**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**