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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 2*x *+ *y *= 7

(ii) We have : π*x *+ *y *= 9. Let *x *= 1/π Then, π × 1/ π + *y *= 9

are all solutions of the equation π*x
*+ *y *= 9.

(iii)
We have : *x *= 4*y *

Let
*x *= 8

Then,
8 = 4*y *⇒ *y *= 2.

∴ (8, 2) is a solution.

Let
*x *= 12. Then, 12 = 4*y*,

⇒
*y *= 3. ∴ (12, 3) is a solution.

Let
*x *= 16. Then, 16 = 4*y*,

⇒
*y *= 4. ∴ (16, 4) is a solution.

Let
*x *= 20. Then, 20 = 4*y*, ⇒ *y *= 5.

∴ (20, 5) is a solution.

Therefore,
(8, 2), (12, 3), (16, 4) and (20, 5) are all solutions of *x *= 4*y*.

**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 *–
2*y *= 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. **NEXT** **HOME PAGE**