Class 8 Notes

Class 8 Mathematics Notes Chapter 2 (Linear Equations in One Variable) – Mathematics Book

Philoid

Philoid

— 8 min read

Mathematics
Alright class, let's get straight into a crucial chapter for your exam preparation: Chapter 2 - Linear Equations in One Variable. Understanding this chapter well builds a strong foundation for more complex algebra later on. We'll cover the concepts thoroughly, focusing on how to solve problems efficiently, which is key for competitive exams.

Chapter 2: Linear Equations in One Variable - Detailed Notes

1. Introduction: Expressions vs. Equations

  • Algebraic Expression: A combination of constants, variables, and mathematical operations (+, -, ×, ÷).
    • Examples: 5x, 2x - 3, y/4 + 1
    • Expressions do not have an equality sign (=). They represent a value that depends on the variable(s).
  • Equation: A statement that asserts the equality of two expressions. It always contains an equality sign (=).
    • Examples: 5x = 25, 2x - 3 = 9, y/4 + 1 = 5
    • An equation has two sides: the Left Hand Side (LHS) and the Right Hand Side (RHS).

2. What is a Linear Equation in One Variable?

  • Linear: The highest power of the variable in the equation is 1.

    • Example: 3x + 5 = 11 (Here, the power of x is 1).
    • Not Linear: x² + 1 = 5 (Power of x is 2), y³ - y = 2 (Highest power of y is 3).

  • In One Variable: The equation contains only one type of variable (like x, y, or z, but not a mix like x and y in the same equation).

    • Example: 2t - 10 = 6 (Only variable is t).
    • Not in One Variable: 2x + 3y = 7 (Contains two variables, x and y).

  • Definition: A linear equation in one variable is an equation that can be written in the form ax + b = 0, where a and b are real numbers, and a ≠ 0. The variable is x. (The form might look different initially, like cx + d = ex + f, but it can be rearranged into ax + b = 0).

3. Solving Linear Equations

The "solution" or "root" of an equation is the value of the variable that makes the equation true (i.e., makes LHS equal to RHS).

Core Principle: To solve an equation, we need to isolate the variable on one side (usually the LHS). We do this by performing the same mathematical operation on both sides of the equation to maintain the balance.

Methods:

(a) Balancing Method (Fundamental Principle):

  • Addition/Subtraction: You can add or subtract the same number from both sides of the equation.

    • Example: Solve x - 5 = 7
      • Add 5 to both sides: x - 5 + 5 = 7 + 5
      • x = 12
    • Example: Solve y + 3 = 10
      • Subtract 3 from both sides: y + 3 - 3 = 10 - 3
      • y = 7

  • Multiplication/Division: You can multiply or divide both sides of the equation by the same non-zero number.

    • Example: Solve 5t = 30
      • Divide both sides by 5: (5t)/5 = 30/5
      • t = 6
    • Example: Solve m/4 = 3
      • Multiply both sides by 4: (m/4) × 4 = 3 × 4
      • m = 12

(b) Transposition Method (Shortcut):

Transposition is essentially a shortcut for the balancing method. When a term moves from one side of the equation to the other, its sign changes.

  • + becomes -

  • - becomes +

  • × becomes ÷ (when moving a factor)

  • ÷ becomes × (when moving a divisor)

  • Example: Solve 2x - 3 = 9

    • Transpose -3 from LHS to RHS (becomes +3): 2x = 9 + 3
    • 2x = 12
    • Transpose 2 (which is multiplying x) from LHS to RHS (becomes division): x = 12 / 2
    • x = 6

  • Example: Solve z/3 + 1 = 7/15

    • Transpose +1 to RHS: z/3 = 7/15 - 1
    • Find common denominator on RHS: z/3 = 7/15 - 15/15
    • z/3 = -8/15
    • Transpose 3 (which is dividing z) to RHS (becomes multiplication): z = (-8/15) × 3
    • z = -24/15
    • Simplify the fraction: z = -8/5

4. Solving Equations with Variables on Both Sides

Strategy:

  1. Collect all terms containing the variable on one side (usually LHS).
  2. Collect all constant terms on the other side (usually RHS).
  3. Simplify and solve for the variable.

  • Example: Solve 2x - 3 = x + 2

    • Transpose x from RHS to LHS (becomes -x): 2x - x - 3 = 2
    • Transpose -3 from LHS to RHS (becomes +3): 2x - x = 2 + 3
    • Simplify both sides: x = 5

  • Example: Solve 5t - 3 = 3t - 5

    • Transpose 3t to LHS: 5t - 3t - 3 = -5
    • Transpose -3 to RHS: 5t - 3t = -5 + 3
    • Simplify: 2t = -2
    • Divide by 2 (or transpose 2): t = -2 / 2
    • t = -1

5. Solving Equations with Brackets

Strategy:

  1. Use the distributive property to remove the brackets first. a(b + c) = ab + ac and a(b - c) = ab - ac.
  2. Simplify the equation.
  3. Solve using the methods described above (transposition, balancing).
  • Example: Solve 3(t - 3) = 5(2t + 1)
    • Expand brackets: 3t - 9 = 10t + 5
    • Transpose 10t to LHS: 3t - 10t - 9 = 5
    • Transpose -9 to RHS: 3t - 10t = 5 + 9
    • Simplify: -7t = 14
    • Divide by -7: t = 14 / (-7)
    • t = -2

6. Solving Equations with Fractions (Rational Coefficients)

Strategy:

  1. Find the Least Common Multiple (LCM) of all the denominators in the equation.
  2. Multiply every term on both sides of the equation by the LCM. This eliminates the fractions.
  3. Solve the resulting simpler linear equation.
  • Example: Solve x/2 - 1/5 = x/3 + 1/4
    • Denominators are 2, 5, 3, 4. LCM(2, 5, 3, 4) = 60.
    • Multiply every term by 60:
      60(x/2) - 60(1/5) = 60(x/3) + 60(1/4)
    • Simplify: 30x - 12 = 20x + 15
    • Now it's a standard linear equation. Transpose 20x to LHS, -12 to RHS:
      30x - 20x = 15 + 12
    • Simplify: 10x = 27
    • Solve for x: x = 27/10

7. Forming Equations from Word Problems (Applications)

This is a very important skill for government exams.

Steps:

  1. Read Carefully: Understand the problem statement completely. Identify what is given and what needs to be found.
  2. Assign Variable: Represent the unknown quantity with a variable (e.g., x, y). If there are multiple related unknowns, try to express them in terms of a single variable (e.g., if one number is 5 more than another, let the smaller be x and the larger be x + 5).
  3. Translate: Convert the statements/conditions given in the problem into mathematical expressions and form an equation relating them. Look for keywords like "is", "equals", "sum", "difference", "product", "times", "more than", "less than", etc.
  4. Solve: Solve the linear equation formed using the techniques learned.
  5. Check: Substitute the solution back into the context of the original word problem (not just the equation you formed) to see if it makes sense and satisfies all conditions. Write the answer with appropriate units if applicable.
  • Example: The sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
    • Let the smaller number be x.
    • The other number exceeds it by 15, so the larger number is x + 15.
    • Their sum is 95: x + (x + 15) = 95 (This is the equation)
    • Solve:
      2x + 15 = 95
      2x = 95 - 15
      2x = 80
      x = 80 / 2
      x = 40
    • The smaller number is x = 40.
    • The larger number is x + 15 = 40 + 15 = 55.
    • Check: Sum = 40 + 55 = 95. Difference = 55 - 40 = 15. Both conditions are met.
    • Answer: The numbers are 40 and 55.

8. Checking Your Solution

Always verify your answer by substituting the value of the variable back into the original equation. If LHS = RHS, your solution is correct. This helps catch calculation errors.

  • Example: We found x = 6 for 2x - 3 = 9.
    • Check: LHS = 2(6) - 3 = 12 - 3 = 9. RHS = 9.
    • Since LHS = RHS, the solution x = 6 is correct.

Key Takeaways for Exams:

  • Speed and Accuracy are crucial. Practice transposition well.
  • Be comfortable with fractions and decimals. The LCM method is very useful for equations with fractions.
  • Word problems are common. Practice translating sentences into equations. Pay attention to keywords.
  • Always check your answer if time permits, especially in word problems.

Multiple Choice Questions (MCQs)

Here are 10 MCQs to test your understanding:

  1. Which of the following is a linear equation in one variable?
    (a) 2x + 5y = 10
    (b) x² - 4 = 0
    (c) 3z - 7 = z + 1
    (d) x/y = 2

  2. The solution of the equation 3x - 4 = 8 is:
    (a) x = 4/3
    (b) x = 12
    (c) x = 4
    (d) x = -4/3

  3. If 5y + 7/2 = 3/2 y - 14, then y equals:
    (a) 5
    (b) -5
    (c) 7/2
    (d) -35/7

  4. Solve for p: 4(p - 2) = 16
    (a) p = 4
    (b) p = 6
    (c) p = 2
    (d) p = 8

  5. If x/3 + 1 = 7/15, the value of x is:
    (a) -8/5
    (b) 8/5
    (c) 5/8
    (d) -5/8

  6. What number should be added to -7/3 to get 3/7?
    (a) 58/21
    (b) -58/21
    (c) 40/21
    (d) -40/21

  7. The sum of three consecutive integers is 51. The middle integer is:
    (a) 16
    (b) 17
    (c) 18
    (d) 15

  8. Solve: (2x + 1) / (3x - 2) = 9/10
    (a) x = 14
    (b) x = 28
    (c) x = 7
    (d) x = -28

  9. The perimeter of a rectangle is 40 cm. If its width is 8 cm, find its length.
    (a) 10 cm
    (b) 12 cm
    (c) 16 cm
    (d) 24 cm

  10. If z/(z+15) = 4/9, find z.
    (a) 12
    (b) 60
    (c) -12
    (d) -60

Answer Key for MCQs:

  1. (c) - Only one variable (z) and its highest power is 1.

  2. (c) - 3x = 8 + 4 => 3x = 12 => x = 12/3 = 4.

  3. (b) - 5y - 3/2 y = -14 - 7/2 => (10y - 3y)/2 = (-28 - 7)/2 => 7y/2 = -35/2 => 7y = -35 => y = -5.

  4. (b) - 4p - 8 = 16 => 4p = 16 + 8 => 4p = 24 => p = 24/4 = 6.

  5. (a) - x/3 = 7/15 - 1 => x/3 = 7/15 - 15/15 => x/3 = -8/15 => x = 3 * (-8/15) => x = -24/15 = -8/5.

  6. (a) - Let the number be x. x + (-7/3) = 3/7 => x = 3/7 + 7/3 => x = (9 + 49)/21 = 58/21.

  7. (b) - Let the integers be x, x+1, x+2. x + (x+1) + (x+2) = 51 => 3x + 3 = 51 => 3x = 48 => x = 16. The integers are 16, 17, 18. Middle is 17.

  8. (b) - Cross-multiply: 10(2x + 1) = 9(3x - 2) => 20x + 10 = 27x - 18 => 10 + 18 = 27x - 20x => 28 = 7x => x = 28/7 = 4. (Correction: Let me recheck Q8 calculation. 10(2x + 1) = 9(3x - 2) => 20x + 10 = 27x - 18 => 10 + 18 = 27x - 20x => 28 = 7x => x = 4. Let me re-evaluate the options or the question itself. Ah, perhaps I made a mistake in the initial calculation or options. Let's assume the question intended to yield one of the options. Let's re-solve carefully. 20x + 10 = 27x - 18. 18 + 10 = 27x - 20x. 28 = 7x. x = 4. None of the options match x=4. Let me assume a typo in the question, maybe (2x+1)/(3x-2) = 5/7? Let's try that: 7(2x+1) = 5(3x-2) => 14x+7 = 15x-10 => 7+10 = 15x-14x => 17 = x. Still not matching. Let's assume the question is correct and the options are wrong. The correct answer for (2x + 1) / (3x - 2) = 9/10 is x=4. Let's pick the closest process/answer type if this were an exam with flawed options, or assume a typo in my question creation. Let's assume the question was meant to be simpler, e.g., 2x+1 = 9 and 3x-2 = 10. First gives 2x=8, x=4. Second gives 3x=12, x=4. So x=4 is consistent. Since 4 is not an option, I will state the calculated answer and note the discrepancy. Let's assume option (b) 28 was intended and check if it works: If x=28, LHS = (228+1)/(328-2) = (56+1)/(84-2) = 57/82. This is not 9/10. There seems to be an issue with Q8 or its options. For the purpose of this exercise, I'll stick with my calculated x=4 and note the options provided are incorrect based on the question as written. Self-correction: It's better to provide a question that works with the options. Let's change Q8 slightly: Solve (x + 1) / (2x + 3) = 3/8. Cross-multiply: 8(x+1) = 3(2x+3) => 8x + 8 = 6x + 9 => 8x - 6x = 9 - 8 => 2x = 1 => x = 1/2. This also doesn't fit typical MCQ options well. Let's try another: Solve x/2 + x/3 = 10. LCM is 6. 6(x/2) + 6(x/3) = 6(10) => 3x + 2x = 60 => 5x = 60 => x = 12. This is better. Let's replace Q8.
    Revised Q8: Solve: x/2 + x/3 = 10
    (a) x = 6
    (b) x = 12
    (c) x = 15
    (d) x = 5
    Revised Answer for Q8: (b)

  9. (b) - Perimeter P = 2(Length + Width). 40 = 2(L + 8). Divide by 2: 20 = L + 8. L = 20 - 8 = 12. Length is 12 cm.

  10. (a) - Cross-multiply: 9z = 4(z + 15) => 9z = 4z + 60 => 9z - 4z = 60 => 5z = 60 => z = 60/5 = 12.

(Final Answer Key with Revised Q8)

  1. (c), 2. (c), 3. (b), 4. (b), 5. (a), 6. (a), 7. (b), 8. (b), 9. (b), 10. (a)

Make sure you practice solving various types of problems from this chapter, especially word problems. Good luck with your preparation!

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