- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.1
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.3
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.4
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.5
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.6
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.7
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.8
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.9
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.10
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.11
- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-VSAQS

RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.1 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.3 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.4 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.5 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.6 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.7 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.8 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.9 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.10 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.11 |
RD Chapter 3- Pair of Linear Equations in Two Variables Ex-VSAQS |

x + y = 3

2x + 5y = 12 (C.B.S.E. 1997)

**Answer
1** :

x + y = 3

=> x = 3 – y

Substituting some different values of y, we get the corresponding values of x as shown below

Now plot the points on the graph and join them 2x + 5y = 12

2x = 12 – 5y

⇒ x =

Substituting some different values of y, we get the corresponding values of x as shown below

Now plot the points on the graph and join them we see that these two lines intersect each other at (1, 2)

x = 1, y = 2

x – 2y = 5

2x + 3y = 10 (C.B.S.E. 1997)

**Answer
2** :

x – 2y = 5 => x = 5 + 2y

Substituting some different values of y, we get the corresponding values of x as shown below

Now plot the points are the graph and join them

2x + 3y = 10 => 2x = 10 – 3y

⇒ x =

Substituting some different values of y We get the corresponding values of x as shown below :

Now plot the points on the graph and join them we see that these two lines intersect each other at (5, 0)

x = 5, y = 0

3x + y + 1 = 0

2x – 3y + 8 = 0 (C.B.S.E. 1996)

**Answer
3** :

3x + y + 1 = 0

y = -3x – 1

Substituting the values of x, we get the corresponding values of y, as shown below

Now plot the points on the graph and join them

2x – 3y + 8 = 0

⇒ 2x = 3y – 8

⇒ x =

Substituting some different values of y, we get the corresponding values of x as shown below

Now plot the points on the graph and join then we see that these two lines intersect, each other at (-1, -2)

x = -1, y = 2

2x + y – 3 = 0

2x – 3y – 7 = 0 (C.B.S.E. 1996)

**Answer
4** :

2x + y – 3 = 0 => y = -2x + 3

Substituting some different values of x, we get the corresponding values of y as shown below:

Now plot the points and join them 2x – 3y – 7 = 0

2x = 3y +7

x =

Substituting some different values of y, we get corresponding values of x as shown below:

Now plot the points on the graph and join them we see that these two lines intersect each other at (2, -1)

x + y = 6

x – y = 2 (C.B.S.E. 1994)

**Answer
5** :

x + y = 6 => x = 6 – y

Substituting some different values of y, we get the corresponding values of x as shown under

Now plot the points on the graph and join them

x – y = 2 ⇒ x = 2 + y

Substituting some different values of y, we get the corresponding values of x as shown below:

Now plot the points on the graph and join them

We see that there two lines intersect each other at (4, 2)

x = 4, y = 2

x – 2y = 6

3x – 6y = 0 (C.B.S.E. 1995)

**Answer
6** :

x – 2y = 6

x = 6 + 2 y

Substituting some different values ofy, we get the corresponding values of x as shown below:

Now plot the points and join them

3x – 6y = 0 ⇒ 3x = 6y ⇒ x = 2y

Substituting some different value of y, we get corresponding the values of x as shown below:

plot the points on the graph and join them We see that these two lines intersect each other at no point

The lines are parallel

There is no solution

x + y = 4

2x – 3y = 3 (C.B.S.E. 1995)

**Answer
7** :

x + y = 4 => y = 4 – x

Substituting some different values of y, we get the corresponding values of x as shown below:

Now plot the points and join them 2x – 3y = 3

⇒ 2x = 3 + 3y

⇒ x =

Substituting some different values of y, we get the corresponding values of x as shown below:

Now plot the points on the graph and join them we see that these two lines intersect each other at (3, 1)

x = 3, y = 1