CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 4 LINEAR EQUATIONS IN TWO VARIABLES EX – 4.2

Linear Equations in Two Variables

Page No: 70

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

Answer

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

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

  • (i) 2x + y = 7            
  • (ii) πx + y = 9               
  • (iii) x = 4y

Answer

(i) 2x + y = 7
⇒ y = 7 – 2x
→ Put x = 0,
y = 7 – 2 × 0 ⇒ y = 7
(0, 7) is the solution.
→ Now, put x = 1
y = 7 – 2 × 1 ⇒ y = 5
(1, 5) is the solution.
→ Now, put x = 2
y = 7 – 2 × 2 ⇒ y = 3
(2, 3) is the solution.
→ Now, put x = -1
y = 7 – 2 × -1 ⇒ y = 9
(-1, 9) is the solution.
The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9
⇒ y = 9 – πx
→ Put x = 0,
y = 9 – π×0 ⇒ y = 9
(0, 9) is the solution.
→ Now, put x = 1
y = 9 – π×1 ⇒ y = 9-π
(1, 9-π) is the solution.
→ Now, put x = 2
y = 9 – π×2 ⇒ y = 9-2π
(2, 9-2π) is the solution.
→ Now, put x = -1
y = 9 – π× -1 ⇒ y = 9+π
(-1, 9+π) is the solution.
The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).

(iii) x = 4y
→ Put x = 0,
0 = 4y ⇒ y = 0
(0, 0) is the solution.
→ Now, put x = 1
1 = 4y ⇒ y = 1/4
(1, 1/4) is the solution.
→ Now, put x = 4
4 = 4y ⇒ y = 1
(4, 1) is the solution.
→ Now, put x = 8
8 = 4y ⇒ y = 2
(8, 2) is the solution.
The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).

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

  • (i) (0, 2)             
  • (ii) (2, 0)            
  • (iii) (4, 0)           
  • (iv) (√2, 4√2)             
  • (v) (1, 1)

Answer

(i) Put x = 0 and y = 2 in the equation x – 2y = 4.
0 – 2×2 = 4
⇒ -4 ≠ 4
∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x – 2y = 4.
2 – 2×0 = 4
⇒ 2 ≠ 4
∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x – 2y = 4.
4 – 2×0 = 4
⇒ 4 = 4
∴ (4, 0) is a solution of the given equation.

(iv) Put x = √2 and y = 4√2 in the equation x – 2y = 4.
√2 – 2×4√2 = 4 ⇒ √2 – 8√2 = 4 ⇒ √2(1 – 8) = 4
⇒ -7√2  ≠ 4
∴ (√2, 4√2) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x – 2y = 4.
1 – 2×1 = 4
⇒ -1 ≠ 4
∴ (1, 1) is not a solution of the given equation.

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

Answer

Given equation = 2x + 3y = k
x = 2, y = 1 is the solution of the given equation.
A/q,
Putting the value of x and y in the equation, we get
2×2 + 3×1 = k
⇒ k = 4 + 3
⇒ k = 7

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