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exercise 4.2 class 9- Linear Equations in Two Variables

Solutions: Exercise 4.2

1. Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions

The correct option is (iii) infinitely many solutions.

Reason: For every real value of x that we choose, we can calculate a corresponding real value of y. Since there are infinite possible values for x, there will be infinite corresponding values for y. Each pair (x, y) forms a unique solution.

For example:
If x = 0, then y = 3(0) + 5 = 5. So, (0, 5) is a solution.
If x = 1, then y = 3(1) + 5 = 8. So, (1, 8) is a solution.
If x = -1, then y = 3(-1) + 5 = 2. So, (-1, 2) is a solution.
This process can continue indefinitely.

2. Write four solutions for each of the following equations:

(i) 2x + y = 7

To find solutions, we can substitute values for x and find the corresponding y.

  • If x = 0: 2(0) + y = 7 => y = 7. Solution: (0, 7)
  • If x = 1: 2(1) + y = 7 => y = 5. Solution: (1, 5)
  • If x = 2: 2(2) + y = 7 => y = 3. Solution: (2, 3)
  • If x = 3: 2(3) + y = 7 => y = 1. Solution: (3, 1)
(ii) πx + y = 9
  • If x = 0: Ï€(0) + y = 9 => y = 9. Solution: (0, 9)
  • If x = 1: Ï€(1) + y = 9 => y = 9 – Ï€. Solution: (1, 9 – Ï€)
  • If x = 2: Ï€(2) + y = 9 => y = 9 – 2Ï€. Solution: (2, 9 – 2Ï€)
  • If y = 0: Ï€x + 0 = 9 => x = 9/Ï€. Solution: (9/Ï€, 0)
(iii) x = 4y
  • If y = 0: x = 4(0) => x = 0. Solution: (0, 0)
  • If y = 1: x = 4(1) => x = 4. Solution: (4, 1)
  • If y = -1: x = 4(-1) => x = -4. Solution: (-4, -1)
  • If y = 2: x = 4(2) => x = 8. Solution: (8, 2)

3. Check which of the following are solutions of the equation x - 2y = 4 and which are not:

We check each point by substituting the x and y values into the Left Hand Side (LHS) of the equation and seeing if it equals the Right Hand Side (RHS), which is 4.

(i) (0, 2): LHS = 0 – 2(2) = -4. Since -4 ≠ 4, it is not a solution.
(ii) (2, 0): LHS = 2 – 2(0) = 2. Since 2 ≠ 4, it is not a solution.
(iii) (4, 0): LHS = 4 – 2(0) = 4. Since 4 = 4, it is a solution.
(iv) (√2, 4√2): LHS = √2 – 2(4√2) = √2 – 8√2 = -7√2. Since -7√2 ≠ 4, it is not a solution.
(v) (1, 1): LHS = 1 – 2(1) = 1 – 2 = -1. Since -1 ≠ 4, it is not a solution.

4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Since (2, 1) is a solution, it must satisfy the equation. We substitute the values of x and y into the equation.

Given equation: 2x + 3y = k

Substitute x = 2 and y = 1:

2(2) + 3(1) = k

4 + 3 = k

7 = k

k = 7

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