Class 7 Mathematics Notes Chapter 4 (Simple Equations) – Mathematics Book

Detailed Notes with MCQs of Chapter 4: Simple Equations. This is a foundational chapter, and understanding it well is crucial not just for your Class 7 exams but also forms the basis for more complex algebraic concepts you'll encounter later, including in various government exams.

Chapter 4: Simple Equations - Detailed Notes for Exam Preparation

1. What is an Equation?

• An equation is a statement of equality involving one or more variables (unknowns) and constants (fixed values).
• The key feature of an equation is the equality sign (=). It indicates that the value of the expression on the left-hand side (LHS) is equal to the value of the expression on the right-hand side (RHS).
• Example: x + 5 = 12. Here, x is the variable, 5 and 12 are constants, and = is the equality sign. LHS is x + 5, RHS is 12.

2. What is a 'Simple' Equation (Linear Equation in One Variable)?

• In this chapter, 'Simple Equations' primarily refer to linear equations in one variable.
• Linear: The highest power of the variable in the equation is 1 (e.g., x, y, p, not x², y³).
• One Variable: The equation contains only one type of unknown (e.g., only x, or only y).
• Example: 2y - 3 = 7 (Linear, one variable 'y')
• Non-Example: x² + 1 = 5 (Not linear because of x²), x + y = 10 (Not one variable, has x and y).

3. Forming Equations from Statements:

• This is a critical skill for solving word problems. You need to translate the information given in words into a mathematical equation.
• Steps:
  • Identify the unknown quantity and represent it with a variable (like x, y, n, etc.).
  • Identify the mathematical operations (addition, subtraction, multiplication, division) described in the statement.
  • Translate the words into mathematical expressions involving the variable and constants.
  • Set up the equality based on the statement.
• Examples:
  • "The sum of a number and 6 is 15." -> Let the number be x. Equation: x + 6 = 15.
  • "5 times a number decreased by 2 equals 18." -> Let the number be y. Equation: 5y - 2 = 18.
  • "One-fourth of a number p is 7." -> Equation: (1/4)p = 7 or p/4 = 7.

4. Solving an Equation:

• Solving an equation means finding the value of the variable that makes the LHS equal to the RHS. This value is called the solution or root of the equation.

• Principle of Balance: An equation is like a balanced scale. To keep it balanced, whatever operation you perform on one side, you must perform the same operation on the other side.

• Methods:

  • Systematic Method (Using Inverse Operations / Balancing):

    • To undo addition, subtract the same number from both sides.
      • Example: x + 3 = 10 -> Subtract 3 from both sides: x + 3 - 3 = 10 - 3 -> x = 7.
    • To undo subtraction, add the same number to both sides.
      • Example: y - 5 = 2 -> Add 5 to both sides: y - 5 + 5 = 2 + 5 -> y = 7.
    • To undo multiplication, divide both sides by the same non-zero number.
      • Example: 4z = 20 -> Divide both sides by 4: 4z / 4 = 20 / 4 -> z = 5.
    • To undo division, multiply both sides by the same non-zero number.
      • Example: p / 3 = 6 -> Multiply both sides by 3: (p / 3) * 3 = 6 * 3 -> p = 18.

  • Transposing Method (Shortcut for Addition/Subtraction):

    • Moving a term (a number or a term with a variable) from one side of the equation to the other.
    • When a term is transposed, its sign changes: + becomes -, and - becomes +.
    • Example 1: x + 3 = 10 -> Transpose +3 to the RHS: x = 10 - 3 -> x = 7.
    • Example 2: 2y - 5 = 9 -> Transpose -5 to the RHS: 2y = 9 + 5 -> 2y = 14. Now, divide by 2 (as per systematic method): y = 14 / 2 -> y = 7.
    • Note: Transposition is essentially applying the addition/subtraction balancing steps more quickly. You still need to use division/multiplication balancing for terms multiplying/dividing the variable.

5. Checking the Solution:

• After finding a solution, substitute the value back into the original equation.
• If LHS = RHS, the solution is correct.
• Example: For 2y - 5 = 9, we found y = 7.
  • Check: LHS = 2(7) - 5 = 14 - 5 = 9. RHS = 9.
  • Since LHS = RHS (9 = 9), the solution y = 7 is correct.

6. Applications (Solving Word Problems):

• This combines forming equations and solving them.
• General Steps:
  1. Read the problem carefully to understand what is given and what needs to be found.
  2. Define the unknown quantity with a variable.
  3. Translate the problem statement into a simple equation.
  4. Solve the equation using the methods learned.
  5. Check if the solution makes sense in the context of the original problem. Write the answer clearly with units if applicable.
• Example: "Ravi's father is 49 years old. He is 4 years older than three times Ravi's age. What is Ravi's age?"
  1. Unknown: Ravi's age. Let it be x years.
  2. Translate:
     • Three times Ravi's age: 3x
     • 4 years older than three times Ravi's age: 3x + 4
     • This age is equal to the father's age (49).
  3. Equation: 3x + 4 = 49
  4. Solve:
     • Transpose +4: 3x = 49 - 4 -> 3x = 45
     • Divide by 3: x = 45 / 3 -> x = 15
  5. Check: Father's age = 3(15) + 4 = 45 + 4 = 49. This matches the given information.
  • Answer: Ravi's age is 15 years.

Key Takeaways for Exams:

• Understand the difference between an expression and an equation (the '=' sign).
• Be proficient in translating word problems into equations.
• Master solving equations using both balancing and transposition methods. Pay close attention to sign changes during transposition.
• Always check your solution by substituting it back into the original equation.
• Practice various types of word problems.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on Chapter 4 concepts:

1. Which of the following is an equation?
   (a) x + 5 > 9
   (b) 2y - 3
   (c) 7p + 1 = 15
   (d) 8 / 4 = 2 (This is an arithmetic fact, not typically what's meant by an algebraic equation with a variable)

2. The statement "A number x divided by 3 gives 5" can be written as:
   (a) 3x = 5
   (b) x / 3 = 5
   (c) x - 3 = 5
   (d) x + 3 = 5

3. What is the solution to the equation y - 7 = 2?
   (a) y = 5
   (b) y = 9
   (c) y = -5
   (d) y = 14

4. If 5m = 30, what is the value of m?
   (a) m = 25
   (b) m = 35
   (c) m = 150
   (d) m = 6

5. Solving 2x + 5 = 11 involves which steps?
   (a) Add 5 to both sides, then divide by 2.
   (b) Subtract 5 from both sides, then divide by 2.
   (c) Divide both sides by 2, then subtract 5.
   (d) Multiply both sides by 2, then add 5.

6. What value of p satisfies the equation p/4 + 3 = 5?
   (a) p = 2
   (b) p = 8
   (c) p = 32
   (d) p = 0.5

7. If you subtract 6 from 4 times a number, you get 10. What is the number?
   (a) 2
   (b) 4
   (c) 16
   (d) 64

8. Transposing -4 from LHS to RHS in the equation 3k - 4 = 8 results in:
   (a) 3k = 8 - 4
   (b) 3k = 8 / 4
   (c) 3k = 8 + 4
   (d) -3k = 8 + 4

9. The value x = 2 is a solution for which equation?
   (a) x + 5 = 3
   (b) 3x - 1 = 7
   (c) x / 2 = 2
   (d) 4x + 1 = 9

10. The length of a rectangle is 3 cm more than its width w. If the perimeter is 26 cm, which equation represents this situation? (Perimeter = 2 * (length + width))
    (a) 2(w + 3 + w) = 26
    (b) w + (w+3) = 26
    (c) 2w + 3 = 26
    (d) w(w+3) = 26

Answer Key for MCQs:

1. (c)
2. (b)
3. (b) [y = 2 + 7 = 9]
4. (d) [m = 30 / 5 = 6]
5. (b)
6. (b) [p/4 = 5 - 3 = 2 => p = 2 * 4 = 8]
7. (b) [Let number be n. 4n - 6 = 10 => 4n = 16 => n = 4]
8. (c)
9. (d) [Check: 4(2) + 1 = 8 + 1 = 9. Matches RHS]
10. (a) [Length = w+3. Perimeter = 2 * ( (w+3) + w ) = 2(2w + 3) = 26]

Study these notes carefully, practice solving various problems from your NCERT book and other resources. Good luck!