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Class Nine Maths Chapter 4 NCERT Solutions
Chapter 4: Linear Equation in Two Variables
TEXTBOOK EXERCISE 4.1
1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be Rs x and that of a pen to be Rs y.)
Sol. Cost of a notebook = x and the cost of a pen = y
Then according to the given statement
x = 2y or x – 2y = 0.
2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case : (i) 2x + 3y =
(ii) x− y/5 − 10 = 0 (iii) –2x + 3y = 6
(iv) x = 3y (v) 2x = – 5y (vi) 3x + 2 = 0
(vii) y – 2 = 0 (viii) 5 = 2x
(iii) –2x + 3y = 6
⇒ –2x + 3y – 6 = 0.
So, a = –2, b = 3, c = – 6.
(iv) x = 3y ⇒ x – 3y + 0 = 0
So, a = 1, b = –3, c = 0.
(v) 2x = – 5y ⇒ 2x + 5y + 0 = 0
So, a = 2, b = 5, c = 0.
(vi) 3x + 2 = 0 ⇒ 3x + 0⋅y + 2 = 0
So, a = 3, b = 0, c = 2.
(vii) y – 2 = 0
⇒ 0⋅x + 1⋅y – 2 = 0
So, a = 0, b = 1, c = – 2.
(viii) 5 = 2x ⇒ 5 – 2x = 0
⇒ –2x + 0⋅y + 5 = 0
So, a = –2, b = 0, c = 5.
TEXTBOOK’S 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
Sol. (iii) Infinitely many solutions. It is because a linear equation in two variables has infinitely many solutions. We keep changing the value of x and solving the linear equation for the corresponding value of y.
2. Write four solutions for each of the following equations :
(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y
Sol. (i) We have : 2x + y = 7
Let x = 1. Then, 2 × 1 + y = 7
⇒ y = 7 – 2 = 5
⇒ (1, 5) is a solution.
Let x = 2. Then, 2 × 2 + y = 7
⇒ y = 7 – 4 = 3 ⇒ (2, 3) is another solution.
Let x = 3. Then, 2 × 3 + y = 7
⇒ y = 7 – 6 = 1 ⇒ (3, 1) is another solution.
Let x = 4. Then, 2 × 4 + y = 7
⇒ y = 7 – 8 = –1 ⇒ (4, –1) is another solution.
Therefore, (1, 5), (2, 3), (3, 1) and (4, –1) are all solutions of 2x + y = 7
(ii) We have : πx + y = 9. Let x = 1/π Then, π × 1/ π + y = 9
are all solutions of the equation πx + y = 9.
(iii) We have : x = 4y
Let x = 8
Then, 8 = 4y ⇒ y = 2.
∴ (8, 2) is a solution.
Let x = 12. Then, 12 = 4y,
⇒ y = 3. ∴ (12, 3) is a solution.
Let x = 16. Then, 16 = 4y,
⇒ y = 4. ∴ (16, 4) is a solution.
Let x = 20. Then, 20 = 4y, ⇒ y = 5.
∴ (20, 5) is a solution.
Therefore, (8, 2), (12, 3), (16, 4) and (20, 5) are all solutions of x = 4y.
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)
Sol. Given equation: x – 2y = 4
(i) When x = 0, y = 2, then
0 – 4 = – 4.
⇒ RHS ≠ LHS.
Therefore, it is not a solution.
(ii) When x = 2, y = 0 then
2 – 0 = 4.
⇒ RHS ≠ LHS.
Therefore, (2, 0) is not a solution.
(iii) When x = 4, y = 0, then 4 – 2 × 0 = 4.
LHS = RHS
Therefore, (0, 2) is a solution.
(iv) When x = √ 2 , y = 4 √ 2 , then √ 2 – 8 √ 2 = 4.
⇒ LHS ≠ RHS
Therefore, (√ 2 , 4 √ 2 ) is not a solution.
(v) When x = 1, y = 1, then 1 – 2 × 1 = 4. LHS ≠ RHS Therefore, (1, 1) is not a solution.
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