golden rule: Whatever you do to one side of the equation, do the same to the other side. The balance must be maintained.

Important Formulas

ax + b = c \implies x = \frac{c - b}{a}

Steps: Transpose constant terms to one side, variable terms to the other, then divide both sides by the coefficient of $x$ .

\frac{p}{q} = \frac{r}{s} \implies ps = qr

Use when both sides are single fractions. Cross-multiply first, then solve the resulting linear equation.

Solved Previous Year Questions

PYQ 1 — Solving a basic equation

Q: Solve $\frac{2x}{3} + 5 = \frac{x}{2} + 4\frac{1}{2}$

Solution:

\frac{2x}{3} + 5 = \frac{x}{2} + \frac{9}{2}

Multiply every term by 6 (LCM of 3 and 2):

4x + 30 = 3x + 27

4x - 3x = 27 - 30

x = -3

Check: LHS $= \frac{2(-3)}{3} + 5 = -2 + 5 = 3$ . RHS $= \frac{-3}{2} + 4.5 = -1.5 + 4.5 = 3$ . ✓

PYQ 2 — Age word problem

Q: The sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Solution: Let the smaller number be $x$ . Then the larger is $x + 15$ .

x + (x + 15) = 95 \implies 2x + 15 = 95 \implies 2x = 80 \implies x = 40

The numbers are 40 and 55.

PYQ 3 — Fraction problem

Q: The numerator of a fraction is 6 less than the denominator. If 3 is added to both numerator and denominator, the fraction becomes $\frac{2}{3}$ . Find the fraction.

Solution: Let denominator = $x$ , numerator = $x - 6$ .

\frac{(x-6)+3}{x+3} = \frac{2}{3} \implies \frac{x-3}{x+3} = \frac{2}{3}

Cross-multiply: $3(x-3) = 2(x+3) \implies 3x - 9 = 2x + 6 \implies x = 15$

Fraction = $\frac{15-6}{15} = \mathbf{\frac{9}{15} = \frac{3}{5}}$

Difficulty Distribution

Difficulty	Expected marks	Question type
Easy	2–3 marks	Direct equation solving
Medium	3 marks	Setting up equation from word problem
Hard	5–6 marks	Two-condition word problems

Expert Strategy

Setting up the equation is 70% of the work in word problems. Students who struggle do so because they can’t translate English to algebra — not because they can’t solve equations.

Read the problem twice. On the first read, identify the unknown and assign a variable. On the second read, identify the condition(s) and write the equation.

For age problems: If the current age is $x$ , then the age $n$ years ago is $x - n$ and after $n$ years is $x + n$ . Write out the table of ages before forming the equation.

For fraction problems: always define the variable as one part (usually the denominator), and express the other part in terms of it.

For perimeter/geometry problems: write expressions for each dimension in terms of $x$ , then use the perimeter/area formula to form the equation.

Common Traps

Trap 1: Forgetting to multiply all terms when clearing fractions. If you multiply by LCM, every single term on both sides must be multiplied — including whole number terms like 5.

Trap 2: Sign errors when transposing. Moving $+5$ from the left to the right gives $-5$ on the right. This is the most common arithmetic slip in this chapter.

Trap 3: Not verifying the answer. Always substitute back. If the original equation has fractions, substituting confirms you haven’t made an error or ended up with division by zero.

Trap 4: Setting up the wrong variable. In age problems, if you let $x$ = the person’s age 10 years ago, make sure all age conditions use $x + 10$ for the current age.