CBSE Class 8 Maths – Chapter 2: Linear Equations in One Variable

EduBack • Class 8 Maths

Chapter 2 — Linear Equations in One Variable

Detailed notes, rules, solved examples, word problems, practice set (Q1–Q20) and separate answer key. MathJax is enabled for clean equations.

Contents

1. What is a Linear Equation?
2. Key Concepts & Rules
3. Solved Examples
4. Word Problems
5. Practice Questions (Q1–Q20)
6. Answer Key

1) What is a Linear Equation?

An equation is a statement that two expressions are equal. A linear equation in one variable has the variable raised to power 1 only.

\[ ax + b = 0 \quad (a \ne 0) \]

Here \(a\) and \(b\) are real numbers. The goal is to find the value of the variable (usually \(x\)) that makes the equality true.

Examples

\(2x + 5 = 0\), \(3y - 7 = 11\), \(\dfrac{x}{2} - 4 = 1\)

Non-examples

\(x^2 + 3 = 0\) (not linear), \(xy + 1 = 0\) (two variables)

2) Key Concepts & Rules

Balancing method

Perform the same operation on both sides to keep the equation balanced. You may add, subtract, multiply or divide (division only by non-zero values).

Transposition

Moving a term across the equals sign changes its sign. This is a shorthand for adding/subtracting on both sides.

\[ 5x - 3 = 2x + 9 \Longrightarrow 5x - 2x = 9 + 3 \Rightarrow 3x = 12 \]

Clearing fractions

Multiply both sides by the LCM of denominators first to eliminate fractions.

\[ \frac{x}{2} + \frac{1}{3} = \frac{7}{6} \Rightarrow {\times 6} 3x + 2 = 7 \]

Special cases

• Identity: an equation true for all \(x\) (infinitely many solutions). Example: \(2(x+1)=2x+2\).
• Contradiction: no solution. Example: \(2x+1=2x+5\) → \(1=5\) (false).

Check solutions: substitute the value back into the original equation to verify correctness.

Solved Examples (step-by-step)

Example 1 — Simple linear equation

Solve: \(2x + 7 = 15\)

1. Subtract 7 from both sides: \(2x = 8\).
2. Divide both sides by 2: \(x = 4\).

Answer: \(x = 4\)

Example 2 — With brackets

Solve: \(3(x + 2) = 18\)

1. Expand: \(3x + 6 = 18\).
2. Subtract 6: \(3x = 12\).
3. Divide by 3: \(x = 4\).

Answer: \(x = 4\)

Example 3 — Variables on both sides

Solve: \(5x - 3 = 2x + 9\)

1. Bring variables left: \(5x - 2x = 9 + 3\).
2. Simplify: \(3x = 12\).
3. Divide: \(x = 4\).

Answer: \(x = 4\)

Example 4 — Clearing fractions

Solve: \(\dfrac{x}{2} + \dfrac{1}{3} = \dfrac{7}{6}\)

1. LCM of denominators = 6. Multiply through by 6: \(3x + 2 = 7\).
2. Subtract 2: \(3x = 5\).
3. Divide by 3: \(x = \dfrac{5}{3}\).

Answer: \(x = \dfrac{5}{3}\)

Example 5 — Mixed linear expression

Solve: \(\dfrac{2x-3}{5} - \dfrac{x+1}{10} = \dfrac{7}{10}\)

1. LCM = 10. Multiply both sides by 10: \(2(2x-3) - (x+1) = 7\).
2. Expand: \(4x - 6 - x - 1 = 7\).
3. Combine like terms: \(3x - 7 = 7\).
4. Add 7: \(3x = 14\).
5. Divide: \(x = \dfrac{14}{3}\).

Answer: \(x = \dfrac{14}{3}\)

\( \text{(Kept on one line — horizontal scroll allowed if necessary)} \)

Example 6 — Identity and no-solution

If after simplification both sides are equal (e.g. \(2x+2=2x+2\)), every value of \(x\) satisfies it — infinitely many solutions.

If simplification leads to a false statement (e.g. \(1=5\)), there is no solution.

Word Problems — translate to equations and solve

WP1 — Two numbers

The sum of two numbers is 95. One number is 15 more than the other. Find the numbers.

1. Let smaller = \(x\), larger = \(x+15\).
2. Equation: \(x+(x+15)=95 \Rightarrow 2x+15=95\).
3. Solve: \(2x=80 \Rightarrow x=40\). Numbers: 40 and 55.

WP2 — Ages

Rina is 6 years older than her brother. In 5 years, sum of ages will be 41. Find present ages.

1. Brother = \(x\), Rina = \(x+6\).
2. In 5 years: \((x+5)+(x+6+5)=41 \Rightarrow 2x+16=41\).
3. \(2x=25 \Rightarrow x=12.5\). Brother = 12.5, Rina = 18.5 years.

WP3 — Cost (money)

3 notebooks and 2 pens cost ₹86. One notebook costs ₹20 more than a pen. Find prices.

1. Let pen = \(p\), notebook = \(p+20\).
2. Equation: \(3(p+20)+2p=86 \Rightarrow 5p+60=86\).
3. \(5p=26 \Rightarrow p=5.2\). Pen ₹5.20, notebook ₹25.20.

WP4 — Rectangle perimeter

Length = breadth + 4. Perimeter = 40 cm. Find length & breadth.

1. Breadth = \(b\). Length = \(b+4\).
2. Perimeter: \(2(l+b)=40 \Rightarrow 2(b+4 + b)=40 \Rightarrow 4b+8=40\).
3. \(4b=32 \Rightarrow b=8\). Length = 12 cm.

WP5 — Speed & time

A car travels 150 km. Increasing speed by 10 km/h reduces time by 0.5 hour. Find original speed.

1. Let original speed = \(v\). Time = \(150/v\).
2. New time = \(150/(v+10)\). Given: \(150/v - 150/(v+10) = 0.5\).
3. Solve: \(150\left(\dfrac{(v+10)-v}{v(v+10)}\right) = 0.5 \Rightarrow \dfrac{1500}{v(v+10)}=0.5\).
4. \(v(v+10)=3000 \Rightarrow v^2+10v-3000=0\). Positive root: \(v=50\) km/h.

Practice Questions (Q1–Q20)

Solve all questions; answers appear in the Answer Key section (separate).

1. Q1. Solve: \(7x + 5 = 19\).
2. Q2. Solve: \(3(x - 4) = 15\).
3. Q3. Solve: \(2x + 7 = x + 13\).
4. Q4. Solve: \(\dfrac{x}{3} + \dfrac{1}{2} = \dfrac{11}{6}\).
5. Q5. Solve: \(\dfrac{2x-3}{5} - \dfrac{x+1}{10} = \dfrac{7}{10}\).
6. Q6. Solve: \(5(2y-3) - 3(y+1) = 7\).
7. Q7. Solve: \(\dfrac{3z}{4} + \dfrac{z}{8} = 7\).
8. Q8. Solve: \(\dfrac{x-2}{3} + \dfrac{x+4}{6} = 5\).
9. Q9. If \(4a - 7 = 3a + 2\), find \(a\).
10. Q10. If \(\dfrac{p-1}{2} - \dfrac{p+5}{3} = 1\), find \(p\).
11. Q11. The sum of two consecutive integers is 43. Find them.
12. Q12. A number is 8 less than twice another. Their sum is 40. Find the numbers.
13. Q13. Perimeter of a triangle is 48 cm. Two sides are 15 cm and 17 cm. Find the third side.
14. Q14. A pen costs ₹12 less than a notebook. 4 pens and 3 notebooks cost ₹306. Find the cost of each.
15. Q15. Three times a number decreased by 11 gives 25. Find the number.
16. Q16. Solve: \(2(x+3) - (x-5) = 20\).
17. Q17. Solve: \(\dfrac{5}{6}(m-2) + \dfrac{1}{3}(m+1) = 4\).
18. Q18. Solve: \(\dfrac{t+4}{5} - \dfrac{t-1}{2} = -1\).
19. Q19. If \(\dfrac{x-1}{4} = \dfrac{2x+5}{7}\), find \(x\).
20. Q20. Determine whether \(2(k+1) = 2k + 3\) has a solution. If yes, find it; else explain why not.

Answer Key (separate)

1. A1. \(x = 2\). (7x+5=19 ⇒ 7x=14 ⇒ x=2)
2. A2. \(x = 9\). (3(x−4)=15 ⇒ x−4=5 ⇒ x=9)
3. A3. \(x = 6\). (2x+7 = x+13 ⇒ x=6)
4. A4. Multiply by 6: \(2x+3=11 ⇒ 2x=8 ⇒ x=4\).
5. A5. Multiply by 10: \(2(2x−3)−(x+1)=7 ⇒ 4x−6−x−1=7 ⇒ 3x−7=7 ⇒ 3x=14 ⇒ x=14/3\).
6. A6. Expand: \(10y−15−3y−3=7 ⇒ 7y−18=7 ⇒ 7y=25 ⇒ y=25/7\).
7. A7. LHS = (6z+z)/8 = 7z/8 = 7 ⇒ z = 8.
8. A8. Multiply by 6: \(2(x−2)+(x+4)=30 ⇒ 3x=30 ⇒ x=10\).
9. A9. \(4a−3a = 2+7 ⇒ a=9\).
10. A10. Multiply by 6: \(3(p−1) −2(p+5) = 6 ⇒ 3p−3 −2p−10 = 6 ⇒ p−13 = 6 ⇒ p=19\).
11. A11. Let n and n+1: n+(n+1)=43 ⇒ 2n+1=43 ⇒ n=21. Numbers: 21, 22.
12. A12. Let one = x, the other = 2x−8. Sum: x + (2x−8)=40 ⇒ 3x−8=40 ⇒ x=16. Numbers: 16 and 24.
13. A13. Third side = 48 − (15+17) = 16 cm.
14. A14. Let notebook = n, pen = n−12. 4(n−12)+3n = 306 ⇒ 7n −48 =306 ⇒ 7n = 354 ⇒ n = 354/7 ≈ 50.5714. Pen ≈ ₹38.5714, Notebook ≈ ₹50.5714.
15. A15. 3x −11 = 25 ⇒ 3x = 36 ⇒ x = 12.
16. A16. 2(x+3) − (x−5) = 20 ⇒ 2x+6 −x +5 =20 ⇒ x+11=20 ⇒ x=9.
17. A17. Multiply by 6: 5(m−2) + 2(m+1) = 24 ⇒ 5m−10 +2m+2 =24 ⇒ 7m −8 =24 ⇒ 7m=32 ⇒ m=32/7.
18. A18. Multiply by 10: 2(t+4) −5(t−1) = −10 ⇒ 2t+8 −5t +5 = −10 ⇒ −3t +13 = −10 ⇒ −3t = −23 ⇒ t = 23/3.
19. A19. Cross-multiply: 7(x−1) = 4(2x+5) ⇒ 7x−7 = 8x+20 ⇒ −x = 27 ⇒ x = −27.
20. A20. 2(k+1) = 2k + 3 ⇒ 2k + 2 = 2k + 3 ⇒ 2 = 3 (false). No solution.

Tip: For fraction problems always clear denominators by multiplying with the LCM first — it avoids mistakes and keeps the work tidy.

↑ Back to top

Published by EduBack. For corrections or suggestions email: report@eduback.in