- RD Chapter 3- Pair of Linear Equations in Two Variables Ex-3.2
- 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-3.2 |
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 |

**Answer
1** :

Let number of rides on the wheel = x

and number of play of Hoopla = y

According to the given conditions x = 2y ⇒ x – 2y = 0 ….(i)

and cost of ride on wheel at the rate of Rs. 3 = 3x

and cost on Hoopla = 4y

and total cost = Rs. 20

3x + 4y = 20 ….(ii)

Now we shall solve these linear equations graphically as under

We take three points of each line and join them to get a line in each case the point of intersection will be the solution

From equation (i)

x = 2y

X | 4 | 0 | 6 |

y | 2 | 0 | 3 |

y = 2, then x = 2 x 2 = 4

y = 0, then x = 2 x 0 = 0

y = 3, then x = 2 x 3 = 6

Now, we plot these points on the graphs and join them to get a line

Similarly in equation (ii)

3x + 4y = 20 ⇒ 3x = 20 – 4y

Now we plot these points and get another line by joining them

These two lines intersect eachother at the point (4, 2)

Its solution is (4, 2)

Which is a unique Hence x = 4, y = 2

**Answer
2** :

Seven years ago

Let age of Aftab’s daughter = x years

and age of Aftab = y years

and 3 years later

Age of daughter = x + 10 years

and age of Aftab = y + 10 years

According to the conditions,

y = 7x ⇒ 7x – y = 0 ……….(i)

y + 10 = 3 (x + 10)

=> y + 10 = 3x + 30

3x – y = 10 – 30 = -20

3x – y = -20 ….(ii)

Equations are

7x – y = 0

3 x – y = -20

Now we shall solve these linear equations graphically as under

7x – y = 0 ⇒ y = 7x

X | 0 | 1 | -1 |

y | 0 | 7 | -7 |

If x = 0, y = 7 x 0 = 0

If x = 1, y = 7 x 1=7

If x = -1, y = 7 x (-1) = -7

Now plot these points on the graph and join

then

3x – y = -20

y = 3x + 20

X | -1 | -2 | -3 |

y | 17 | 14 | 11 |

If x = -1, y = 3 x (-1) + 20 = -3 + 20= 17

If x = -2, y = 3 (-2) + 20 = -6 + 20 = 14

If x = -3, y = 3 (-3) + 20 = -9 + 20= 11

Now plot the points on the graph and join them we see that lines well meet at a point on producing at (5, 35).

**Answer
3** :

Path of A train is 3x + 4y – 12 = 0

and path of B train is 6x + 8y – 48 = 0

Graphically, we shall represent these on the graph as given under 3x + 4y- 12 = 0

**Answer
4** :

Plot the points (-2, 3) and (2, -2) and join them to get a line

and also plot the points (0, 5), (4, 0) and joint them to get another line as shown on the graph

We see that these two lines are parallel to each other

On comparing the ratios , and without drawing them, find out whether the lines representing following pairs of linear equations intersect at a point, are parallel or coincide :

(i) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

(ii) 9x + 3y +12 = 0

18x + 6y + 24 = 0

(iii) 6x – 3y +10 = 0

2x – y + 9 = 0

**Answer
5** :