Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.1 pdf || UP Board
Pair of Linear Equations in Two Variables
Important Points
- Two lines a1x + b1y + c1
= 0, a2x + b2y + c = 0 Intersecting, Coincident
or Parallel lines if,
Exercise – 3.1
1. Aftab tells his daughter, “Seven
years ago, I was seven times as old as you were then. Also, three years from
now, I shall be three times as old as you will be.” (Isn’t this interesting?)
Represent this situation algebraically and graphically.
Solution:
Let Aftab's present age = x years
Let the present age of the daughter
= y years
Aftab's age 7 years ago = x – 7
years
and daughter’s age 7 years ago = y –
7 years
x – 7 = 7(y – 7)
x -7 = 7y – 49
x – 7y = -42 ………………(1)
Aftab's present age after 3 years =
x + 3 years
Present age of daughter after 3
years = y + 3 years
x + 3 = 3(y + 3)
x + 3 = 3y + 9
x – 3y = 6 ………………….…(2)
So, in algebraic form,
x – 7y = -42
x – 3y = 6
Now to express graphically three
solutions,
From equation (1),
x – 7y = -42
From equation (2),
x – 3y = 6
2. The coach of a cricket team buys
3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 2 more balls of
the same kind for ₹ 1300. Represent this situation algebraically and
geometrically.
Solution:
Let cost of one bat = ₹x
Let cost of one ball = ₹y
according to the first condition,
3x + 6y = 3900 ……………. (1)
according to the second condition,
x + 2y = 1300 …………….(2)
So, in algebraic form,
3x + 6y = 3900
x + 2y = 1300
Now to express graphically three
solutions,
From equation (1),
3x + 6y = 3900
x |
300 |
100 |
-100 |
y |
500 |
600 |
700 |
From equation (2),
x + 2y = 1300
x |
300 |
100 |
-100 |
y |
500 |
600 |
700 |
3. The cost of 2 kg of apples and 1
kg of grapes on a day was found to be ₹ 160. After month, the cost of 4 kg of
apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and
geometrically.
Solution:
Let cost of 1 kg apple = ₹x
Let cost of 1 kg of grapes = ₹ y
according to the first condition,
2x + y = 160 ……………. (1)
according to the second condition,
4x + 2y = 300 …………….(2)
So, in algebraic form,
2x + y = 160
4x + 2y = 300
Now to express graphically three
solutions,
From equation (1),,
2x + y = 160
x |
50 |
60 |
70 |
y |
60 |
40 |
20 |
From equation (2),
4x + 2y = 300
x |
70 |
80 |
75 |
y |
10 |
-10 |
0 |
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