Class 9 NCERT Solutions Maths Chapter 4 - Linear Equations in Two Variables

Ex. 4.1

Ex. 4.2

Linear Equations in Two Variables Exercise Ex. 4.1

Solution 1

Let cost of notebook and a pen be x and y respectively.

Cost of note book = 2

cost of pen
x = 2y
x - 2 y = 0

Solution 2

(i)

Comparing this equation with ax + by + c = 0

(ii)

Comparing this equation with ax + by + c = 0

a = 1, b = -

, c = -10

(iii) - 2x + 3 y = 6
- 2x + 3 y - 6 = 0

Comparing this equation with ax + by + c = 0
a = - 2, b = 3, c = - 6

(iv) x = 3y
       1x - 3y + 0 = 0
       Comparing this equation with ax + by + c = 0
       a = 1, b = - 3, c = 0

(v) 2x = - 5y
2x + 5y + 0 = 0

Comparing this equation with ax + by + c = 0
a = 2, b = 5, c = 0

(vi) 3x + 2 = 0
3x + 0.y + 2 = 0

Comparing this equation with ax + by + c = 0
a = 3, b = 0, c = 2

(vii) y - 2 = 0
0.x + 1.y - 2 = 0

Comparing this equation with ax + by + c = 0
a = 0, b = 1, c = - 2

(viii) 5 = 2x
       - 2x + 0.y + 5 = 0
        Company this equation with ax + by + c = 0
        a = - 2, b = 0, c = 5

Linear Equations in Two Variables Exercise Ex. 4.2

Solution 1

y = 3x + 5 is a linear equation in two variables and it has infinite solutions. As for every value of x there will be a value of y satisfying above equation and vice versa.
Hence, the correct answer is (iii).

Solution 2

(i) 2x + y = 7
For x = 0
2(0) + y = 7

y = 7

So, (0, 7) is a solution of this equation
For x = 1

2(1) + y = 7

y = 5

So, (1, 5) is a solution of this equation

For x = -1
2(-1) + y = 7

y = 5

So, (-1, 9) is a solution of this equation

            For x = 2
       2(2) + y = 7

y = 3

So (2, 3) is a solution of this equation.

(ii)

+ y = 9

For x = 0

(o) + y = 9

y = 9

      So (0, 9) is a solution of this equation
      For x = 1

(1) + y =9

y = 9 -

So, (1, 9 -

) is a solution of this equation

For x = 2

(2) + y = 9

y = 9 - 2

So, (2, 9 -2

) is a solution of this equation

For x = -1

(-1) + y = 9

y = 9 +

So, (-1, 9 +

) is a solution of this equation

(iii) x = 4y
       For x = 0
       0 = 4y

y = 0

So, (0, 0) is a solution of this equation
       For y = 1
       x = 4(1) = 4
       So, (4, 1) is a solution of this equation
       For y = - 1
       x = 4(-1)
       x = -4
       So, (-4, - 1) is a solution of this equation
       For x = 2
       2 = 4y

y =

So,

is a solution of this equation.

Solution 3

(i)   (0, 2)
    Putting x = 0, and y = 2 in the L.H.S of given equation
      x - 2y = 0 - (2

2 )

= - 4

As -4 # 4
    L.H.S # R.H.S
      So (0, 2) is not a solution of this equation.

(ii) (2, 0)

Putting x = 2, and y = 0 in the L.H.S of given equation
x - 2y = 2 - (2

= 2

As 2 # 4
L.H.S

R.H.S
So (2, 0) is not a solution of this equation.

(iii) (4, 0)
       Putting x = 4, and y = 0 in the L.H.S of given equation
   x - 2y = 4 - 2(0)
       = 4 = R.H.S
       So (4, 0) is a solution of this equation.

(iv)

Putting x =

and y = 4

in the L.H.S of given equation.

L.H.S

R.H.S
So

is not a solution of this equation.

(v) (1, 1)
Putting x = 1, and y = 1 in the L.H.S of given equation
x - 2y = 1 - 2(1)

= 1 - 2

= - 1

As -1 # 4
L.H.S

R.H.S
So (1, 1) is not a solution of this equation.

Solution 4

Putting x = 2, and y = 1 in the given equation

2x + 3y = k
2(2) + 3(1) = k
4 + 3 = k
k = 7

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Class 9 NCERT Solutions Maths Chapter 4 - Linear Equations in Two Variables

Linear Equations in Two Variables Exercise Ex. 4.1

Solution 1

Solution 2

Linear Equations in Two Variables Exercise Ex. 4.2

Solution 1

Solution 2

Solution 3

Solution 4

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