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# Ex 3.1 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

## Class 10 : Ex 3.1 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

The document Ex 3.1 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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Ques 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.
Sol: At present:
Let Aftab’s age = x years
His daughter’s age = y years
Seven years ago:
Aftab’s age = (x - 7) years
Daughter’s age = (y - 7) years
According to the condition,
[Aftab’s age] = 7 [His daughter’s age]
⇒ [x - 7] = 7 [y - 7]
⇒ x - 7 = 7y - 49
⇒ x - 7y - 7 + 49 = 0
⇒ x - 7y + 42 = 0 ...(1)
After three years:
Aftab’s age = (x + 3) years
His daughter’s = (y + 3) years
According to the condition,
[Aftab’s age] = 3 [His daughter’s age]
⇒ [x + 3] = 3 [y + 3]
⇒ x + 3 = 3y + 9
⇒ x - 3y + 3 - 9 = 0
⇒ x - 3y - 6 = 0 ...(2)
Graphical Representation of equation (1) and (2):
From equation (1), we have:
x - 7y + 42 = 0  ⇒ x 0 7 35 y 6 7 11 (x, y) (0. 6) (7, 7) (35, 11)

From equation (2), we have
x - 3y - 6 = 0    ⇒ x 0 6 3 y -2 0 8 (x, y) (0. -2) (6, 0) (30, 8) Plotting the points (0, 6), (7, 7) and (35, 11), we obtain a straight line l1.
Plotting the points (0, - 2), (6, 0), (30, 8), we obtain a straight line l2. The lines l1 and l2 intersect at (42, 12).

Ques 2: The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for 1300. Represent this situation algebraically and geometrically.
Sol:
Let the cost of a bat = ₹ x
And the cost of a ball = ₹ y
∴ Cost of 3 bats = ₹ 3x
Cost of 6 balls = ₹ 6y
Again,
Cost of 1 bat = ₹ x
Cost of 2 balls = ₹ 2y
Algebraic representation:
[Cost of 3 bats] + [Cost of 6 balls] = ₹ 3900
⇒ 3x + 6y = 3900
⇒ x + 2y = 1300 ...(1)
Also,
[Cost of 1 bat] + [Cost of 3 balls] = ₹ 1300
⇒ x + 3y = 1300 ...(2)
Thus, (1) and (2) are the algebraic representation of the given situation.
Geometrical representation:
We have, for equation (1):
x + 2y = 1300  or  y = 1300-x/2

 x 0 1300 100 y 650 0 600 (x, y) (0, 650) (1300, 0) (100, 600)

⇒ and for equation (2):
x + 3y = 1300 ⇒ x 400 100 1000 y 300 400 100 (x, y) (400, 300) (100, 400) (1000, 000) Plotting the points (0, 650), (100, 600) and (0, 1300), we get a straight line l1 and plotting points (100, 400), (400, 300) and (1000, 100), we get another straight line l2. The straight lines l1 and l2 are the graphical representation of the given situation. We also see from the obtained figures that the straight lines representing the two equations intersect at (1300, 0).

Ques 3: The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is 300. Represent the situation algebraically and geometrically.
Sol:
Let the cost of 1 kg of apples = ₹ x
And the cost of 1 kg of grapes = ₹ y
Algebraic Representation:
2x + y = 160 ...(1)
and 4x + 2y = 300 ⇒ 2x + y = 150 ...(2)
Graphical Representation:
From equation (1), we have
2x + y = 160 ⇒ y = 160 - 2x

 x 50 40 30 y 60 80 100 (x, y) (50, 60) (40, 80) (30, 100)

From the equation (2), we have:
2x + y =150 ⇒ y = 150 - 2x

 x 50 30 25 y 50 90 100 (x, y) (50, 50) (30, 90) (25, 100) Plotting the points (30, 100), (40, 80) and (50, 60), we get a straight line l1.
Also, joining the points (50, 50), (30, 90) and (25, 100), we get a straight line l2.
The straight lines l1 and l2 are the graphical representation of the equations 1 and 2 respectively. The lines are parallel.

• Graphical Method (Solution of a Pair of Linear Equations)

Since, the general form of a pair of linear equations is:
a1 x + b1 y + c1 = 0 ...(1)
and a2 x + b2 y + c2 = 0 ...(2)
When we draw their graphs on the same graph paper then we get two straight lines.
∴ Let us suppose that AB and CD represent the graphs of (1) and (2) respectively. Now, three cases are possible:
Case-I: When the lines AB and CD intersect at a point (m, n). In this case the given pair of linear equations has a unique solution. The solution is given by:
x = m; y = n
Such pair of linear equations is called consistent system of linear equations.
Case-II: When the lines AB and CD are parallel. In this case there is no common point, i.e. There is no solution of the given system of linear equations. Such pair of linear equations is called inconsistent system.
Case-III: When lines AB and CD are coincident. In this case every point is a solution i.e., the system has an infinite number of solutions.
Such pair of linear equations is called consistent and dependent.

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