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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

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 - (22 )
= - 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  0)
= 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