cbse ncert solutions for class 10 mathematics chapter 3
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Class- X-CBSE-Mathematics Pair of Linear Equations in Two Variables
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CBSE NCERT Solutions for Class 10 Mathematics Chapter 3 Back of Chapter Questions
1. Aftab tells his daughter, βSeven years ago, I was seven times as old as you werethen. Also, three years from now, I shall be three times as old as you will be.β(Isnβt this interesting?) Represent this situation algebraically and graphically.
Solution:
Let the present age of Aftab be x and present age of daughter be y.
Hence, seven years ago,
Age of Aftab = π₯π₯ β 7
Age of daughter = π¦π¦ β 7
Hence, as per the given condition,(π₯π₯ β 7) = 7(π¦π¦ β 7)
βΉ x β 7 = 7y β 49
βΉ x β 7y = β42β¦β¦β¦β¦β¦β¦β¦β¦β¦(i)
Three years later,
Age of Aftab = π₯π₯ + 3
Age of daughter = π¦π¦ + 3
Hence, as per the given condition, (π₯π₯ + 3) = 3(π¦π¦ + 3)
β π₯π₯ + 3 = 3π¦π¦ + 9
β π₯π₯ β 3π¦π¦ = 6 β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(ii)
Hence, equation (i) and (ii) represent given conditions algebraically as:
π₯π₯ β 7π¦π¦ = β42
π₯π₯ β 3π¦π¦ = 6
Graphical Representation:
x β 7y = β42
β x = β42 + 7y
Two solutions of this equation are:
π₯π₯ β7 0
π¦π¦ 5 6
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x β 3y = 6
β x = 6 + 3y
Two solutions of this equation are:
π₯π₯ 6 0
π¦π¦ 0 β2
The graphical representation is as follows:
2. The coach of a cricket team buys 3 bats and 6 balls for βΉ 3900. Later, she buys another bat and 3 more balls of the same kind for βΉ 1300. Represent this situation algebraically and geometrically.
Solution:
Let the price of a bat be βΉ x and a ball be βΉ y.
Hence, we can represent algebraically the given conditions as:
3π₯π₯ + 6π¦π¦ = 3900
π₯π₯ + 3π¦π¦ = 1300
3π₯π₯ + 6π¦π¦ = 3900 βΉ π₯π₯ =3900 β 6π¦π¦
3
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Two solutions of this equation are:
π₯π₯ 300 100
π¦π¦ 500 600
π₯π₯ + 3π¦π¦ = 1300 βΉ π₯π₯ = 1300 β 3π¦π¦
Two solutions of this equation are:
π₯π₯ 100 β200
π¦π¦ 400 500
The graphical representation is as follows:
3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be βΉ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is βΉ 300. Represent the situation algebraically and geometrically.
Solution:
Let the cost of 1 kg of apples be βΉ π₯π₯ and 1 kg grapes be βΉ y.
The given conditions can be algebraically represented as:
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2π₯π₯ + π¦π¦ = 160 (1)
4π₯π₯ + 2π¦π¦ = 300 (2)
2π₯π₯ + π¦π¦ = 160 βΉ π¦π¦ = 160 β 2π₯π₯
Two solutions of this equation are:
π₯π₯ 50 80
π¦π¦ 60 0
4π₯π₯ + 2π¦π¦ = 300 βΉ π¦π¦ =300 β 4π₯π₯
2
Two solutions of this equation are:
π₯π₯ 60 75
π¦π¦ 30 0
The graphical representation is as follows:
β¦ β¦ β¦
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EXERCISE 3.2
1. Form the pair of linear equations in the following problems and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 Pencils and 7 pens together cost βΉ 50, whereas 7 pencils and 5 pens together cost βΉ 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Let the number of girls in the class be x and number of boys in the class be y.
As per the given conditions
π₯π₯ + π¦π¦ = 10 (1)
π₯π₯ β π¦π¦ = 4 (2)
π₯π₯ + π¦π¦ = 10 βΉ π₯π₯ = 10 β π¦π¦
Two solutions of this equation are:
π₯π₯ 0 10
π¦π¦ 10 0
π₯π₯ β π¦π¦ = 4 βΉ π₯π₯ = 4 + π¦π¦
Two solutions of this equation are:
π₯π₯ 2 4
π¦π¦ β2 0
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The graphical representation is as follows:
From the graph, it can be observed that the two lines intersect each other at the point (7, 3).
So,π₯π₯ = 7 and y = 3.
Hence, the number of girls and boys in the class are 7 and 3 respectively.
(ii) Let the cost of one pencil be βΉ π₯π₯ and one pen be βΉ π¦π¦ respectively.
As per the given conditions,
5π₯π₯ + 7π¦π¦ = 50 (1)
7π₯π₯ + 5π¦π¦ = 46 (2)
5π₯π₯ + 7π¦π¦ = 50 βΉ π₯π₯ =50 β 7π¦π¦
5
Two solutions of this equation are:
π₯π₯ 3 10
π¦π¦ 5 0
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7π₯π₯ + 5π¦π¦ = 46 βΉ π₯π₯ =46 β 5π¦π¦
7
Two solutions of this equation are:
π₯π₯ 8 3
π¦π¦ β2 5
The graphical representation is as follows:
From the graph, it can be observed that the two lines intersect each other at the point (3, 5).
So,π₯π₯ = 3 and π¦π¦ = 5.
Therefore, the cost of one pencil and one pen are βΉ3 and βΉ5 respectively.
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2. On comparing the ratios ππ1ππ2
, ππ1ππ2
and ππ1ππ2
, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5π₯π₯ β 4π¦π¦ + 8 = 0 7π₯π₯ + 6π¦π¦ β 9 = 0
(ii) 9π₯π₯ + 3π¦π¦ + 12 = 0 18π₯π₯ + 6π¦π¦ + 24 = 0
(iii) 6π₯π₯ β 3π¦π¦ + 10 = 0 2π₯π₯ β π¦π¦ + 9 = 0
Solution:
(i) 5π₯π₯ β 4π¦π¦ + 8 = 0
7π₯π₯ + 6π¦π¦ β 9 = 0
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get
ππ1 = 5, ππ1 = β4, ππ1 = 8
ππ2 = 7, ππ2 = 6, ππ2 = β9
ππ1ππ2
=57
ππ1ππ2
=β46
=β23
Since, ππ1ππ2β ππ1
ππ2, the given pair of equations intersect at exactly one point.
(ii) 9π₯π₯ + 3π¦π¦ + 12 = 0
18π₯π₯ + 6π¦π¦ + 24 = 0
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 9, ππ1 = 3, ππ1 = 12
ππ2 = 18, ππ2 = 6, ππ2 = 24
ππ1ππ2
=9
18=
12
ππ1ππ2
=36
=12
ππ1ππ2
= 1224
= 12Since,ππ1
ππ2= ππ1
ππ2= ππ1
ππ2 the given pair of equation are coincident.
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(iii) 6x β 3y + 10 = 0
2x β y + 9 = 0
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 6, ππ1 = β3, ππ1 = 10
ππ2 = 2, ππ2 = β1, ππ2 = 9
ππ1ππ2
=62
=31
ππ1ππ2
=β3β1
=31
ππ1ππ2
=109
Since, ππ1ππ2
= ππ1ππ2β ππ1
ππ2 the given pair of equation are parallel to each other.
3. On comparing the ratios ππ1ππ2
, ππ1ππ2
and ππ1ππ2
, find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3π₯π₯ + 2π¦π¦ = 5 2π₯π₯ β 3π¦π¦ = 7
(ii) 2π₯π₯ β 3π¦π¦ = 8 4π₯π₯ β 6π¦π¦ = 9
(iii) 32π₯π₯ +
53π¦π¦ = 7 9π₯π₯ β 10π¦π¦ = 14
(iv) 5π₯π₯ β 3π¦π¦ = 11 β10π₯π₯ + 6π¦π¦ = β22
(v) 43π₯π₯ + 2π¦π¦ = 8 2π₯π₯ + 3π¦π¦ = 12
Solution:
(i) 3π₯π₯ + 2π¦π¦ = 5
2π₯π₯ β 3π¦π¦ = 7
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 3, ππ1 = 2, ππ1 = β5
ππ2 = 2, ππ2 = β3, ππ2 = β7
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ππ1ππ2
=32
,ππ1ππ2
=2β3
= β23
,ππ1ππ2
=β5β7
=57
Since, ππ1ππ2β ππ1
ππ2 the given pair of equation has only one solution.
Hence, the pair of linear equations is consistent.
(ii) 2π₯π₯ β 3π¦π¦ = 8
4π₯π₯ β 6π¦π¦ = 9
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 2, ππ1 = β3, ππ1 = β8
ππ2 = 4, ππ2 = β6, ππ2 = β9
ππ1ππ2
=24
=12
,ππ1ππ2
=β3β6
=12
,ππ1ππ2
=β8β9
=89
Since, ππ1ππ2
= ππ1ππ2β ππ1
ππ2 the given pair of equation has no solution.
Hence, the pair of linear equations is inconsistent.
(iii) 32π₯π₯ + 5
3π¦π¦ = 7
9π₯π₯ β 10π¦π¦ = 14
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 =32
, ππ1 =53
, ππ1 = β7
ππ2 = 9, ππ2 = β10, ππ2 = β14
ππ1ππ2
=329
=16
,ππ1ππ2
=53
β10= β
16
,ππ1ππ2
=β7β14
=12
Since, ππ1ππ2β ππ1
ππ2 the given pair of equation has only one solution
Hence, the pair of linear equations is consistent.
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(iv) 5π₯π₯ β 3π¦π¦ = 11
β10π₯π₯ + 6π¦π¦ = β22
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 5, ππ1 = β3, ππ1 = β11
ππ2 = β10, ππ2 = 6, ππ2 = 22
ππ1ππ2
=5
β10= β
12
,ππ1ππ2
=β36
= β12
,ππ1ππ2
=β1122
= β12
Since, ππ1ππ2
= ππ1ππ2
= ππ1ππ2
the given pair of equation has infinite number of solution.
Hence, the pair of linear equations is consistent.
(iv) 43π₯π₯ + 2π¦π¦ = 8
2π₯π₯ + 3π¦π¦ = 12
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 43
, ππ1 = 2, ππ1 = β8
ππ2 = 2, ππ2 = 3, ππ2 = β12
ππ1ππ2
=432
=23
,ππ1ππ2
=23
,ππ1ππ2
=β8β12
=23
Since, ππ1ππ2
= ππ1ππ2
= ππ1ππ2
the given pair of equation has infinite number of Solution.
Hence, the pair of linear equations is consistent.
4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) π₯π₯ + π¦π¦ = 5 2π₯π₯ + 2π¦π¦ = 10
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(ii) π₯π₯ β π¦π¦ = 8 3π₯π₯ β 3π¦π¦ = 16
(iii) 2π₯π₯ + π¦π¦ β 6 = 0 4π₯π₯ β 2π¦π¦ β 4 = 0
(iv) 2π₯π₯ β 2π¦π¦ β 2 = 0 4π₯π₯ β 4π¦π¦ β 5 = 0
Solution:
(i) π₯π₯ + π¦π¦ = 5
2π₯π₯ + 2π¦π¦ = 10
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 1, ππ1 = 1, ππ1 = β5
ππ2 = 2, ππ2 = 2, ππ2 = β10
ππ1ππ2
=12
,ππ1ππ2
=12
,ππ1ππ2
=β5β10
=12
Since, ππ1ππ2
= ππ1ππ2
= ππ1ππ2
the given pair of linear equation has infinite number of Solution.
Hence, the pair of linear equations is consistent.
Now, x + y = 5 βΉ x = 5 β y
Two solutions of this equation are:
π₯π₯ 5 0
π¦π¦ 0 5
2π₯π₯ + 2π¦π¦ = 10 β π₯π₯ =10 β 2π¦π¦
2
Two solutions of this equation are:
π₯π₯ 4 3
π¦π¦ 1 2
Thus, the graphical representation is as follows:
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In the graph, it is observed that the two lines coincide. Hence, the given pair of equations has infinite solutions.
Let x = t, then y = 5 β t. So, the ordered pair (t, 5 β t), where t is a constant, is the solution of the given pair of linear equations.
(ii) π₯π₯ β π¦π¦ = 8
3π₯π₯ β 3π¦π¦ = 16
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 1, ππ1 = β1, ππ1 = β8
ππ2 = 3, ππ2 = β3, ππ2 = β16
ππ1ππ2
=13
,ππ1ππ2
=β1β3
=13
,ππ1ππ2
=β8β16
=12
Since, ππ1ππ2
= ππ1ππ2β ππ1
ππ2 the given pair of linear equation has no solution.
Hence, the pair of linear equations is inconsistent.
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(iii) 2π₯π₯ + π¦π¦ β 6 = 0
4π₯π₯ β 2π¦π¦ β 4 = 0
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 2, ππ1 = 1, ππ1 = β6
ππ2 = 4, ππ2 = β2, ππ2 = β4
ππ1ππ2
=24
=12
,ππ1ππ2
=1β2
= β12
,ππ1ππ2
=β6β4
=32
Since ππ1ππ2β ππ1
ππ2, the given pair of linear equation has only one solution.
Hence, the pair of linear equations is consistent.
Now, 2π₯π₯ + π¦π¦ β 6 = 0 β π¦π¦ = 6 β 2π₯π₯
Two solutions of this equation are:
π₯π₯ 0 3
π¦π¦ 6 0
4π₯π₯ β 2π¦π¦ β 4 = 0 β π¦π¦ =4π₯π₯ β 4
2
Two solutions of this equation are:
π₯π₯ 1 0
π¦π¦ 0 β2
Hence, the graphical representation is as follows:
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In the graph, it is observed that the two lines intersect each other at the point (2, 2). Hence, the solution of the given pair of equations is (2, 2).
(iv) 2π₯π₯ β 2π¦π¦ β 2 = 0
4π₯π₯ β 4π¦π¦ β 5 = 0
Comparing pair of equations with ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ +ππ2 = 0, we get:
ππ1 = 2, ππ1 = β2, ππ1 = β2
ππ2 = 4, ππ2 = β4, ππ2 = β5
ππ1ππ2
=24
=12
,ππ1ππ2
=β2β4
=12
,ππ1ππ2
=β2β5
=25
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Since ππ1ππ2
= ππ1ππ2β ππ1
ππ2, the given pair of linear equation has no solution
Hence, the pair of linear equations is inconsistent.
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let the length of the rectangular garden be π₯π₯ and width be π¦π¦.
As per the given conditions,
π₯π₯ β π¦π¦ = 4
π₯π₯ + π¦π¦ = 36
π₯π₯ β π¦π¦ = 4 β π₯π₯ = π¦π¦ + 4
Two solutions of this equation are:
π₯π₯ 0 4
π¦π¦ β4 0
π₯π₯ + π¦π¦ = 36
Two solutions of this equation are:
π₯π₯ 0 36
π¦π¦ 36 0
Hence, the graphical representation is as follows:
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From the graph, it can be observed that the two lines intersect each other at the point (20, 16). So, π₯π₯ = 20 and π¦π¦ = 16.
Thus, the length and width of the rectangular garden is 20 m and 16 m respectively.
6. Given the linear equation 2π₯π₯ + 3π¦π¦ β 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
(i) For the two lines ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and a2x + b2y + c2 = 0, to be intersecting, we must have ππ1
ππ2β ππ1
ππ2
Hence, the other linear equation can be 3x + 7y β 15 = 0
ππ1ππ2
=23
,ππ1ππ2
=37
,ππ1ππ2
=β8β15
=8
15
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(ii) For the two lines ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ + ππ2 = 0, to be parallel, we must have ππ1
ππ2= ππ1
ππ2β ππ1
ππ2
Hence, the other linear equation can be 4π₯π₯ + 6π¦π¦ + 9 = 0, as
ππ1ππ2
=24
=12
,ππ1ππ2
=36
=12
,ππ1ππ2
=β89
(iii) For the two lines ππ1π₯π₯ + ππ1π¦π¦ + ππ1 = 0 and ππ2π₯π₯ + ππ2π¦π¦ + ππ2 = 0, to be parallel, we must have ππ1
ππ2= ππ1
ππ2= ππ1
ππ2
So, the other linear equation can be 6π₯π₯ + 9π¦π¦ β 24 = 0,
as ππ1ππ2
= 26
= 13
, ππ1ππ2
= 39
= 13
, ππ1ππ2
= β8β24
= 13
7. Draw the graphs of the equations π₯π₯ β π¦π¦ + 1 = 0 and 3π₯π₯ + 2π¦π¦ β 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis and shade the triangular region.
Solution:
π₯π₯ β π¦π¦ + 1 = 0 β π₯π₯ = π¦π¦ β 1
Two solutions of this equation are:
π₯π₯ 0 β1
π¦π¦ 1 0
3x + 2y β 12 = 0 β x =12 β 2y
3
Two solutions of this equation are:
π₯π₯ 4 0
π¦π¦ 0 6
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Now, we have drawn these two equations on graph. Shaded part represents the triangle formed by given two lines and the x βaxis:
In the graph, we can observe that the coordinates of the vertices of the triangle are (2, 3), (β1, 0), and (4, 0).
β¦ β¦ β¦
EXERCISE 3.3
1. Solve the following pair of linear equations by the substitution method.
(i) π₯π₯ + π¦π¦ = 14 π₯π₯ β π¦π¦ = 4
(ii) π π β π‘π‘ = 3 π π 3
+π‘π‘2
= 6
(iii) 3π₯π₯ β π¦π¦ = 3 9π₯π₯ β 3π¦π¦ = 9
(iv) 0.2π₯π₯ + 0.3π¦π¦ = 1.3 0.4π₯π₯ + 0.5π¦π¦ = 2.3
(v) β2 π₯π₯ + β3 π¦π¦ = 0 β3 π₯π₯ β β8 π¦π¦ = 0
(vi) 3π₯π₯2β
5π¦π¦3
= β2 π₯π₯3
+π¦π¦2
=136
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Solution;
(i) π₯π₯ + π¦π¦ = 14 β¦ (i)
π₯π₯ β π¦π¦ = 4 β¦ (ii)
From (i), we get:
π₯π₯ = 14 β π¦π¦β¦ (iii)
Substituting the value of π₯π₯ in equation (ii), we get:
(14 β π¦π¦) β π¦π¦ = 4
β 14 β 2π¦π¦ = 4
β 10 = 2π¦π¦
β π¦π¦ = 5
Substituting the value of π¦π¦ in equation (iii), we get:
π₯π₯ = 9 β΄ π₯π₯ = 9,π¦π¦ = 5
(ii) π π β π‘π‘ = 3 β¦ (i) π π 3
+π‘π‘2
= 6 β¦ (ii)
From (i), we get:
(iii) π π = π‘π‘ + 3 β¦ (iii)
Substituting the value of π π in equation (iii), we get:π‘π‘+33
+ π‘π‘2
= 6
β 2π‘π‘ + 6 + 3π‘π‘ = 36
β 5π‘π‘ = 30
β π‘π‘ = 6
Substituting the value of t in equation (iii), we get:
π π = 9
β΄ π π = 9, π‘π‘ = 6
(iv) 3π₯π₯ β π¦π¦ = 3 β¦ (i)
9π₯π₯ β 3π¦π¦ = 9 β¦ (ii)
From (I), we get
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π¦π¦ = 3π₯π₯ β 3 β¦ (iii)
Substituting the value of π¦π¦ in equation (ii), we get:
(v) 9π₯π₯ β 3(3π₯π₯ β 3) = 9
β 9π₯π₯ β 9π₯π₯ = 9
β9 = 9
This is always true.
Hence, the given pair of equations has infinitely many solutions and variables are related as
π¦π¦ = 3π₯π₯ β 3
(vi) 0.2π₯π₯ + 0.3π¦π¦ = 1.3 β¦ (i)
0.4π₯π₯ + 0.5π¦π¦ = 2.3 β¦ (ii)
From (i), we get:
π₯π₯ =1.3 β 0.3π¦π¦
0.2 β¦ (iii)
Substituting the value of π₯π₯ in equation (ii), we get:
0.4 οΏ½1.3 β 0.3π¦π¦
0.2οΏ½ + 0.5π¦π¦ = 2.3
β 2.6 β 0.6π¦π¦ + 0.5π¦π¦ = 2.3
β 2.6 β 2.3 = 0.1π¦π¦
β 0.3 = 0.1π¦π¦
β π¦π¦ = 3
Substituting the value of π¦π¦ in equation (iii), we get:
π₯π₯ =1.3 β 0.3 Γ 3
0.2=
0.40.2
= 2
β΄ π₯π₯ = 2,π¦π¦ = 3
(v) β2π₯π₯ + β3π¦π¦ = 0. . . (i)
β3π₯π₯ β β8π¦π¦ = 0. . . (ii)
From equation (i), we get:
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π₯π₯ =ββ3π¦π¦β2
. . . (iii)
Substituting the value of π₯π₯ in equation (ii), we get:
β3οΏ½ββ3π¦π¦β2
οΏ½ β β8π¦π¦ = 0
β β3π¦π¦β2
β 2β2π¦π¦ = 0
β π¦π¦ οΏ½β3β2
β 2β2οΏ½ = 0
β π¦π¦ = 0
Substituting the value of π¦π¦ in equation (iii),we get:
π₯π₯ = 0
β΄ π₯π₯ = 0,π¦π¦ = 0
(vi) 32π₯π₯ β 5
3π¦π¦ = β2. . . (i)
π₯π₯3
+π¦π¦2
=136
. . . (ii)
From equation (i), we get:
9π₯π₯ β 10π¦π¦ = β12
β π₯π₯ =β12 + 10π¦π¦
9 β¦ (iii)
Substituting the value of π₯π₯ in equation (ii), we get: β12+10π¦π¦
93
+π¦π¦2
=136
ββ24 + 20π¦π¦ + 27y
54=
136
β β24 + 47π¦π¦ = 13 Γ 9
β 47π¦π¦ = 117 + 24
β 47π¦π¦ = 141
β π¦π¦ = 3
Substituting the value of y in equation (iii), we get:
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π₯π₯ =β12 + 10 Γ 3
9= 2
β΄ π₯π₯ = 2,π¦π¦ = 3
2. Solve 2π₯π₯ + 3π¦π¦ = 11 and 2π₯π₯ β 4π¦π¦ = β24 and hence find the value of βππβ for which π¦π¦ = πππ₯π₯ + 3.
Solution:
2π₯π₯ + 3π¦π¦ = 11 . . . (i)
2π₯π₯ β 4π¦π¦ = β24. . . (ii)
Form equation (i), we get:
π₯π₯ =11 β 3π¦π¦
2 β¦ (iii)
Substituting the value of π₯π₯ in equation (ii), we get:
2 οΏ½11 β 3π¦π¦
2οΏ½ β 4π¦π¦ = β24
β 11 β 3π¦π¦ β 4π¦π¦ = β24
β β7π¦π¦ = β35
β π¦π¦ = 5
Substituting the value of π¦π¦ in equation (iii), we get:
π₯π₯ =11 β 3 Γ 5
2= β2
β΄ π₯π₯ = β2,π¦π¦ = 5
Given, π¦π¦ = πππ₯π₯ + 3
β 5 = β2ππ + 3
β β2ππ = 2
β ππ = β1
Hence, ππ = β1.
3. Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
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(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(ii) The coach of a cricket team buys 7 bats and 6 balls for βΉ 3800. Later, she buys 3 bats and 5 balls for βΉ 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is βΉ 105 and for a journey of 15 km, the charge paid is βΉ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(vii) A fraction becomes 911
, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5
6. Find the fraction.
(viii) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacobβs age was seven times that of his son. What are their present ages?
Solution:
(i) Let two numbers are x and y such that y > x.
As per the question:
π¦π¦ = 3π₯π₯ . . . . (1)
π¦π¦ β π₯π₯ = 26 . . . . (2)
On substituting the value of π¦π¦ from equation (1) into equation (2), we get
26 + π₯π₯ = 3π₯π₯
β 3π₯π₯ β π₯π₯ = 26
β x = 13 . . . (3)
Substituting the value of π₯π₯ in equation (1), we get
y = 39
Hence, two numbers are 13 and 39.
(ii) Let larger angle be x and smaller angle be π¦π¦
As we know, the sum of supplementary angles is always 180Β°.
Hence, as per the question,
π₯π₯ + π¦π¦ = 180 β¦ (1)
π₯π₯ β π¦π¦ = 18 β¦ (2)
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From (1), we get
π₯π₯ = 180Β° β π¦π¦β¦ (3)
Substituting π₯π₯ in equation (2), we obtain
180o β π¦π¦ β π¦π¦ = 18Β°
β 162o = 2π¦π¦
β 81o = π¦π¦ β¦ (4)
Putting this in equation (3), we obtain
π₯π₯ = 180ππ β 81ππ = 99ππ
Hence, the angles are 99and 81Β°.
(iii) Let the cost of a bat be π₯π₯ and a ball be π¦π¦
As per the given information,
7π₯π₯ + 6π¦π¦ = 3800 β¦ (1)
3π₯π₯ + 5π¦π¦ = 1750 β¦ (2)
From (1), we get
π¦π¦ =3800 β 7π₯π₯
6 β¦ (3)
Substituting π₯π₯ in equation (2), we get
3π₯π₯ + 5 οΏ½3800 β 7π₯π₯
6οΏ½ = 1750
β 3π₯π₯ +9500
3β
35π₯π₯6
= 1750
β 3π₯π₯ β35π₯π₯
6= 1750 β
95003
β18π₯π₯ β 35π₯π₯
6=
5250 β 95003
β β17π₯π₯
6=β4250
3
β β17π₯π₯ = β8500
β π₯π₯ = 500 β¦ (4)
Substituting this in equation (3), we get
π¦π¦ =3800 β 7 Γ 500
6= 50
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Hence, the cost of a bat is βΉ 500 and that of a ball is βΉ 50.
(iv) Let the fixed charge be βΉ x and charge per Km be βΉ y.
π₯π₯ + 10π¦π¦ = 105 β¦ (1)
π₯π₯ + 15π¦π¦ = 155 β¦ (2)
From (1), we get
π₯π₯ = 105 β 10π¦π¦ β¦ (3)
Substituting this in equation (2), we get
105 β 10π¦π¦ + 15π¦π¦ = 155
β 5π¦π¦ = 50
β π¦π¦ = 10 β¦ (4)
Putting π¦π¦ in equation (3), we get
π₯π₯ = 105 β 10 Γ 10
π₯π₯ = 5
Hence, fixed charge = βΉ 5
And per Km charge = βΉ 10
Hence, charge for 25 km = π₯π₯ + 25π¦π¦
= 5 + 250 = βΉ 255
(v) Let the fraction be π₯π₯π¦π¦
As per the given information, π₯π₯+2π¦π¦+2
= 911
β 11π₯π₯ + 22 = 9π¦π¦ + 18
β 11π₯π₯ β 9π¦π¦ = β4 β¦ (1)
π₯π₯ + 3π¦π¦ + 3
=56
β 6π₯π₯ + 18 = 5π¦π¦ + 15
β 6π₯π₯ β 5π¦π¦ = β3 β¦ (2)
From equation (1), we get π₯π₯ = β4+9π¦π¦11
β¦ (3)
Substituting π₯π₯ in equation (2), we get
6 οΏ½β4 + 9π¦π¦
11οΏ½ β 5π¦π¦ = β3
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β β24 + 54π¦π¦ β 55π¦π¦ = β33
β βπ¦π¦ = β9
β π¦π¦ = 9 β¦ (4)
Substituting π¦π¦ in equation (3), we obtain
π₯π₯ =β4 + 81
11= 7
Hence, the fraction is 79.
(vi) Let the age of Jacob be π₯π₯ and the age of his son be π¦π¦
As per the given information,
(π₯π₯ + 5) = 3(π¦π¦ + 5)
β π₯π₯ β 3π¦π¦ = 10 β¦ (1)
(π₯π₯ β 5) = 7(π¦π¦ β 5)
β π₯π₯ β 7π¦π¦ = β30 β¦ (2)
From (1), we get
π₯π₯ = 3π¦π¦ + 10 β¦ (3)
Substituting π₯π₯ in equation (2), we get
3π¦π¦ + 10 β 7π¦π¦ = β30
β β4π¦π¦ = β40
β π¦π¦ = 10 β¦ (4)
Substituting π¦π¦ in equation (3), we get
π₯π₯ = 3 Γ 10 + 10 = 40
Hence, the present age of Jacob is 40 years and the present age of his son is 10 years.
β¦ β¦ β¦
EXERCISE 3.4
1. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) π₯π₯ + π¦π¦ = 5 and 2π₯π₯ β 3π¦π¦ = 4
(ii) 3π₯π₯ + 4π¦π¦ = 10 and 2π₯π₯ β 2π¦π¦ = 2
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(iii) 3π₯π₯ β 5π¦π¦ β 4 = 0 and 9π₯π₯ = 2π¦π¦ + 7
(iv) π₯π₯2
+ 2π¦π¦3
= β1 and π₯π₯ β π¦π¦3
= 3
Solution:
(i) Elimination method:
π₯π₯ + π¦π¦ = 5 β¦ (1)
2π₯π₯ β 3π¦π¦ = 4 β¦ (2)
Multiplying equation (1) by 3, we get:
3π₯π₯ + 3π¦π¦ = 15 β¦ (3)
Adding equation (2) and (3), we get:
5π₯π₯ = 19
β π₯π₯ =195
Substituting π₯π₯ in equation (1), we get:
π¦π¦ = 5 β195
=65
β΄ π₯π₯ =195
,π¦π¦ =65
Substitution method:
From equation (1), we get:
π₯π₯ = 5 β π¦π¦ β¦ (4)
Putting π₯π₯ in equation (2), we get:
2(5 β π¦π¦) β 3π¦π¦ = 4
β5π¦π¦ = β6
π¦π¦ =65
Putting π¦π¦ in equation (4), we get:
π₯π₯ = 5 β65
=195
Hence, π₯π₯ = 65
,π¦π¦ = 195
(ii) Elimination method:
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3π₯π₯ + 4π¦π¦ = 10 β¦ (1)
2π₯π₯ β 2π¦π¦ = 2 β¦ (2)
Multiplying equation (2) by 2, we get:
4π₯π₯ β 4π¦π¦ = 4 . . . (3)
Adding equation (1) and (3), we get:
7π₯π₯ = 14
β π₯π₯ = 2
Putting x in equation (1), we get:
6 + 4π¦π¦ = 10
β 4π¦π¦ = 4
β π¦π¦ = 1
Hence, π₯π₯ = 2,π¦π¦ = 1
Substitution method:
From equation (2), we get:
π₯π₯ = π¦π¦ + 1 β¦ (4)
Putting π₯π₯ in equation (1), we get:
3(π¦π¦ + 1) + 4π¦π¦ = 10
β 7π¦π¦ = 7
β π¦π¦ = 1
Putting y in equation (4), we get:
π₯π₯ = 1 + 1 = 2
β΄ π₯π₯ = 2,π¦π¦ = 1
(iii) Elimination method:
3π₯π₯ β 5π¦π¦ β 4 = 0 β¦ (1)
9π₯π₯ = 2π¦π¦ + 7
β 9π₯π₯ β 2π¦π¦ β 7 = 0 β¦ (2)
Multiplying equation (1) by 3, we get:
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9π₯π₯ β 15π¦π¦ β 12 = 0 β¦ (3)
Subtracting equation (2) from equation (3), we get:
13π¦π¦ = β5
β π¦π¦ = β5
13
Putting y in equation (1), we get:
3π₯π₯ +2512
β 4 = 0
β π₯π₯ =9
13
β΄ π₯π₯ =9
13,π¦π¦ = β
513
Substitution method:
From equation (1), we get:
π₯π₯ =5π¦π¦ + 4
3 β¦ (4)
Putting π₯π₯ in equation (2), we get:
9 οΏ½5π¦π¦ + 4
3οΏ½ β 2π¦π¦ β 7 = 0
13π¦π¦ = β5
π¦π¦ = β5
13
Putting y in equation (4), we get:
π₯π₯ =5 οΏ½β 5
13οΏ½ + 4
3
π₯π₯ =9
13
β΄ π₯π₯ =9
13,π¦π¦ =
β513
(iv) Elimination method:
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π₯π₯2
+2π¦π¦3
= β1 β¦ (1)
π₯π₯ βπ¦π¦3
= 3 β¦ (2)
Multiplying equation (2) by 2, we get:
2π₯π₯ β2π¦π¦3
= 6 β¦ (3)
Adding equation (1) and (3), we get:
5π₯π₯2
= 5
β π₯π₯ = 2
Putting π₯π₯ in equation (1), we get:
1 +2π¦π¦3
= β1
β π¦π¦ = β3
β΄ π₯π₯ = 2,π¦π¦ = β3
Substitution method:
From equation (2), we get:
π¦π¦ = 3π₯π₯ β 9 β¦ (3)
Putting π¦π¦ in equation (1), we get:
π₯π₯2
+2(3π₯π₯ β 9)
3= β1
β 3π₯π₯ + 4(3π₯π₯ β 9) = β6
β 15π₯π₯ = 30
β π₯π₯ = 2
Putting x in equation (3), we get:
π¦π¦ = 6 β 9 = β3
β΄ π₯π₯ = 2,π¦π¦ = β3
Note: Solution must be same in both the cases.
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2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1
2 if we only add 1 to the denominator.
What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iv) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw βΉ 2000. She asked the cashier to give her βΉ 50 and βΉ 100 notes only. Meena got 25 notes in all. Find how many notes of βΉ 50 and βΉ 100 she received.
(iv) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid βΉ 27 for a book kept for seven days, while Susy paid βΉ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
(i) Let the fraction be π₯π₯π¦π¦
π₯π₯ + 1π¦π¦ β 1
= 1
β π₯π₯ β π¦π¦ = β2 β¦ (1)
π₯π₯π¦π¦ + 1
=12
β 2π₯π₯ β π¦π¦ = 1 β¦ (2)
As per the question, subtracting equation (1) from equation (2), we get:π₯π₯ = 3
Putting x in equation (2), we get:
6 β π¦π¦ = 1
β βπ¦π¦ = β5
β π¦π¦ = 5
Hence, the fraction is 35.
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(ii) Let present age of Nuri be x and Sonu be y.
As per the question,
(π₯π₯ β 5) = 3(π¦π¦ β 5)
β π₯π₯ β 3π¦π¦ = β10. . . (1)
(π₯π₯ + 10) = 2(π¦π¦ + 10)
β π₯π₯ β 2π¦π¦ = 10 β¦ (2)
Subtracting equation (1) from equation (2), we get:
y = 20
Putting the value of y in equation (2), we get:
π₯π₯ β 40 = 10
β π₯π₯ = 50
Hence, the age of Nuri is 50 years and the age of Sonu is 20 years.
(iii) Let the unit digit of the number be π₯π₯ and tens digit of the number be π¦π¦ respectively.
Hence, number = 10π¦π¦ + π₯π₯
After reversing the digits, number = 10π₯π₯ + π¦π¦
As per the question,
π₯π₯ + π¦π¦ = 9 β¦ (1)
9(10π¦π¦ + π₯π₯) = 2(10π₯π₯ + π¦π¦)
β 88π¦π¦ β 11π₯π₯ = 0
β β π₯π₯ + 8π¦π¦ = 0 β¦ (2)
Adding equations (1) and (2), we get:
9π¦π¦ = 9
β π¦π¦ = 1
Putting π¦π¦ in equation (1), we get:
π₯π₯ = 8
Hence, the number is 10π¦π¦ + π₯π₯ = 10 Γ 1 + 8 = 18
(v) Let the number of βΉ 50 notes be π₯π₯ and number of βΉ 100 notes be y.
(vi) As per the question,
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π₯π₯ + π¦π¦ = 25 β¦ (1)
50π₯π₯ + 100π¦π¦ = 2000
β π₯π₯ + 2π¦π¦ = 40 β¦ (2)
Subtracting equation (1) from equation (2), we get:
π¦π¦ = 15
Putting y in equation (1), we get:
π₯π₯ = 10
Hence, Meena received 10 notes of βΉ 50 and 15 notes of βΉ 100.
(vii) Let the fixed charge for first three days be βΉ π₯π₯ and each day charge thereafter be βΉ π¦π¦.
As per the question,
π₯π₯ + 4π¦π¦ = 27 . . . (1)
π₯π₯ + 2π¦π¦ = 21 . . . (2)
Subtracting equation (2) from equation (1), we get:
2π¦π¦ = 6
β π¦π¦ = 3
Putting y in equation (2), we get:
π₯π₯ + 6 = 21
β π₯π₯ = 15
Hence, the fixed charge is βΉ 15 and each day charge thereafter is βΉ 3.
β¦ β¦ β¦
EXERCISE 3.5
1. Which of the following pair of linear equation has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) π₯π₯ β 3π¦π¦ β 3 = 0 3π₯π₯ β 9π¦π¦ β 2 = 0
(ii) 2π₯π₯ + π¦π¦ = 5 3π₯π₯ + 3π¦π¦ = 8
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(iii) 3π₯π₯ β 5π¦π¦ = 20 6π₯π₯ β 10π¦π¦ = 40
(iv) π₯π₯ β 3π¦π¦ β 7 = 0 3π₯π₯ β 3π¦π¦ β 15 = 0
Solution:
(i) π₯π₯ β 3π¦π¦ β 3 = 0
3π₯π₯ β 9π¦π¦ β 2 = 0
Here, ππ1 = 1, ππ1 = β3, ππ1 = β3
ππ2 = 3, ππ2 = β9, ππ2 = β2
ππ1ππ2
=13
,ππ1ππ2
=β3β9
=13
,ππ1ππ2
=β3β2
=32
Since, ππ1ππ2
= ππ1ππ2β ππ1
ππ2
Hence, the given pair of equations has no solution.
(ii) 2π₯π₯ + π¦π¦ = 5
β 2π₯π₯ + π¦π¦ β 5 = 0 β¦ (1)
3π₯π₯ + 2π¦π¦ = 8
β 3π₯π₯ + 2π¦π¦ β 8 = 0 β¦ (2)
Here, ππ1 = 2, ππ1 = 1, ππ1 = β5
ππ2 = 3, ππ2 = 2, ππ2 = β8
ππ1ππ2
=23
,ππ1ππ2
=12
,ππ1ππ2
=β5β8
=58
Since, ππ1ππ2β ππ1
ππ2
Hence, the given pair of equations has unique solution.
Now, using cross-multiplication method we know that,
π₯π₯ππ1ππ2 β ππ2ππ1
=π¦π¦
ππ1ππ2 β ππ2ππ1=
1ππ1ππ2 β ππ2ππ1
βπ₯π₯
β8 β (β10) =π¦π¦
β15 β (β16) =1
4 β 3
βπ₯π₯2
=π¦π¦1
=11
βπ₯π₯2
= 1,π¦π¦1
= 1
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β΄ π₯π₯ = 2,π¦π¦ =1
(iii) 3π₯π₯ β 5π¦π¦ = 20
β 3π₯π₯ β 5π¦π¦ β 20 = 0 β¦ (1)
6π₯π₯ β 10π¦π¦ = 40
β 6π₯π₯ β 10π¦π¦ β 40 = 0 β¦ (2)
Here, ππ1 = 3, ππ1 = β5, ππ1 = β20
ππ2 = 6, ππ2 = β10, ππ2 = β40
ππ1ππ2
=36
=12
,ππ1ππ2
=β5β10
=12
,ππ1ππ2
=β20β40
=12
Since, ππ1ππ2
= ππ1ππ2
= ππ1ππ2
Hence, the given pair of equations has infinite solutions.
The solution will be found by assuming the value of π₯π₯ to be ππ. Hence the ordered pair οΏ½ππ, 3ππβ20
5οΏ½, where ππ is a constant, are the solutions of the
given pair of equations.
(vi) π₯π₯ β 3π¦π¦ β 7 = 0
3π₯π₯ β 3π¦π¦ β 15 = 0
Here, ππ1 = 1, ππ1 = β3, ππ1 = β7
ππ2 = 3, ππ2 = β3, ππ2 = β15
ππ1ππ2
=13
,ππ1ππ2
=β3β3
= 1,ππ1ππ2
=β7β15
=7
15,
Since, ππ1ππ2β ππ1
ππ2
Hence, the given pair of equations has unique solution.
Now, using cross-multiplication method we know that,
π₯π₯ππ1ππ2 β ππ2ππ1
=π¦π¦
ππ1ππ2 β ππ2ππ1=
1ππ1ππ2 β ππ2ππ1
βπ₯π₯
45 β (21) =π¦π¦
β21 β (β15) =1
β3 β (β9)
βπ₯π₯
24=
π¦π¦β6
=16
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β π₯π₯24
= 16 and π¦π¦
β6= 1
6
β΄ π₯π₯ = 4,π¦π¦ = β1
2. (i) For which values of ππ and ππ does the following pair of linear equations have an infinite number of solutions?
2π₯π₯ + 3π¦π¦ = 7
(ππ β ππ)π₯π₯ + (ππ + ππ)π¦π¦ = 3ππ + ππ β 2
(ii) For which value of ππ will the following pair of linear equations have no solution?
3π₯π₯ + π¦π¦ = 1
(2ππ β 1)π₯π₯ + (ππ β 1)π¦π¦ = 2ππ + 1
Solution:
(i) 2π₯π₯ + 3π¦π¦ = 7
β 2π₯π₯ + 3π¦π¦ β 7 = 0 β¦ (1)
(ππ β ππ)π₯π₯ + (ππ + ππ)π¦π¦ = 3ππ + ππ β 2
β (ππ β ππ)π₯π₯ + (ππ + ππ)π¦π¦ β (3ππ + ππ β 2) = 0 β¦ (2)
Here, ππ1 = 2, ππ1 = 3, ππ1 = β7
ππ2 = (ππ β ππ), ππ2 = (ππ + ππ), ππ2 = β(3ππ + ππ β 2)
ππ1ππ2
=2
ππ β ππ,β―ππ1ππ2
=3
ππ + ππ,ππ1ππ2
=β7
β(3ππ + ππ β 2)=
7(3ππ + ππ β 2)
As we know that, for the equations to have infinitely many solutions:
ππ1ππ2
=ππ1ππ2
=ππ1ππ2
Hence, 2ππβππ
= 73ππ+ππβ2
β 6ππ + 2ππ β 4 = 7ππ β 7ππ
β ππ β 9ππ + 4 = 0 β¦ (3)
And 2ππβππ
= 3ππ+ππ
β 2ππ + 2ππ = 3ππ β 3ππ
β ππ β 5ππ = 0 β¦ (4)
Now, using cross-multiplication method in equation (3) and (4),
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ππ0 β (β20)
=ππ
4 β 0=
1β5 β (β9)
βππ
20=ππ4
=14
β ππ = 5, ππ = 1
Hence, the values of ππ and ππ are 5 and 1 respectively.
(ii) 3π₯π₯ + π¦π¦ = 1
β 3π₯π₯ + π¦π¦ β 1 = 0 β¦ (1)
(2ππ β 1)π₯π₯ + (ππ β 1)π¦π¦ = (2ππ + 1)
β (2ππ β 1)π₯π₯ + (ππ β 1)π¦π¦ β (2ππ + 1) = 0 β¦ (2)
Here, ππ1 = 3, ππ1 = 1, ππ1 = β1
ππ2 = (2ππ β 1), ππ2 = (ππ β 1), ππ2 = β(2ππ + 1)
ππ1ππ2
=3
2ππ β 1,ππ1ππ2
=1
ππ β 1,ππ1ππ2
=β1
β2ππ β 1=
12ππ + 1
,
As we know that, for the equations to have no solutions:
ππ1ππ2
=ππ1ππ2β ππ1ππ2
β3
2ππ β 1=
1ππ β 1
β 1
2ππ + 1
β3
2ππ β 1=
1ππ β 1
β 3ππ β 3 = 2ππ β 1
β ππ = 2
Again for ππ = 2, 32ππβ1
= 1ππβ1
= 1 but 12ππ+1
= 15
Since, 1 β 15
Hence, the values of ππ is 2 for no solution.
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3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8π₯π₯ + 5π¦π¦ = 9
3π₯π₯ + 2π¦π¦ = 4
Solution:
Substitution method:
8π₯π₯ + 5π¦π¦ = 9 β¦ (1)
3π₯π₯ + 2π¦π¦ = 4 β¦ (2)
From equation (2), we get:
π¦π¦ =4 β 3π₯π₯
2 β¦(3)
Putting y in equation (1), we get:
8π₯π₯ + 5 οΏ½4 β 3π₯π₯
2οΏ½ = 9
β 16π₯π₯ + 20 β 15π₯π₯ = 18
So, π₯π₯ = β2
Putting π₯π₯ = β2 in (3), we get
π¦π¦ = οΏ½4 β 3 Γ β2
2οΏ½ = 5
β΄ π₯π₯ = β2,π¦π¦ = 5
Cross-multiplication method:
8π₯π₯ + 5π¦π¦ β 9 = 0
3π₯π₯ + 2π¦π¦ β 4 = 0
Here, ππ1 = 8, ππ1 = 5, ππ1 = β9
ππ2 = 3, ππ2 = 2, ππ2 = β4
π₯π₯β20 β (β18) =
π¦π¦β27 β (β32) =
116 β 15
βπ₯π₯β2
=π¦π¦5
=11
β π₯π₯β2
= 1 and π¦π¦5
= 1
β π₯π₯ = β2 and π¦π¦ = 5
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4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay βΉ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays βΉ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
(ii) A fraction becomes 13 when 1 is subtracted from the numerator and it
becomes 14 when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(iv) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
(i) Let the fixed charge of the food be x and the charge for food per day be y.
As per the question,
x + 20y = 1000 . . . (1)
x + 26y = 1180 . . . (2)
Subtracting equation (1) from equation (2), we get:
6y = 180
β y = 30
Putting y in equation (2), we get:
x + 26 Γ 30 = 1180
β x = 1180 β 780
β x = 400
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Hence, the fixed charge of the food and the charge per day are βΉ 400 and βΉ 30 respectively.
(ii) Let the fraction be xy
As per the question, xβ1y
= 13
β 3x β y = 3 . . . (1) x
y+8= 1
4
β 4x β y = 8 . . . (2)
Subtracting equation (1) from equation (2), we get:
x = 5
Putting x in equation (2), we get:
20 β y = 8
β y = 12
Hence, the fraction is 512
(iii) Let the number of right answers be x and number of wrong answers be y.
As per the question,
3x β y = 40 β¦ (1)
4x β 2y = 50
β 2π₯π₯ β π¦π¦ = 25 . . . . (2)
Subtracting equation (2) from equation (1), we get:
x = 15
Putting x in equation (2), we get:
30 β y = 25
y = 5
Hence, the number of right answers and the number of wrong answers is 15 and 5 respectively.
Hence, the total number of questions is 20.
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(iv) Let the speed of first car be u km/h and speed of second car be v km/h respectively.
Speed of both cars while they are travelling in same direction = (u βv) km/h
Speed of both cars while they are travelling in opposite directions i.e., when they are travelling towards each other = (u + v)km/h
Distance travelled = speed Γ Time
As per the question,
5(u β v) = 100
β u β v = 20 . . . (1)
1(u + v) = 100
β u + v = 100 . . . (2)
Adding equations (1) and (2), we get:
2u = 120
β u = 60
Putting u in equation (2), we get:
v = 40
Hence, speed of the first car is 60 km/h and speed of the second car is 40 km/h.
(v) Let length of rectangle be x units and breadth of rectangle be y units.
Hence, area = xy
As per the question,
(x β 5)(y + 3) = xy β 9
β 3π₯π₯ β 5π¦π¦ β 6 = 0 β¦ (1)
(x + 3)(y + 2) = xy + 67
β 2π₯π₯ + 3π¦π¦ β 61 = 0 β¦ (2)
Using cross-multiplication method,
π₯π₯ππ1ππ2 β ππ2ππ1
=π¦π¦
ππ1ππ2 β ππ2ππ1=
1ππ1ππ2 β ππ2ππ1
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we get:
π₯π₯305 β (β18)
=π¦π¦
β12 β (β183)=
19 β (β10)
βπ₯π₯
323=
π¦π¦171
=1
19
β π₯π₯ = 17,π¦π¦ = 9
Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.
EXERCISE 3.6
1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 12π₯π₯
+1
3π¦π¦= 2
13π₯π₯
+1
2π¦π¦=
136
(ii) 2βπ₯π₯
+3οΏ½π¦π¦
= 2 4βπ₯π₯
β9οΏ½π¦π¦
= β1
(iii) 4π₯π₯
+ 3π¦π¦ = 14 3π₯π₯β 4π¦π¦ = 23
(iv) 5π₯π₯ β 1
+1
π¦π¦ β 2= 2
6π₯π₯ β 1
β3
π¦π¦ β 2= 1
(v) 7π₯π₯ β 2π¦π¦π₯π₯π¦π¦
= 5 8π₯π₯ + 7π¦π¦π₯π₯π¦π¦
= 15
(vi) 6π₯π₯ + 3π¦π¦ = 6π₯π₯π¦π¦ 2π₯π₯ + 4π¦π¦ = 5π₯π₯π¦π¦
(vii) 10π₯π₯ + π¦π¦
+2
π₯π₯ β π¦π¦= 4
15π₯π₯ + π¦π¦
β5
π₯π₯ β π¦π¦= β2
(viii) 13π₯π₯ + π¦π¦
+1
3π₯π₯ β π¦π¦=
34
1
2(3π₯π₯ + π¦π¦) β1
2(3π₯π₯ β π¦π¦) =β18
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Solutions:
(i) 12π₯π₯
+ 13π¦π¦
= 2
13π₯π₯
+ 12π¦π¦
= 136
Let 1π₯π₯
= ππ ππππππ 1π¦π¦
= ππ
Hence both equations convert to: ππ2
+ ππ3
= 2
β 3p + 2q β 12 = 0. . . (1) ππ3
+ ππ2
= 136
β 2ππ + 3ππ β 13 = 0 . . . (2)
Using cross-multiplication method, we get: ππ
β26β(β36)= ππ
β24β(β39)= 1
9β4
β ππ10
= ππ15
= 15
β ππ = 2, ππ = 3
β΅ 1π₯π₯
= ππ = 2, 1π¦π¦
= ππ = 3
Hence, π₯π₯ = 12
,π¦π¦ = 13
(ii) 2βπ₯π₯
+ 3
βπ¦π¦= 2
4βπ₯π₯β 9
βπ¦π¦= β1
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let 1βπ₯π₯
= ππ and 1βπ¦π¦
= ππ
Hence both equations convert to:
2ππ + 3ππ = 2 . . . (1)
4ππ β 9ππ = β1 . . . (2)
Multiplying equation (1) by 2, we get:
4ππ + 6ππ = 4 . . . (3)
Subtracting equation (2) from equation (3), we get:
15ππ = 5
β ππ = 13
Putting the value of ππ in equation (1), we get:
2ππ + 3 Γ 13
= 2
β 2p = 1
β ππ = 12
Since, ππ = 1βπ₯π₯
= 12
β βπ₯π₯ = 2
β x = 4
ππ = 1
βπ¦π¦= 1
3
β οΏ½π¦π¦ = 3
β y = 9
β΄ π₯π₯ = 4,π¦π¦ = 9
(iii) 4π₯π₯
+ 3π¦π¦ = 14
3π₯π₯β 4π¦π¦ = 23
Let 1π₯π₯
= ππ
Hence both equations convert to:
4p + 3y = 14
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β 4ππ + 3π¦π¦ β 14 = 0 . . . (1)
3p β 4y = 23
β 3ππ β 4π¦π¦ β 23 = 0 . . . (2)
Using cross-multiplication method, we get: ππ
β69β56= π¦π¦
β42β(β92)= 1
β16β9
β ππβ125
= π¦π¦50
= β125
β ππβ125
= β125
, π¦π¦50
= β125
β p = 5, y = β2
Since, ππ = 1π₯π₯
= 5
β π₯π₯ = 15 and y = β2
(iv) 5π₯π₯β1
+ 1π¦π¦β2
= 2
6π₯π₯β1
β 3π¦π¦β2
= 1
Let 1π₯π₯β1
= ππ ππππππ 1π¦π¦β2
= ππ
Hence both equations convert to:
5p + q = 2 β¦ (1)
6p β 3q = 1 β¦ (2)
Multiplying equation (1) by 3, we get:
15p + 3q = 6 β¦ (3)
Adding (2) and (3), we get:
21p = 7
β ππ = 13
Putting ππ in equation (1), we get:
5 Γ 13
+ ππ = 2
β ππ = 2 β 53
= 13
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Since, ππ = 1π₯π₯β1
= 13
β π₯π₯ β 1 = 3
β π₯π₯ = 4
ππ = 1π¦π¦β2
= 13
β y β 2 = 3
β y = 5
β΄ π₯π₯ = 4,π¦π¦ = 5
(v) 7π₯π₯β2π¦π¦π₯π₯π¦π¦
= 5
β 7π¦π¦β 2
π₯π₯= 5 . . . (1)
8π₯π₯+7π¦π¦π₯π₯π¦π¦
= 15
β 8π¦π¦
+ 7π₯π₯
= 15 . . . (2)
Let 1π₯π₯
= ππ and 1π¦π¦
= ππ
Hence both equations convert to:
β2p + 7q = 5
β β2ππ + 7ππ β 5 = 0. . . (3)
7p + 8q = 15
β 7ππ + 8ππ β 15 = 0 . . . (4)
using cross-multiplication method, we get: ππ
β105β(β40)= ππ
β35β30= 1
β16β49
β ππβ65
= ππβ65
= 1β65
β ππβ65
= 1β65
, ππβ65
= 1β65
β p = 1, q = 1
Since, ππ = 1π₯π₯
= 1, ππ = 1π¦π¦
= 1
Hence, x = 1, y = 1
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(vi) 6x + 3y = 6xy
β 6π¦π¦
+ 3π₯π₯
= 6. . . (1)
2x + 4y = 5xy
β 2π¦π¦
+ 4π₯π₯
= 5 . . . (2)
Let 1π₯π₯
= ππ and 1π¦π¦
= ππ
Hence both equations convert to:
3p + 6q β 6 = 0
4p + 2q β 5 = 0
Using cross-multiplication method, we get: ππ
β30β(β12)= ππ
β24β(β15)= 1
6β24
β ππβ18
= ππβ9
= 1β18
β ππβ18
= 1β18
, ππβ9
= 1β18
β ππ = 1, ππ = 12
Since, ππ = 1π₯π₯
= 1, ππ = 1π¦π¦
= 12
Hence, x = 1, y = 2
(vii) 10π₯π₯+π¦π¦
+ 2π₯π₯βπ¦π¦
= 4
15π₯π₯+π¦π¦
β 5π₯π₯βπ¦π¦
= β2
Let 1π₯π₯+π¦π¦
= ππ and 1π₯π₯βπ¦π¦
= ππ
Hence both equations convert to:
10p + 2q = 4
β 10ππ + 2ππ β 4 = 0 . . . (1)
15p β 5q = β2
β 15ππ β 5ππ + 2 = 0. . . (2)
Using cross-multiplication method, we get:
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ππ4β20
= ππβ60β20
= 1β50β30
β ππβ16
= 1β80
ππππππ ππβ80
= 1β80
β ππ = 15
ππππππ ππ = 1
Since, ππ = 1π₯π₯+π¦π¦
= 15
ππππππ ππ = 1π₯π₯βπ¦π¦
= 1
x + y = 5. . . . (3)
x β y = 1. . . . . (4)
Adding both equations (3) and (4), we get:
2x = 6
β x = 3
Putting x in equation (3), we get:
y = 2
β΄ π₯π₯ = 3,π¦π¦ = 2
(viii) 13π₯π₯+π¦π¦
+ 13π₯π₯βπ¦π¦
= 34
12(3π₯π₯+π¦π¦)
β 12(3π₯π₯βπ¦π¦)
= β18
let 13π₯π₯+π¦π¦
= ππ and 13π₯π₯βπ¦π¦
= ππ
Hence both equations convert to:
ππ + ππ = 34
β¦ (1)
ππ2β ππ
2= β1
8
β ππ β ππ = β14
β¦ (2)
Adding (1) and (2), we get:
2ππ = 34β 1
4
β 2ππ = 12
β ππ = 14
Putting p in (1), we get:
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14
+ ππ = 34
β ππ = 34β 1
4= 1
2
Since, ππ = 13π₯π₯+π¦π¦
= 14
β 3x + y = 4 β¦ (3)
ππ = 13π₯π₯βπ¦π¦
= 12
β 3x β y = 2 β¦ (4)
Adding equations (3) and (4), we get:
6x = 6
β x = 1
Putting the value of x in (3), we get:
3(1) + y = 4
β y = 1
β΄ π₯π₯ = 1,π¦π¦ = 1
2. Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(v) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
(i) Let the speed of Ritu be x km/h in still water and the speed of stream be y km/h.
Hence, speed of Ritu while rowing upstream = (x β y) km/h
And speed of Ritu while rowing downstream = (x + y) km/h
According to the question,
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2(x + y) = 20
β x + y = 10 β¦ (1)
2(x β y) = 4
β x β y = 2 β¦ (2)
Adding equations (1) and (2), we get:
2x = 12
β x = 6
Putting the value of x in equation (2), we get:
y = 4
Thus, Rita's speed is 6km/h in still water and the speed of the current is 4 km/h.
(ii) Let the number of days taken by a woman to finish the work be π₯π₯ and a man to finish the work be π¦π¦.
Work done by a woman in 1 day = 1π₯π₯
Work done by a man in 1 day = 1π¦π¦
According to the question,
4 οΏ½2π₯π₯
+ 5π¦π¦οΏ½ = 1
β2π₯π₯
+5π¦π¦
=14
β¦ (1)
3 οΏ½3π₯π₯
+6π¦π¦οΏ½ = 1
β 3π₯π₯
+ 6π¦π¦
= 13
β¦ (2)
Let 1π₯π₯
= ππ ππππππ 1π¦π¦
= ππ
Hence given equations convert to:
2ππ + 5ππ = 14
β 8ππ + 20ππ β 1 = 0 β¦ (3)
3ππ + 6ππ = 13
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β 9ππ + 18ππ β 1 = 0 β¦ (4)
Using cross-multiplication method, we get: ππ
β20β(β18)= ππ
β9β(β8)= 1
144β180
β ππβ2
= ππβ1
= 1β36
β ππβ2
= 1β36
, ππβ1
= 1β36
β ππ = 118
, ππ = 136
Since, ππ = 1π₯π₯
= 118
, ππ = 1π¦π¦
= 136
Hence, π₯π₯ = 18,π¦π¦ = 36
Hence, the number of days taken by a woman and a man to finish the work is 18 and 36.
(iii) Let the speed of train be u km/h and speed of bus be v km/h.
Now as we know, Time taken = Distanceβ―β―travelledSpeed
Hence, As per the question, 60π’π’
+ 240π£π£
= 4. . . (1)
100π’π’
+ 200π£π£
= 256
. . . (2)(Since, 10 minutes= 16 hours)
Let 1π’π’
= ππ and 1π£π£
= ππ
Hence given equations convert to:
60p + 240q = 4 . . . (3)
100ππ + 200ππ = 256
β 600ππ + 1200ππ = 25. . . . (4)
Multiplying equation (3) by 10, we get:
600p + 2400q = 40. . . . . (5)
Subtracting equation (4) from equation (5), we get:
1200q = 15
β ππ = 151200
= 180
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Putting q in equation (3), we get:
60p + 3 = 4
β 60p = 1
β ππ = 160
Since, ππ = 1π’π’
= 160
, ππ = 1π£π£
= 180
β u = 60 km/h, v = 80 km/h
Hence, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.
β¦ β¦ β¦
EXERCISE 3.7 (Optional)*
1. The ages of two friends Ani and Biju differ by 3 years. Aniβs father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
The difference of ages of Ani and Biju is 3 years. Hence, either Biju is 3 years older than Ani or Ani is 3 years older than Biju.
Let the age of Ani and Biju be x years and y years respectively.
Age of Dharma = 2x years
Age of cathy = π¦π¦2 years
Case I:
Ani is older than Biju by 3 years
π₯π₯ β π¦π¦ = 3 β¦ (1)
2x βy2
= 30
β 4x β y = 60 . . . (2)
Subtracting (1) from (2), we get:
3x = 60 β 3 = 57 β x = 19
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Hence, age of Ani = 19 years
And age of Biju = 19 β 3 = 16 years
Case II: Biju is older than Ani by 3 years
y β x = 3 β¦ (3)
2x βy2
= 30
β 4x β y = 60 β¦ (4)
Adding (3) and (4), we get:
3x = 63
β x = 21
Hence, age of Ani = 21 years
And age of Biju = 21 + 3 = 24 years
2. One says, βGive me a hundred, friend! I shall then become twice as rich as youβ. The other replies, βIf you give me ten, I shall be six times as rich as youβ. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara πΌπΌπΌπΌ]
[Hint: π₯π₯ + 100 = 2(π¦π¦ β 100),π¦π¦ + 10 = 6(π₯π₯ β 10)].
Solution:
Let the money with the first person be βΉ x and money with the second person be βΉ y.
As per the question,
x + 100 = 2(y β 100)
β x + 100 = 2y β 200
β x β 2y = β300 β¦ (1)
6(x β 10) = (y + 10)
β 6x β 60 = y + 10
β 6x β y = 70 β¦ (2)
Multiplying equation (2) by 2, we get:
12x β 2y = 140 β¦ (3)
Subtracting equation (1) from equation (3), we get:
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11x = 140 + 300
β 11x = 440
β x = 40
Putting x in equation (1), we get:
40 β 2y = β300
β 40 + 300 = 2y
β 2y = 340
β y = 170
Hence, the two friends has βΉ 40 and βΉ 170 respectively.
3. A train covered a certain distance at a uniform speed. If the train would have been 10 km hβ faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km hβ ; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let total distance travel by train be d km, the speed of the train be x km/h and the time taken by train to travel d km be t hours.
Since, Speed= Distanceβ―travlledTime takenβ―toβ―tranvelβ―thatβ―distance
β π₯π₯ =πππ‘π‘
β ππ = π₯π₯π‘π‘. . . (1)
As per the question, (π₯π₯ + 10) = ππ(π‘π‘β2)
β (π₯π₯ + 10)(π‘π‘ β 2) = ππ
β π₯π₯π‘π‘ + 10π‘π‘ β 2π₯π₯ β 20 = ππ
By using equation (1), we get:
β2π₯π₯ + 10π‘π‘ = 20 . . . (2)
(π₯π₯ β 10) =ππ
(π‘π‘ + 3)
β (π₯π₯ β 10)(π‘π‘ + 3) = ππ
By using equation (1), we get:
3x β 10t = 30 β¦ (3)
Adding equations (2) and (3), we get:
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π₯π₯ = 50
Substituting x in equation (2), we get:
(β2) Γ (50) + 10t = 20
β β100 + 10t = 20
β 10t = 120
β t = 12
From equation (1), we get:
d = xt = 50 x 12 = 600
Hence, the distance covered by the train is 600 km.
4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of rows be x and number of students in a row be y.
Total number of students in the class = Number of rows Γ Number of students in a row = xy
As per the question,
Total number of students = (x β 1)(y + 3)
β xy = (x β 1)(y + 3)
β π₯π₯π¦π¦ = xy β y + 3x β 3
β 3x β y β 3 = 0
β 3x β y = 3 β¦ (1)
Again, Total number of students = (x + 2)(y β 3)
β xy = xy + 2y β 3x β 6
β 3x β 2y = β6 β¦ (2)
Subtracting equation (2) from (1), we get:
y = 9
Substituting the value of y in equation (1), we get:
3x β 9 = 3
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β 3x = 9 + 3 = 12
β x = 4
Hence, number of rows, x = 4
And number of students in a row, y = 9
Hence, total number of students in a class = π₯π₯π¦π¦ = 4 Γ 9 = 36
5. In a βABC,β C = 3 β B = 2(β A + β B). Find the three angles.
Solution:
Given,
β C = 3 β B = 2(β A + β B)
β 3β B = 2(β A + β B)
β β B = 2 β A β 2 β A β β B = 0 . . . (1)
We know that the sum of all angles of a triangle is 180Β°
Hence, β A + β B + β C = 180o
β β π΄π΄ + β π΅π΅ + 3β B = 180o
β β A + 4β B = 180o β¦ (2)
Multiplying equation (1) by 4, we get:
8β π΄π΄ β 4 β π΅π΅ = 0 β¦ (3)
Adding equations (2) and (3), we get:
9 β A = 180o
β β A = 20o
From equation (2), we get:
20o + 4 β B = 180o
β 4β B = 160o
β β B = 40o
Hence, β C = 3 β B = 3 Γ 40o = 120o
Thus, the value of β A,β B and β C are 20o, 40o and 120o respectively.
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6. Draw the graphs of the equations 5π₯π₯ β π¦π¦ = 5 and 3π₯π₯ β π¦π¦ = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the π¦π¦ axis.
Solution:
5x β y = 5
β y = 5x β 5
Two solutions of this equation are:
π₯π₯ 0 1
π¦π¦ β5 0
3x β y = 3
β y = 3x β 3
Two solutions of this equation are:
π₯π₯ 0 1
π¦π¦ β3 0
The graphical representation of the two lines will be as follows:
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We have shown required triangle β³ ABC in the graph. The coordinates of its vertices are A (1, 0), B (0,β3), C (0,β5).
7. Solve the following pair of linear equations:
Solution:
(i) Both equations can be written as:
πππ₯π₯ + πππ¦π¦ β (ππ β ππ) = 0 β¦ (1)
πππ₯π₯ β πππ¦π¦ β (ππ + ππ) = 0 β¦ (2)
By using cross multiplication method, we get:
π₯π₯βππ(ππ + ππ) β ππ(ππ β ππ)
=π¦π¦
βππ(ππ β ππ) + ππ(ππ + ππ)=
1βππ2 β ππ2
βπ₯π₯
βππ2 β ππ2=
π¦π¦ππ2 + ππ2
=1
βππ2 β ππ2
β π₯π₯ = 1,π¦π¦ = β1
(ii) Both equations can be written as:
πππ₯π₯ + πππ¦π¦ β ππ = 0 β¦ (1)
πππ₯π₯ + πππ¦π¦ β (1 + ππ) = 0 . . . (2)
By using cross multiplication method, we get:
π₯π₯βππ(1 + ππ) + ππππ
=π¦π¦
βππππ + ππ(1 + ππ)=
1ππ2 β ππ2
(i) πππ₯π₯ + πππ¦π¦ = ππ β ππ πππ₯π₯ β πππ¦π¦ = ππ + ππ
(ii) πππ₯π₯ + πππ¦π¦ = ππ πππ₯π₯ + πππ¦π¦ = 1 + ππ
(iii) π₯π₯ππβπ¦π¦ππ
= 0 πππ₯π₯ + πππ¦π¦ = ππ2 + ππ2
(iv) (ππ β ππ)π₯π₯ + (ππ + ππ)π¦π¦= ππ2 β 2ππππ β ππ2 (ππ + ππ)(π₯π₯ + π¦π¦) = ππ2 + ππ2
(v) 152π₯π₯ β 378π¦π¦ = β74 β378π₯π₯ + 152π¦π¦ = β604
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βπ₯π₯
c(a β b) β b=
π¦π¦ππ(ππ β ππ) + ππ
=1
ππ2 β ππ2
β π₯π₯ =ππ(ππ β ππ) β ππππ2 β ππ2
, π¦π¦ =ππ(ππ β ππ) + ππππ2 β ππ2
(iii) π₯π₯ππβ π¦π¦
ππ= 0
πππ₯π₯ β πππ¦π¦ = 0 β¦ (1)
πππ₯π₯ + πππ¦π¦ = ππ2 + ππ2 β¦ (2)
Multiplying equation (1) and (2) by b and a respectively, we get:
ππ2π₯π₯ β πππππ¦π¦ = 0 β¦ (3)
ππ2π₯π₯ + πππππ¦π¦ = ππ3 + ππππ2 β¦ (4)
Adding equations (3) and (4), we get:
ππ2π₯π₯ + ππ2π₯π₯ = ππ3 + ππππ2
β π₯π₯(ππ2 + ππ2) = ππ(ππ2 + ππ2)
β π₯π₯ = ππ
Substituting π₯π₯ in equation (1), we get:
ππ(ππ) β πππ¦π¦ = 0
ππππ β πππ¦π¦ = 0
πππ¦π¦ = ππππ
π¦π¦ = ππ
Hence, π₯π₯ = ππ,π¦π¦ = ππ
(iv) (ππ β ππ)π₯π₯ + (ππ + ππ)π¦π¦ = ππ2 β 2ππππ β ππ2 β¦ (1)
(ππ + ππ)(π₯π₯ + π¦π¦) = ππ2 + ππ2
β (ππ + ππ)π₯π₯ + (ππ + ππ)π¦π¦ = ππ2 + ππ2. . . (2)
Subtracting equation (2) from (1), we get:
(ππ β ππ)π₯π₯ β (ππ + ππ)π₯π₯ = (ππ2 β 2ππππ β ππ2) β (ππ2 + ππ2)
(ππ β ππ β ππ β ππ)π₯π₯ = β2ππππ β 2ππ2 β 2πππ₯π₯ = β2ππ(ππ + ππ)
π₯π₯ = ππ + ππ
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Substituting x in equation (1), we get:
(ππ β ππ)(ππ + ππ) + (ππ + ππ)π¦π¦ = ππ2 β 2ππππ β ππ2
ππ2 β ππ2 + (ππ + ππ)π¦π¦ = ππ2 β 2ππππ β ππ2
(v) (ππ + ππ)π¦π¦ = β2ππππ
π¦π¦ =β2ππππππ + ππ
Hence, π₯π₯ = ππ + ππ,π¦π¦ = β2ππππππ+ππ
(vi) 152π₯π₯ β 378π¦π¦ = β74. . (1)
β378π₯π₯ + 152π¦π¦ = β604 β¦ (2)
Adding the equations (1) and (2), we get:
β226π₯π₯ β 226π¦π¦ = β678
β π₯π₯ + π¦π¦ = 3 β¦ (3)
Subtracting the equation (2) from equation (1), we get:
530π₯π₯ β 530π¦π¦ = 530
β π₯π₯ β π¦π¦ = 1 β¦ (4)
Adding equations (3) and (4), we get:
2π₯π₯ = 4
β π₯π₯ = 2
Substituting x in equation (3), we get:
π¦π¦ = 1
Hence, π₯π₯ = 2,π¦π¦ = 1
8. ABCD is a cyclic quadrilateral (see Fig.) Find the angles of the cyclic quadrilateral.
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Solution:
As we know, the sum of opposite angles in a cyclic quadrilateral is 180Β°.
β A + β C = 180
β 4π¦π¦ + 20Β° β 4π₯π₯ = 180
β β4π₯π₯ + 4π¦π¦ = 160Β°
β π₯π₯ β π¦π¦ = β40 β¦ (1)
Also, β B + β D = 180Β°
β 3π¦π¦ β 5Β° β 7π₯π₯ + 5Β° = 180Β°
β β7π₯π₯ + 3π¦π¦ = 180 β¦ (2)
Multiplying equation (1) by 3, we get:
3π₯π₯ β 3π¦π¦ = β120 β¦ (3)
Adding equations (2) and (3), we get:
β4π₯π₯ = 60Β°
β π₯π₯ = β15Β°
Substituting π₯π₯ in equation (1), we get:
β15Β° β π¦π¦ = β40Β°
β π¦π¦ = β15Β° + 40Β° = 25
Hence,
β A = 4π¦π¦ + 20Β° = 4(25Β°) + 20Β° = 120ππ
β B = 3π¦π¦ β 5Β° = 3(25Β°) β 5Β° = 70ππ
β C = β4π₯π₯ = β4(β15Β°) = 60ππ
β D = β7π₯π₯ + 5Β° = β7(β15Β°) + 5Β° = 110ππ
β¦ β¦ β¦
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