# Linear Equations in Two Variables| NCERT Solutions| Class 9

NCERT Solutions of Chapter: Linear Equations in Two Variables. NCERT Solutions along with worksheets and notes for Class 9.

Exercise 4.1

(1) The cost of the note book 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).

Ans-

Let, cost of a notebook = Rs x

Cost of a pen = Rs y

According to given condition

Cost of a notebook = 2 the cost of a pen

x = 2y

x – 2y = 0

Hence, x – 2y = 0 is required statement.

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

Ans-

y = 3x + 5 has infinitely many solutions because a linear equations of two variables has many solutions.

(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) (  , 4 )  (v) (1, 1)

Ans-

(i) (0, 2)

Put x = 0 and y = 2 in equation x – 2y = 4

x- 2y = 4

LHS

=0 – 2(2)

= – 4

≠RHS

Therefore, (0, 2) is not a solution of given equation.

(ii) (2, 0)

Put x = 2 and y = 0 in equation x – 2y = 4

x- 2y = 4

LHS

=2 – 2(0)

= 2 – 0

= 2

≠RHS

Therefore, (2, 0) is not a solution of given equation.

(iii) (4, 0)

Put x = 4 and y = 0 in equation x – 2y = 4

x- 2y = 4

LHS

=4 – 2(0)

= 4 – 0

= 4

=RHS

Therefore, (4, 0) is a solution of given equation.

(iv) (√2  , 4√2 )

Put x =√2  and y = 4√2 )   in equation x – 2y = 4

x- 2y = 4

LHS

x – 2y = 4

= √2  – 2(4√2 )

= √2 (1 – 8)

= -7

≠RHS

Therefore, ( √2 , 4√2 ) is not a solution of given equation.

(v) (1, 1)

Put x = 1 and y = 1 in equation x – 2y = 4

x- 2y = 4

LHS

=1 – 2(1)

= 1 – 2

= -1

≠RHS

Therefore, (1, 1) is not a solution of given equation.

Hence, only (4, 0) is 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.

Ans-

Put x = 2 and y = 1 in equation 2x + 3y = k

2(2) + 3(1) = k

4 + 3 = k

7 = k

or

k = 7

Hence, value of k is 7.

Exercise 4.3

(1) Draw the graph of each of the following linear equations in two variables:

(i) x + y = 4   (ii) x – y = 2   (iii) y = 3x   (iv) 3 = 2x + y

Ans-

(i) x + y = 4

Solutions of the equation

Put x = 0

0 + y = 4

y = 4

Therefore (0, 4) is a solution of given equation.

Put y = 0

x + 0 = 4

x = 4

Therefore (4, 0) is a solution of given equation.

Hence, (0, 4) and (4, 0) are the two solutions of the given equation.

Solutions of the equations are tabulated below to draw a graph-

 x 0 4 y 4 0

AB is required linear equation in two variables.

(ii) x – y = 2

Solutions of the equation

Put x = 0

0 – y = 2

y = -2

Therefore (0, -2) is a solution of given equation.

Put y = 0

x – 0 = 2

x = 2

Therefore (2, 0) is a solution of given equation.

Hence, (0, -2) and (2, 0) are the two solutions of the given equation.

Solutions of the equations are tabulated below to draw a graph-

 x 0 2 y -2 0

AB is required linear equation in two variables.

(iii) y = 3x

Solutions of the equation

Put x = 0

y = 3(0)

y = 0

Therefore (0, 0) is a solution of given equation.

Put y = 3

3x = 3

x = 1

Therefore (1, 3) is a solution of given equation.

Hence, (0, 0) and (1, 3) are the two solutions of the given equation.

Solutions of the equations are tabulated below to draw a graph-

 x 0 1 y 0 3

AB is required linear equation in two variables.

(iv) 3 = 2x + y

2x + y = 3

Solutions of the equation

Put x = 0

2(0) + y = 3

y = 3

Therefore (0, 3) is a solution of given equation.

Put y = 1

2x + 1 = 3

2x = 2

x = 1

Therefore (1, 1) is a solution of given equation.

Hence, (0, 3) and (1, 1) are the two solutions of the given equation.

Solutions of the equations are tabulated below to draw a graph-

 x 0 1 y 3 1

AB is required linear equation in two variables.

(2) Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Ans-

The solutions of equations are (2, 14).

Equations which satisfy these solutions are

x + y = 16

2x + y = 18

7x = y

These are some equations which can satisfy given solutions. There can be infinite lines or equations which will satisfy the given solutions or coordinates.

(4) The taxi fare in a city is as follows: For the first kilometer, the fare is Rs 8and for the subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information and draw its graph.

Ans-

Total distance covered = x km

Fare for first kilometer = Rs 8

Fare for the subsequent distance = Rs 5

Total fare = Rs y

Subsequent distance = Total distance – First kilometer

= x – 1

Fare for subsequent distance = (x – 1) 5

Total fare = fare for first kilometer + fare for subsequent distance

y = 8 + (x – 1) 5

y = 8 + 5x – 5

y = 5x + 3

-5x + y = 3

or multiply equation with -1

5x – y = -3

5x – y + 3 = 0

Hence, required equation is 5x – y + 3 = 0.

5x – y + 3 = 0

Solutions of the equation

Put x = 0

5(0) – y + 3= 0

-y + 3 = 0

y = 3

Therefore (0, 3) is a solution of given equation.

5x – y + 3 = 0

Put x = -1

5(-1) – y + 3 = 0

-5 –y + 3 = 0

-2 – y = 0

y = -2

Therefore (-1, -2) is a solution of given equation.

Hence, (0, 3) and (-1, -2) are the two solutions of the given equation.

Solutions of the equations are tabulated below to draw a graph-

 x 0 -1 y 3 -2

AB is required linear equation in two variables.

(5) From the choices given below, choose the equation whose graphs are given in Fig below (a) and (b).

For fig (a)

(i) y = x

(ii) x + y = 0

(iii) y = 2x

(iv) 2 + 3y = 7x

For fig (b)

(i) y = x + 2

(ii) y = x – 2

(iii) y = -x + 2

(iv) x + 2y = 6

Ans-

For fig (a)

The coordinates of the line of graph are (0, 0), (-1, 1) and(1, -1).

By inspection, value of x and y are equal or equal with different signs. Equation x + y = 0 is the corresponding to the graph and coordinates.

{ Note: we can check equations also

(i) y = x

Put x = 0

y = 0

Therefore, coordinate (0, 0) is satisfied.

(-1, 1)

Put x = -1

y= -1

Therefore, coordinates (-1,1 ) are not satisfied.

Hence, equation y = x is not satisfied.

(ii) x + y = 0

(0, 0)

Put x = 0

0 + y = 0

y = 0

Therefore, coordinates (0, 0) are satisfied.

(-1, 1)

Put x = -1

-1 + y = 0

y = 1

Therefore, coordinates (-1, 1)are satisfied.

(1, -1)

Put x = 1

1 + y = 0

y = -1

Therefore, coordinates (1, -1)are satisfied.

Hence, equation x + y = 0 is satisfied.

We can check all these equations.}

For Fig (b)

The coordinates of the line of graph are (-1, 3), (0, 2) and(2, 0).

By inspection, coordinates satisfy equation y = – x + 2.

{Note: Check all equations as above. }

AB is required linear equation in two variables.

(i) The work done is 10 units when distance travelled is 2 units.

(ii) The work done is 0 units when distance travelled is 0 units.

(7) Yamini and Fatima, two students of Class IX of a school, together contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

Ans-

Let contribution of Yamini is Rs x and contribution of Fatima is Rs y.

According to condition

x + y = 100

The values of the x and y are tabulated below-

 x 0 50 100 y 100 50 0

The graph from above coordinates plotted below-

AB is required linear equation in two variables.

Exercise 4.4

(1) Give the geometric representations of y = 3 as an equation

(i) in one variable

(ii) In two variable

Ans-

(i) The geometric representations of y = 3 as an equation in one variable is presenting below-

(ii) y = 3 is a straight line parallel to x- axis.

Therefore, graph of linear equation is given below-

AB is required linear equation in two variables.

AB is required linear equation in two variables.

Helping Topics

Linear Equations in Two Variables

Worksheet Class 9