Linear Equation Examples for Beginners (CBSE)

The fastest way to get comfortable with linear equations is to work through plenty of examples. In this beginner guide we cover the main types you will meet in CBSE maths — simple one-variable equations, equations with fractions, two-variable systems and real-world word problems — each solved step by step so the method sticks.

Try each one yourself first, then check with the free linear equation calculator. For the underlying method, see our guide on how to solve linear equations.

One-variable examples

Example 1. Solve 2x + 3 = 11. Subtract 3: 2x = 8. Divide by 2: x = 4. Check: 2(4) + 3 = 11. Correct.

Example 2. Solve 5x − 7 = 3x + 5. Bring variables together: 5x − 3x = 5 + 7, so 2x = 12 and x = 6.

Example 3. Solve 4(x − 2) = 12. Expand: 4x − 8 = 12, so 4x = 20 and x = 5.

Equations with fractions

Example 4. Solve (x/3) + 2 = 5. Subtract 2: x/3 = 3. Multiply by 3: x = 9.

Example 5. Solve (x + 1)/2 = (x − 1)/3. Multiply both sides by 6 (the LCM): 3(x + 1) = 2(x − 1), so 3x + 3 = 2x − 2, giving x = −5. Clearing fractions first is the key move here.

Key takeaway: Whenever an equation contains fractions, multiply every term by the LCM of the denominators before you start — it turns a messy problem into a clean one.

Two-variable examples

Example 6 (elimination). Solve x + y = 12 and x − y = 2. Add the equations: 2x = 14, so x = 7 and y = 5.

Example 7 (substitution). Solve y = 2x and x + y = 9. Substitute: x + 2x = 9, so 3x = 9, x = 3 and y = 6.

Word-problem examples

Example 8 — Coins. A piggy bank has ₹2 and ₹5 coins, 20 coins in all, worth ₹73. Let the number of ₹2 coins be x and ₹5 coins be y: x + y = 20 and 2x + 5y = 73. Solving gives x = 9 and y = 11.

Example 9 — Ages. Ravi is 4 years older than his sister; the sum of their ages is 26. Let the sister be x: x + (x + 4) = 26, so 2x = 22 and x = 11. The sister is 11 and Ravi is 15.

Example 10 — Numbers. The sum of two numbers is 30 and their difference is 8. Let them be x and y: x + y = 30 and x − y = 8. Adding gives 2x = 38, so x = 19 and y = 11.

Benefits of learning through examples

Worked examples make the abstract concrete and build pattern recognition, so you quickly learn which method suits which problem. They train the crucial skill of translating a real situation into an equation, which is where most marks are won or lost in word problems. Seeing verification at the end of each example instils the habit of checking answers. And steadily progressing from one-variable sums to systems and word problems builds the confidence needed for CBSE exams and competitive tests alike.

Challenges and limitations

Examples show the method, but real exams often reword problems in unfamiliar ways, so you must practise varied questions, not just memorise these. Setting up the equation from a word problem remains the hardest step and takes deliberate practice. Sign errors and mishandled fractions are common pitfalls even when the approach is understood. And a two-variable system can be inconsistent or dependent, which a beginner might not expect. Use the examples as a springboard for wider practice rather than a complete substitute for it.

Common mistakes beginners make

• Not defining the variable clearly. State what x represents before writing the equation.
• Forgetting to clear fractions. Multiply by the LCM first to avoid messy arithmetic.
• Sign errors when transposing. Moving a term across the equals sign flips its sign.
• Choosing an inefficient method. Elimination is quicker when coefficients match; substitution when a variable is isolated.
• Skipping verification. Always substitute your answer back into the original equation.
• Misreading the word problem. Re-read to capture every condition before forming equations.

Best practices and expert recommendations

• Attempt before checking. Solve each example yourself, then verify with a calculator.
• Define variables explicitly. Write down what each letter stands for in word problems.
• Clear fractions early. Multiply through by the LCM at the start.
• Pick the right method. Match elimination or substitution to the structure of the system.
• Verify every answer. Substitute back to confirm correctness.
• Practise widely. Work varied problems beyond these examples for exam readiness.

Extra practice examples

Example 11 — Fractions on both sides. Solve (2x + 1)/3 = (x + 4)/2. Multiply both sides by 6, the LCM: 2(2x + 1) = 3(x + 4), so 4x + 2 = 3x + 12, giving x = 10. Clearing the denominators first keeps the arithmetic clean.

Example 12 — A three-step equation. Solve 7x − 3(x − 2) = 22. Expand the bracket: 7x − 3x + 6 = 22, so 4x + 6 = 22, then 4x = 16 and x = 4. Working left to right and simplifying at each stage avoids mistakes.

Example 13 — A speed and distance word problem. A train covers a distance in 5 hours at a steady speed; travelling 10 km/h faster it would take 4 hours. Let the speed be x km/h; distance = 5x = 4(x + 10). So 5x = 4x + 40, giving x = 40 km/h and a distance of 200 km. This shows how a real situation collapses neatly into a single linear equation.

Example 14 — A money-mixture problem. A person invests a total of ₹10,000 in two parts; one part earns simple interest at 5% and the other at 8%, giving ₹680 in a year. Let the first part be x: 0.05x + 0.08(10000 − x) = 680. Solving, 0.05x + 800 − 0.08x = 680, so −0.03x = −120 and x = 4,000. The two parts are ₹4,000 and ₹6,000. Practising varied word problems like these is the surest route to confidence in exams.

Even more practice for exams

Example 15 — Consecutive numbers. The sum of three consecutive integers is 72. Let the middle one be x; then (x − 1) + x + (x + 1) = 72, so 3x = 72 and x = 24. The integers are 23, 24 and 25. Consecutive-number problems reduce neatly to a single linear equation once you name the unknowns well.

Example 16 — A perimeter problem. The length of a rectangle is 5 cm more than its breadth, and the perimeter is 50 cm. Let the breadth be x; then 2(x + x + 5) = 50, so 4x + 10 = 50, giving x = 10. The breadth is 10 cm and the length 15 cm. Geometry word problems frequently hide a simple linear equation inside a familiar formula.

Example 17 — A ratio problem. Two numbers are in the ratio 3:5 and their sum is 64. Let them be 3x and 5x; then 3x + 5x = 64, so 8x = 64 and x = 8, making the numbers 24 and 40. Practising a wide spread of problem types — numbers, ages, geometry, ratios, money and motion — is the most reliable way to walk into a CBSE exam confident that whatever form the question takes, you can translate it into an equation and solve it cleanly.

Frequently asked questions

What is a simple linear equation example?
A classic example is 2x + 3 = 11. Subtracting 3 gives 2x = 8, and dividing by 2 gives x = 4. Substituting back confirms 2(4) + 3 = 11, so x = 4 is correct.

How do I solve a linear equation with fractions?
Multiply every term by the least common multiple of the denominators to clear the fractions, then solve the resulting equation normally. For (x/3) + 2 = 5, multiplying appropriately gives x = 9.

How do I set up a word problem as an equation?
First define a variable for the unknown, then translate each condition in the problem into a mathematical statement. Combine these into one or more equations and solve, checking that the answer fits every condition.

Which method is best for two-variable examples?
Use elimination when the coefficients of one variable match or are easy to match, and substitution when one variable is already isolated or simple to isolate. Both give the same correct answer.

Do these examples match the CBSE syllabus?
Yes. They cover one-variable equations (Class 8), two-variable equations (Class 9) and pairs of linear equations (Class 10), along with the word problems commonly asked in CBSE and state-board exams.