NCERT Solutions for Class 10 Maths Chapter 3 - Pair of Linear Equations in Two Variables

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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.

Sol. Let be present age of Aftab = x
Present age of Aftab's daughter = y
According to question,
(x – 7) = 7(y – 7)
x – 7 = 7y – 49
x – 7y = – 49 + 7
x – 7y = – 42 ...(i)
Again, according to question
(x + 3) = 3(y + 3)
x + 3 = 3y + 9 x – 3y = 6 ...(ii)
By equation (i)

Put x = 0 in equation (i)
0 – 7y = – 42
– 7y = – 42
y = 6
Put x = 7 in equation (i)
7 – 7y = – 42
– 7y = – 42 – 7
– 7y = – 49
y = 7
Put x = 14 in equation (i)
14 – 7y = – 42
– 7y = – 42 – 14
– 7y = – 56
y = 8
By equation (ii)

x	0	7	14
y	6	7	8

x	0	3	6
y	– 2	– 1	0

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 be
Cost of a bat = ₹ x
Cost of a ball = ₹ y
According to question,
3x + 6y = 3900
3(x + 2y) = 3900
x + 2y = 1300 ...(i)
Again according to question
x + 3y = 1300 ...(ii)
By equation (i)

x	100	300	500
y	600	500	400

Put x = 100 in equation (i)
100 + 2y = 1300
2y = 1200
y = 600
Put x = 300 in equation (i)
300 + 2y = 1300
2y = 1000
y = 500
Put x = 500 in equation (i)
500 + 2y = 1300
2y = 800
y = 400
By equation (ii)

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 be
Cost of 1 kg apples = ₹ x
Cost of 1 kg grapes = ₹ y
According to question
2x + y = 160 ...(i)
Again according to question,
4x + 2y = 300
2x + y = 150 ...(ii)
By equation (i)

x	0	20	40
y	160	120	80

Put x = 0 in equation (i)
2 × 0 + y = 160
y = 160
Put x = 20 in equation (i)
2 × 20 + y = 160
y = 160 – 40
y = 120
Put x = 40 in equation (i)
2 × 40 + y = 160
y = 160 – 80
y = 80
By equation (ii)

x	0	30	50
y	150	90	50

Put x = 0 in equation (ii)
2 × 0 + y = 150
y = 150
Put x = 30 in equation (ii)
2 × 30 + y = 150
60 + y = 150
y = 90
Put x = 50 in equation (ii)
2 × 50 + y = 150
y = 150 – 100
y = 50

Exercise 3.2

1. From the pair of linear equations in the following problems, and find their solution 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 ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Sol. (i) Suppose that
Number of girls = x
Number of boys = y
According to question
x + y = 10 ...(i)
x – y = 4 ...(ii)
By equation (i)

x	3	5	7
y	7	5	3

Put the value x = 3 in equation (i) 3 + y = 10
y = 10 – 3 = 7
Put x = 5 in equation (i)
5 + y = 10
y = 5
Put x = 7 in equation (i)
7 + y = 10
y = 10 – 7
y = 3
By equation (ii)

x	3	5	7
y	-1	1	3

Put x = 3 in equation (ii)
3 – y = 4
– y = 1
y = – 1
Put x = 5 in equation (ii)
5 – y = 4
– y = 4 – 5
– y = – 1
y = 1
Put x = 7 in equation (ii)
7 – y = 4
– y = 4 – 7
– y = – 3
y = 3
Lines are intersect each other at (7, 3)
[In graph]
Hence,
Number of girls = 7
Number of boys = 3

(ii) Suppose that cost of one pencil and one pen are respectively ₹ x and ₹ y.
According to question
5x + 7y = 50 ...(i)
7x + 5y = 46 ...(ii)
By equation (i)

x	3	5	7
y	7	5	3

Put x = – 4 in equation (i)
5 × (– 4) + 7y = 50
– 20 + 7y = 50
7y = 50 + 20
7y = 70
y = 10
Put x = 3 in equation (i)
5 × 3 + 7y = 50
7y = 50 – 15
7y = 35
y = 357
y = 5
Put x = 10 in equation (i)
5 × 10 + 7y = 50
50 + 7y = 50
y = 0
By equation (ii)

x	-2	3	8
y	12	5	-2

Put x = – 2 in equation (ii)
7 × (– 2) + 5y = 46
5y = 46 + 14
5y = 60
y=60/5
y = 12
Put x = 3 in equation (ii)
7 × 3 + 5y = 46
5y = 46 – 21
5y = 25
y = 5
Put x = 8 in equation (ii)
7 × 8 + 5y = 46
5y = 46 – 56
5y = – 10
y = – 2
Lines are intersect (3, 5) (In graph)
cost of one pencil = 3
cost of one pen = 5

2. On comparing the ratios \text{(a1/a2),(b1/b2), and (c1/c2);}find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident :
(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

Sol. (i) 5x – 4y + 8 = 0 ...(i)
7x + 6y – 9 = 0 ...(ii)
Comparing equations (i) and (ii) with
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 respectively.
a1 = 5, b1 = – 4, c1 = 8
a2 = 7, b2 = 6, c2 = – 9

$$\frac{a_1}{a_2}=\frac{5}{7}\\\frac{b_1}{b_2}=\frac{-4}{6}=\frac{-2}{3}\\\text{Here,}\\\frac{5}{7}≠\frac{-2}{3}\\\frac{a_1}{a_2}≠\frac{b_1}{b_2}\\\text{The lines are intersect each other at a point on graph.}$$

(ii) 9x + 3y + 12 = 0 ...(i)
18x + 6y + 24 = 0 ...(ii)
Compare with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.
Then,
a1 = 9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24

$$\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}\\\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\\\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\\\text{Here,}\\\frac{1}{2}=\frac{1}{2}=\frac{1}{2}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\text{So, lines are coincident.}$$

(iii) 6x – 3y + 10 = 0 ...(i)
2x – y + 9 = 0 ...(ii)
Compare with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a1 = 6, b1 = – 3, c1 = 10
a2 = 2, b2 = – 1, c2 = 9

$$\frac{a_1}{a_2}=\frac{6}{2}=\frac{3 }{1}\\\frac{b_1}{b_2}=\frac{-3}{-1}=\frac{3}{1}\\\frac{c_1}{c_2}=\frac{10}{9}\\\text{Here,}\\\frac{3}{1}=\frac{3}{1}≠\frac{10}{1}\\\text{So, lines are parallel.}$$

$$\textbf{3. On comparing the ratios}\\\frac{a_1}{a_2},\frac{b_1}{b_2}\textbf{and}\frac{c_1}{c_2}\textbf{find out whether the following}\space\text{pair of linear equations are consistent or inconsistent.}\\(i)3x + 2y = 5; 2x – 3y = 7\\(ii) 2x – 3y = 8; 4x – 6y = 9\\(iii)\frac{3}{2}x+\frac{5}{3}y= 7; 9x – 10y = 14\\(iv) 5x – 3y = 11; – 10x + 6y = – 22\\(v)\frac{4}{3}x+2y= 8; 2x + 3y = 12$$

Sol. (i) 3x + 2y = 5
2x – 3y = 7
Rearranging then
3x + 2y – 5 = 0 ...(i)
2x – 3y – 7 = 0 ...(ii)
Compare with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a1 = 3, b1 = 2, c1 = – 5
a2 = 2, b2 = – 3, c2 = – 7
$$\\\frac{a_1}{a_2}=\frac{3}{2}\\\frac{b_1}{b_2}=\frac{2}{-3}\\\frac{c_1}{c_2}=\frac{-5}{-7}\\\frac{3}{2}≠\frac{2}{-3}\\\frac{a_1}{a_2}≠\frac{b_1}{b_2}\\\text{So, pair of linear equations are consistant.}$$

(ii) 2x – 3y = 8
4x – 6y = 9
Rearranging
2x – 3y – 8 = 0 ...(i)
4x – 6y – 9 = 0 ...(ii)
Compare with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
a1 = 2, b1 = – 3, c1 = – 8
a2 = 4, b2 = – 6, c2 = – 9
$$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\\\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\\\frac{c_1}{c_2}=\frac{-8}{-9}=\frac{8}{9}\\\frac{1}{2}=\frac{1}{2}≠\frac{8}{9}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\\\text{Equations have no solution.}\\\text{So pair of equations are inconsistant.}$$

$$(iii)\frac{3}{2}x+\frac{3}{2}y=7\\9x – 10y = 14 \\\text{Rearranging both}\\\frac{3}{2}x+\frac{3}{2}y-7=0...(i)\\9x – 10y – 14 = 0 ...(ii)\\ \text{Compare with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0}\\a_1 =\frac{3}{2},b_1=\frac{5}{3},c_1=-7\\a_2 = 9, b_2 = – 10, c_2 = – 14\\\frac{a_1}{a_2}=\frac{3/2}{9}=\frac{3}{9×2}=\frac{1}{6}\\\frac{b_1}{b_2}=\frac{5/3}{-10}=\frac{5}{-10×3}=-\frac{1} {6}\\\text{Here,}\\\frac{1}{6}≠-\frac{1}{6}\\\frac{a_1}{a_2}≠\frac{b_1}{b_2}\\\text{The lines may intersect in a single point.}\\\text{In this case pair of equations are consistant.}$$

(iv) 5x –3y = 11
– 10x + 6y = – 22
Rearranging them
5x –3y – 11 = 0 ...(i)
– 10x + 6y + 22 = 0 ...(ii)
Compare with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Then,
a1 = 5, b1 = – 3, c1 = – 11
a2 = – 10, b2 = 6, c2 = 22
$$\frac{a_1}{a_2}=\frac{5}{-10}=\frac{-1}{2}\\\frac{b_1}{b_2}=\frac{-3}{6}=\frac{-1}{2}\\\frac{c_1}{c_2}=\frac{-11}{22}=\frac{-1}{2}\\\text{Here,}\\\frac{-1}{2}=\frac{-1}{2}=\frac{-1} {2}\\\text{Here,}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\text{The lines may be coincident. In this case}\\\text{pair of equations are dependent consistent.}$$
$$\textbf{(v)}\frac{4}{3}x+2y=8\\2x + 3y = 12\\\text{Rearranging them}\\\frac{4}{3}x+2y-8=0...(i)\\2x + 3y – 12 = 0 ...(ii)\\\text{Compare with}\space a_1x + b_1y + c_1 = 0\text{ and}\space a_2x + b_2y + c2 = 0\\a_1=\frac{4}{3},b_1 = 2, c_1 = – 8\\a_2 = 2, b_2 = 3, c_2 = – 12\\\frac{a_1}{a_2}=\frac{4}{3×2}=\frac{2}{3}\\\frac{b_1}{b_2}=\frac{2}{3}\\\frac{c_1}{c_2}=\frac{-8}{-12}=\frac{2}{3}\\⇒\frac{2}{3}=\frac{2}{3}=\frac{2}{3}\\⇒\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\text{Pair of equations are dependent consistant.}$$

4. Which of the following pairs of linear equations are consistent or inconsistent? If consistent obtain the solution graphically?
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Sol. (i) x + y = 5
2x + 2y = 10
Rearranging them
x + y – 5 = 0 ...(i)
2x + 2y – 10 = 0 ...(ii)
Compare with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
a₁ = 1, b₁ = 1, c₁ = – 5
a₂ = 2, b₂ = 2, c₂ = – 10
Then
$$\frac{a_1}{a_2}=\frac{1}{2}\\\frac{b_1}{b_2}=\frac{1}{2}\\⇒\frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}\\⇒\frac{1}{2}=\frac{1}{2}=\frac{1}{2}\\\text{Pair of equations are consistent.}$$
By equation (i)

x	1	2	3
y	4	3	2

Put x = 1 in equation (i)
1 + y – 5 = 0
y = 4
Put x = 2 in equation (i)
2 + y – 5 = 0
y = 3
Put x = 3 in equation (i)
3 + y – 5 = 0
y = 2
By equation (ii)

x	1	2	3
y	4	3	2

Put x = 1 in equation (ii)
2 × 1 + 2y – 10 = 0
2y = 8
y = 4
Put x = 2 and x = 3 in equation (ii) Then finding respectively.
y = 3, y = 2.

The lines may be coincident and the equations have infinitely many solutions.

(ii) x – y = 8 ...(i)
3x – 3y = 16 ...(ii)
Compare with
a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂ = 0
Then
a₁ = 1, b₁ = – 1, c₁ = – 8
a₂ = 3, b₂ = – 3, c₂ = – 16
$$\frac{a_1}{a_2}=\frac{1}{3}\\\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\\⇒\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}\\\text{Here,}\\\frac{1}{3}=\frac{1}{3}≠\frac{1}{2}\\\text{Pair of equations are consistent.}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\\\text{Pair of equations have no solution.}\\\text{Pair of equations are inconsistent.}$$

(iii) 2x + y – 6 = 0 ...(i)
4x – 2y – 4 = 0 ...(ii)
Compare with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
a₁ = 2, b₁ = 1, c₁ = – 6
a₂ = 4, b₂ = – 2, c₂ = – 4
$$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\\\frac{b_1}{b_2}=\frac{-1}{2}\\⇒\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}\\\text{Here,}\\\frac{1}{2}≠\frac{-1}{2}\\\frac{a_1}{a_2}≠\frac{b_1}{b_2}\\\text{Pair of equations are consistent.}$$
By equation (i)

x	0	2	4
y	6	2	-2

Put x = 0 in equation (i)
2 × 0 + y – 6 = 0
y = 6
Put x = 2 in equation (i)
2 × 2 + y – 6 = 0
y – 2 = 0
y = 2
Put x = 4 in equation (i)
2 × 4 + y – 6 = 0
y = – 2
By equation (ii)

x	0	2	4
y	-2	2	6

Put x = 0 in equation (ii)
4 × 0 – 2y – 4 = 0
– 2y = 4
y = – 2
Put x = 2 in equation (ii)
4 × 2 – 2y – 4 = 0
8 – 2y – 4 = 0
– 2y = – 4
y = 2
Put x = 4 in equation (ii)
4 × 4 – 2y – 4 = 0
16 – 2y – 4 = 0
– 2y = – 12
y = 6

Hence
The two lines intersect at the point (2, 2). So, x = 2, y = 2 is the required solution of the pair of linear equations.

(iv) 2x – 2y – 2 = 0 ...(i)
4x – 4y – 5 = 0 ...(ii)
Compare with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c2 = 0
Then,
a₁ = 2, b₁ = – 2, c₁ = – 2
a₂ = 4, b₂ = – 4, c₂ = – 5

$$⇒\frac{a_1}{a_2}=\frac{2}{4}\\⇒\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{2}{4}\\⇒\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}\\\text{Here,}\\⇒\frac{2}{4}=\frac{2}{4}≠\frac{2}{5}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\\\text{Pair of equations are inconsistent.}$$

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.

Sol. Let be width of rectangular garden = x
Length = x + 4
(According to question)
Half of the perimeter of rectangular = 36 m
1/2[2x + 2(x + 4)] = 36
⇒ 2x + 2x + 8 = 72
⇒ 4x + 8 = 72
⇒ 4x = 72 – 8
⇒ 4x = 64
⇒ x =64/4=16
Width x = 16 m
Length x + 4 = 16 + 4 = 20 m.

6. Given that linear equation 2x + 3y – 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

Sol. Given linear equations
2x + 3y – 8 = 0 ...(i)
Let another linear equation
2x + 4y – 12 = 0 ...(ii)
Comparing with a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂ = 0
Then,
a₁ = 2, b₁ = 3, c₁ = – 8
a₂ = 2, b₂ = 4, c₂ = – 12
$$⇒\frac{a_1}{a_2}=\frac{2}{2}=\frac{1}{1}\\⇒\frac{b_1}{b_2}=\frac{3}{4}\\1≠\frac{3}{4}\\⇒\frac{a_1}{a_2}≠\frac{b_1}{b_2}\\\text{The lines may intersect in a single point. Then the required equation is}\\2x + 4y – 12 = 0$$
(ii) Given linear equation
2x + 3y – 8 = 0 ...(i)
Let another linear equation,
4x + 6y + 12 = 0 ...(ii)
Comparing
with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Then,
a₁ = 2, b₁ = 3, c₁ = – 8
a₂ = 4, b₂ = 6, c₂ = 12

$$⇒\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{1}\\⇒\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\\⇒\frac{c_1}{c_2}=\frac{-8}{12}=\frac{-2}{3}\\⇒\frac{1}{2}=\frac{1}{2}≠\frac{2}{3}\\⇒\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\\\text{So, the lines may be parallel. Then the required equation}\space 4x + 6y + 12 = 0$$

(iii) Given linear equation
2x + 3y – 8 = 0 ...(i)
Let another linear equation
4x + 6y – 16 = 0 ...(ii)
Comparing
with a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂ = 0
Then,
a₁ = 2, b₁ = 3, c₁ = – 8
a₂ = 4, b₂ = 6, c₂ = – 16
Then,
$$⇒\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}\\⇒\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\\⇒\frac{c_1}{c_2}=\frac{-8}{16}=\frac{1}{2}\\⇒\frac{1}{2}=\frac{1}{2}=\frac{1}{2}\\⇒\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\text{So, equations (i) and (ii) are coincident lines. Hence the required equation is}\\4x + 6y – 16 = 0$$

7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and x-axis and shade the triangular region.

Sol. x – y + 1 = 0 ...(i)
3x + 2y – 12 = 0 ...(ii)
By equation (i)

x	-1	2	4
y	0	3	5

Put x = 0 in equation (i)
– 1 – y + 1 = 0
y = 0
Put x = 2 in equation (i)
2 – y + 1 = 0
y = 3
Put x = 4 in equation (i)
4 – y + 1 = 0
y = 5
By equation (ii)

x	0	2	4
y	6	3	0

Put x = 0
3 × 0 + 2y – 12 = 0
2y = 12
y = 6
Put x = 2
3 × 2 + 2y – 12 = 0
2y = 6
y = 3
Put x = 4
3 × 4 + 2y –12 = 0
y = 0
Required D is DABC and its vertices are A(– 1, 0), B(4, 0) and C(2, 3).

Exercise 3.3

1. Solve the following pair of linear equations by the substitution method.
$$(i) x + y = 14, x – y = 4\\(ii) s – t=3,\frac{s}{3}+\frac{t}{2}=6\\(iii) 3x – y = 3, 9x – 3y = 9\\(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3\\(v)\sqrt{2x}+\sqrt{3y}=0,\sqrt{3x}-\sqrt{8y}=0\\\frac{3x}{2}-\frac{5y}{3}=-2,\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$$

Sol. (i) x + y = 14 ...(i)
x – y = 4 ...(ii)
By equation (i)
y = 14 – x ...(iii)
Put the value of y in equation (ii)
⇒ x – (14 – x) = 4
⇒ x – 14 + x = 4
⇒ 2x = 4 + 14
⇒ 2x = 18
⇒ x = 9
Put x = 9 in equation (iii)
y = 14 – 9
y = 5
Hence,
x = 9, y = 5.
$$(ii) s – t = 3 ...(i)\\\frac{s}{3}+\frac{t}{2}=6...(ii)\\\text{By equation (i)}\\s = 3 + t ...(iii)\\\text{Put the value s in equation (ii)}\\⇒\frac{3+t}{2}+\frac{t}{2}=6\\⇒\frac{6+2t+3t}{6}=6\\⇒ 6 + 5t = 36\\⇒ 5t = 30\\⇒t=\frac{30}{5}\\t = 6\\\text{Put the value of t in equation (iii)}\\⇒\frac{s}{3}+\frac{6}{2}= 6\\⇒\frac{s}{3}=3\\⇒ s = 9\\\text{Hence,}\\s = 9, t = 6.$$

(iii) 3x – y = 3 ...(i)
Multiply by 3 on both sides
9x – 3y = 9 ...(ii)
a₁ = 3, b₁ = – 1, c₁ = – 3
a₂ = 9, b₂ = – 3, c₂ = – 9
Then
$$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\text{Pair of equation have infinitely many solutions.Then}\\y = 3x – 3\\\text{Where x can take any value.}$$

(iv) 0.2x + 0.3y = 1.3 ...(i)
0.4x + 0.5y = 2.3 ...(ii)
By equation (i)
0.2x = 1.3 – 0.3y
Multiply by 10 on both sides
⇒ 2x = 13 – 3y
$$x=\frac{13-3y}{2}...(iii)\\\text{Put the value of x in equation (ii)}\\\frac{0.4(13-3y)}{2}+0.5y=2.3\\⇒[\frac{0.4(13-3y)}{2}+0.5y]×10== 2.3 ×1 0\\⇒\frac{4(13-3y)}{2}+5y=23\\⇒ 26 – 6y + 5y = 23\\⇒ 26 – y = 23\\⇒ – y = 23 – 26\\⇒ y = 3\\\text{Put the value of y in equation (iii)}\\⇒x =\frac{1333}{2}\\⇒ x = 2\\\text{Hence,}\\x = 2, y = 3$$

$$(v)\sqrt{2x}+\sqrt{3y}=0...(i)\\\sqrt{3x}-\sqrt{8y}=0...(ii)\\\text{By equation (i)}\\\sqrt{3x}=\sqrt{8y}\\⇒ x =\frac{\sqrt{8y}}{\sqrt{3}}...(iii)\\\text{Put the value of x in equation (i)}\\\sqrt{2}×\frac{\sqrt{8y}}{\sqrt{3}}+\sqrt{3y}=0\\⇒ 4y + 3y = 0\\⇒ 7y = 0\\⇒ y = 0\\\text{Put the value of y in equation (iii)}\\=\frac{\sqrt{8}×10}{\sqrt{3}}\\⇒ x = 0\\\text{Hence,}\\x = 0, y = 0$$

$$(vi)\frac{3x}{2}-\frac{5y}{3}= – 2 ...(i)\\\frac{x}{2}+\frac{y}{3}=\frac{13}{6}...(ii)\\\text{By equation (i)}\\\frac{3x}{2}-\frac{5y}{3}= – 2\\\\⇒\frac{3x}{2}+2=\frac{5y}{3}\\\frac{3x+4}{2}=\frac{5y}{3}\\⇒ 9x + 12 = 10y\\⇒y=\frac{9x+12}{10}...(iii)\\\text{Put the value of y in equation (ii)}\\⇒\frac{x}{3}+\frac{9x+12}{2×10}=\frac{13}{6}\\⇒\frac{20x+27x+36}{60}=\frac{13}{6}\\⇒ 47x + 36 = 130\\⇒ 47x = 130 – 36\\⇒ 47x = 94\\⇒ x = 2\\\text{Put the value of x in equation (iii)}\\⇒ y =\frac{9×2+12}{10}\\⇒y=\frac{18+12}{10}\\⇒ y =\frac{30}{10}\\⇒ y = 3\\\text{Hence,}\\x = 2, y = 3$$

2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of m for which y = mx + 3.

Sol. Given
2x + 3y = 11 ...(i)
2x – 4y = – 24 ...(ii)
By equation (ii)
2x = 4y – 24
$$x=\frac{2(2y-12)}{2}$$
Put the value of x in equation (i)
2(2y – 12) + 3y = 11
4y – 24 + 3y = 11
7y – 24 = 11
7y = 35
y = 5
Put the value of y in equation (iii)
x = 2 × 5 – 12
x = – 2
Again given
y = mx + 3
Put values x and y.
5 = – 2m + 3
2 = – 2m
m =-2/2
m = – 1
Hence,
x = – 2, y = 5, m = – 1.

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.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 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 ₹ 105 and for a journey of 15 km, the charge paid is ₹ 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?
(v) A fraction becomes 9/11, 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.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

Sol. (i) Suppose that two numbers are x and y.
(Here x > y)
According to question
x – y = 26 ...(i)
x = 3y
x – 3y = 0 ...(ii)
By equation (ii)
x = 3y
Put the value of x in equation (i) 3y – y = 26
2y = 26
y = 13
Put the value of y in equation (ii) x – 3 × 13 = 0
x = 39
Required numbers are 39 and 13.

(ii) Let be two supplementary angles x and y.
According to question,
x – y = 18 ...(i)
(Here x > y)
We know that sum of supplementary angles are 180°
x + y = 180 ...(ii)
By equation (i)
x = 18 + y ...(iii)
Put the value of x in equation (ii) 18 + y + y = 180
2y = 180 – 18
2y = 162
y = 81
Put the value of y in equation (iii)
x = 18 + 81
x = 99
Hence,
x = 99, y = 81
(measures of the two angles in degrees)

(iii) Let the cost of each bat and each ball respectively x and y.
According to question,
7x + 6y = 3800 ...(i)
Again,
3x + 5y = 1750 ...(ii)
By equation (i)
7x = 3800 – 6y
$$x=\frac{3800-6y}{7}...(iii)\\\text{Put the value of x in equation (ii)}\\3(\frac{3800-6y}{7})+5y = 1750\\11400 – 18y = 7(1750 – 5y)\\11400 – 18y = 12250 – 35y\\35y – 18y = 12250 – 11400\\17y = 850\\y = 50\\\text{Put the value of y in equation (iii)}\\x=\frac{3800-6×50}{7}\\x=\frac{3500}{7}\\x = 500\\\text{Hence, Cost of one bat x = ₹ 500 Cost of one ball y = ₹ 50.}$$

(iv) Let the fixed charge is ₹ x and per km charge is ₹ y.
Then according to question,
x + 10y = 105 ...(i)
x + 15y = 155 ...(ii)
By equation (i)
x = 105 – 10y
Put the value of x in equation (ii)
105 – 10y + 15y = 155
5y = 50
y = ₹ 10
Put the value of y in equation (i)
x + 10 × 10 = 105
x = ₹ 5
Hence
Fixed charge = ₹ 5
Per km charge = ₹ 10 /km
Charge of 25 km
= ₹ (5 + 25 × 10)
= ₹255

$$\textbf{(v)}\text{ Let be A Fraction}\frac{x}{y}\\\text{According to question}\\\frac{x+2}{y+2}=\frac{9}{11}\\11x + 22 = 9y + 18\\11x – 9y = – 4 ...(i)\\\text{Again}\\\frac{x+3}{y+3}=\frac{5}{6}\\6x + 18 = 5y + 15\\6x – 5y = – 3 ...(ii)\\\text{By equation (ii)}\\6x + 3 = 5y\\y=\frac{6x+3}{5}...(iii)\\\text{Put the value of y in equation (i)}\\11x-9(\frac{6x+3}{5})= – 4\\55x – 54x – 27 = – 20\\x = – 20 + 27\\x = 7\\\text{Put the value of x in equation (iii)}\\y=\frac{45}{5}\\y = 9\\\text{Fraction is}\frac{7}{9}$$

(vi) Let be present age of Jacob's is x and present age of Jacob's son is y.
Then according to question
(x + 5) = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 10 ...(i)
Five years ago
x – 5 = 7(y – 5)
x – 5 = 7y – 35
x – 7y = – 30
x – 7y + 30 = 0 ...(ii)
By equation (i)
x = 10 + 3y
Put the value of x in equation (ii) 10 + 3y – 7y + 30 = 0
– 4y + 40 = 0
y = 10 years
Put the value of y in equation (i) x – 3 × 10 = 10
x = 40 years
Hence,
Jacob's present age = 40 years
Jacob's son present age = 10 years.

Exercise 3.4

1. Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
$$(iv)\frac{x}{2}+\frac{2y}{3}= – 1 \text{and} x-\frac{y}{3}= 3$$

Sol. (i) x + y = 5 ...(i)
2x – 3y = 4 ...(ii)
Multiply equation (i) by 2
2x + 2y = 10
2x – 3y = 4
– + – By subtracting
5y = 6
$$y=\frac{6}{5}\\\text{Put the value of y in equation (i)}\\x+\frac{6}{5}=5\\x=5-\frac{6}{5}\\x =\frac{25-6}{5}\\x=\frac{19}{5}\\\text{Hence,}\\x=\frac{19}{5},y=\frac{6}{5}$$

(ii) 3x + 4y = 10 ...(i)
2x – 2y = 2 ...(ii)
Multiply
equation (i) by 2 and equation (ii) by 3 then,
6x + 8y = 20
6x – 6y = 6
$$\frac{-+-}{}\text{By Subtracting}\\14y = 14\\y = 1\\\text{Put the value of y in equation (i)}\\3x + 4 × 1 = 10\\3x = 10 – 4\\3x = 6\\x = 2\\\text{Hence,}\\x = 2, y = 1$$

(iii) 3x – 5y – 4 = 0 ...(i)
9x = 2y + 7
Rearranging
9x – 2y – 7 = 0 ...(ii)
by equation (i) and (ii)
3x – 5y – 4 = 0
9x – 2y – 7 = 0
Multiply
equation (i) by 2 and equation (ii) by 5, then
6x – 10y – 8 = 0
45x – 10y – 35 = 0
$$\frac{-++}{– 39x + 27 = 0}\\x =\frac{-27}{-39}\\x=\frac{9}{13}\\\text{Put the value of x in equation (i)}\\3×\frac{9}{13}-5y-4=0\\27 – 65y – 52 = 0\\– 65y – 25 = 0\\y =\frac{25}{-65}\\y =\frac{-5}{13}\\\text{Hence,}\\x=\frac{9}{13}\text{and}y =\frac{-5}{13}$$

$$\textbf{(iv)} \text{Given}\\\frac{x}{2}+\frac{2y}{2}= – 1\\x-\frac{y}{3}=3\\\text{Rearranging them}\\3x + 4y = – 6 ...(i)\\3x – y = 9 ...(ii)\\\frac{-+-}{5y = – 15}\text{Subtracting}\\y = – 3\\\text{Put the value of y in equation (i)}\\3x + 4 × – 3 = – 6\\3x – 12 = – 6\\3x = 6\\x = 2\\\text{Hence}, x = 2, y = – 3$$

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?
(iii) 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 ₹2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

$$\textbf{(Sol. i)} \text{Let be the fraction is}\frac{x}{y}\\\text{Then according to question}\\\frac{x+1}{y-1}=1\\x + 1 = y – 1\\x – y = – 2 ...(i)\\\text{Again according to question}\\\frac{x}{y+1}=\frac{1}{2}\\2x = y + 1\\2x – y = 1 ...(ii)\\\text{By equations (i) and (ii)}\\x – y = – 2\\2x – y = 1\\\frac{-+-}{– x = – 3}\text{By subtracting}\\x = 3\\\text{Put x = 3 in equation (i)}\\3 – y = – 2\\– y = – 2 – 3\\– y = – 5\\y = 5\\\text{Hence,}\\\text{The fraction is}=\frac{3}{5}$$

(ii) Let be x and y are the present ages (in years) of Nuri and Sonu respectively.
Five years ago
(x – 5) = 3(y – 5)
x – 5 = 3y – 15
x – 3y = – 10 ...(i)
Ten years later
(x + 10) = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 10 ...(ii)
By equations (i) and (ii)
x – 3y = – 10
x – 2y = 10
$$\frac{-+-}{– y = – 20}$$
y = 20 years.
Put the value of y in equation (i)
x – 3 × 20 = – 10
x – 60 = – 10
x = – 10 + 60
x = 50 years
Hence
Present age of Nuri (x) = 50 years.
Present age of Sonu (y) = 20 years.

(iii) Suppose that x and y are respectively the tens and unit digits of the number.
Then
x + y = 9 ...(i)
Number = 10x + y
According to question
9(10x + y) = 2(10y + x)
90x + 9y = 20y + 2x
90x – 2x + 9y – 20y = 0
88x – 11y = 0
8x – y = 0 ...(ii)
By equations (i) and (ii)
x + y = 9
8x – y = 0 by adding
9x = 9
x = 1
Put the value of x in equation (i)
1 + y = 9
y = 8
Number = 10x + y
= 10 × 1 + 8
= 18

(iv) Suppose that x and y are respectively the number of ₹ 50 and ₹ 100 notes.
Total notes = 25
x + y = 25 ...(i)
According to question
50x + 100y = 2000 ...(ii)
By equations (i) and (ii)
x + y = 25 × 50
50x + 100y = 2000 × 1
50x + 50y = 1250
50x + 100y = 2000
$$\frac{---}{– 50y = – 750}\text{By subtracting}\\y=\frac{75}{5}$$
y=15
Put the value of y in equation (i)
x + 15 = 25
x = 25 – 15
x = 10
Hence,
Notes of 50(x) = 10
Notes of 100(y) = 15

(v) Let be the fixed charge is x and the charge for each extra day is y.
According to question
Saritha paid = 27
x + 4y = 27 ...(i)
Again,
Susy paid = 21
x + 2y = 21 ...(ii)
By equations (i) and (ii)
x + 4y = 27
x + 2y = 21
$$\frac{---}{2y = 6}\text{subtracting}$$
y = 3
Put the value of y in equation (i)
x + 4 × 3 = 27
x = 27 – 12
x = 15
Hence,
Fixed charge for first three days (x) = ₹ 15
Additional charge of extra day (y) = ₹ 3.

Exercise 3.5

1. Which of the following pairs of linear equations has unique soluiton, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0 and 3x – 9y – 2 = 0
(ii) 2x + y = 5 and 3x + 2y = 8
(iii) 3x – 5y = 20 and 6x – 10y = 40
(iv) x – 3y – 7 = 0 and 3x – 3y – 15 = 0

Sol. (i) x – 3y – 3 = 0 ...(i)
3x – 9y – 2 = 0 ...(ii)
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Then,
a₁ = 1, b₁ = – 3, c₁ = – 3
a₂ = 3, b₂ = – 9, c₂ = – 2
$$⇒\frac{a_1}{a_2}=\frac{1}{3}\\⇒\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}\\⇒\frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}\\\frac{1}{3}=\frac{1}{3}≠\frac{3}{2}\\⇒\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\\\text{Pair of linear equations have no solution.}$$

(ii) 2x + y = 5
3x + 2y = 8
Rearranging equations
2x + y – 5 = 0 ...(i)
3x + 2y – 8 = 0 ...(ii)
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Then,
a₁ = 2, b₁ = 1, c₁ = – 5
a₂ = 3, b₂ = 2, c₂ = – 8
$$\frac{a_1}{a_2}=\frac{2}{3}\\\frac{b_1}{b_2}=\frac{1}{2}\\\frac{2}{3}≠\frac{1}{2}\\\text{The pair of equations has a unique solution}\\x=\frac{b_1c_1-b_2c_2}{a_1b_1-a_2b_2}\\x=\frac{1×(−8)-2×(-5)}{2×2-3×1}\\x=\frac{-8+10}{4-3}\\x=2\\\text{and}y=\frac{c_1a_2-c_2a_2}{a_1b_2-a_2b_1}\\y=\frac{-5×3-(-8)×2}{4-3}\\y=\frac{-15+16}{1}=1\\\text{Therefore, x = 2, y = 1 is the solution of the given pair of equations.}$$

(iii) 3x – 5y = 20
6x – 10y = 40
Rearranging equations
3x – 5y – 20 = 0 ...(i)
6x – 10y – 40 = 0 ...(ii)
Compare with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Then,
a₁ = 3, b₁ = – 5, c₁ = – 20
a₂ = 6, b₂ = – 10, c₂ = – 40
$$⇒\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}\\⇒\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}\\⇒\frac{c_1}{c_2}=\frac{-20}{-40}=\frac{1}{2}\\\frac{1}{2}=\frac{1}{2}=\frac{1}{2}\\\text{Clearly}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\text{Hence},\\\text{Pair of linear equations has infinitely many solutions.}$$

(iv) x – 3y – 7 = 0 ...(i)
3x – 3y – 15 = 0 ...(ii)
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Then,
a₁ = 1, b₁ = – 3, c₁ = – 7
a₂ = 3, b₂ = – 3, c₂ = – 15
$$⇒\frac{a_1}{a_2}=\frac{1}{3}\\⇒\frac{b_1}{b_2}=\frac{-3}{-3}=1\\⇒\frac{c_1}{c_2}=\frac{-7}{-15}\\\frac{1}{3}≠1\\⇒\frac{a_1}{a_2}≠\frac{b_1}{b_2}\\\text{Pair of linear equations has unique solution.}\\x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}\\⇒x=\frac{(-3)×(-15)-(-3)×(-7)}{1×(-3)-3×(-3)}\\⇒x=\frac{45-21}{-3+9}\\⇒x=\frac{24}{6}\\⇒ x = 4\\y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\\⇒y=\frac{-7×3-(-15)×1}{-3+9}\\⇒y=\frac{-21+15}{16}=\frac{-6}{-6}=-1\\\text{Therefore, x = 4, y = – 1 is the solution of the given pair of equations.}$$

2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1

Sol. (i) 2x + 3y = 7
(a – b)x + (a + b)y = 3a + b – 2
Rearranging equations
2x + 3y – 7 = 0
(a – b)x + (a + b)y – (3a + b – 2) = 0
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Then,
a1 = 2, b1 = 3, c1 = – 7
a2 = (a – b), b2 = (a + b), c2 = – (3a + b – 2)
Given that Pair of linear equations have an infinite number of solutions.
$$\text{So, we need}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\⇒\\\frac{2}{a-b}=\frac{3}{a+b}=\frac{-7}{-(3a+b-2)}\\\text{From I and II, we have}\\\frac{2}{a-b}=\frac{3}{a+b}\\⇒ 2a + 2b = 3a – 3b\\⇒ 2a + 2b – 3a + 3b = 0\\⇒ – a + 5b = 0\\\text{From II and III, we have}\\\frac{3}{a+b}=\frac{7}{(3a+b-2)}\\⇒ 9a + 3b – 6 = 7a + 7b\\⇒ 2a – 4b – 6 = 0 ...(ii)\\By equations (i) and (ii)\\– a + 5b = 0\\2a – 4b – 6 = 0\\\text{Multiply equation(i) by 2, then}\\– 2a + 10b = 0\\2a – 4b – 6 = 0\\\text{By sum}\\6b – 6 = 0\\b = 1\\\text{Put the value b in equation (i)}\\– a + 5 × 1 = 0\\a = 5\\\text{Therefore, the value of a and b, that satisfies the conditions, are a = 5 and b = 1. For this values pair of linear equations has infinitely many solutions.}$$

(ii) 3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
Here a₁ = 3, b₁ = 1, c₁ = – 1
a₂ = 2k – 1, b₂ = k – 1, c₂ = – (2k + 1)
$$⇒\frac{a_1}{a_2}=\frac{3}{2k-1}\\⇒\frac{b_1}{b_2}=\frac{1}{k-1}\\⇒\frac{c_1}{c_2}=\frac{-1}{-(2k-1}\\\text{For a pair of linear equations to have no solutions.}\\\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}\\\frac{3}{2k-1}=\frac{1}{k-1}≠\frac{3}{2k-1}\\\text{From I and II, we have}\\\frac{3}{2k-1}=\frac{1}{k-1}\\3k – 3 = 2k – 1\\3k – 2k = – 1 + 3\\\text{Which gives k = 2}$$

3. Solve the following pair of linear equations by the substitution and cross multiplication methods :
8x + 5y = 9
3x + 2y = 4

Sol. (A) Substitution method :
8x + 5y = 9 ...(i)
3x + 2y = 4 ...(ii)
By equation (i)
8x = 9 – 5y

$$x = \frac{9-5y}{8}...(iii)\\\text{Put the value of x in equation (ii)}\\2(\frac{9-5y}{8})+2y=4\\27 – 15y + 16y = 32\\27 + y = 32\\y = 32 – 27\\y = 5\\x=\frac{9-5×5}{8}\\x=\frac{9-25}{8}\\\\x=\frac{-16}{8}= – 2\\\text{Therefore, x = – 2 and y = 5 is the solution of the given pair of equations.}$$
(B) Cross-multiplication method :
8x + 5y = 9
3x + 2y = 4
Rearranging them
8x + 5y – 9 = 0 ...(i)
3x + 2y – 4 = 0 ...(ii)

$$\frac{x}{-20+18}=\frac{y}{-27+32}=\frac{1}{16-15}\\\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}\\\text{Take}\frac{x}{-2}=\frac{1}{1}\\x = – 2\\\text{Again take}\\\frac{y}{5}=\frac{1}{1}\\\text{x = – 2, y = 5 is the solution of the given pair of equations.}$$

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 ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 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?

(v) The area of a rectangle gets reduced by 9 sq 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.

Sol. (i) Let be ₹ x is the fixed charges and ₹ y is the charges for food per day.
From the given information
x + 20y = 1000 ...(i)
x + 26y = 1180 ...(ii)
$$\frac{---}{– 6y = – 180}\text{By subtracting}\\y = ₹ 30\\\text{Put the value of y in equation (i)}\\x + 20 × 30 = 1000\\x + 600 = 1000\\x = ₹ 400\\\text{Therefore,}\\\text{Fixed charges = ₹ 400}\\\text{Food per day charges = ₹ 30}$$
(ii) Let be x and y are numerator and denominator of the fraction.
$$\text{Then the fraction is}\frac{x}{y}\\\text{From the given information.}\\\frac{x-1}{y}=\frac{1}{3}\\3x – 3 = y\\3x – y – 3 = 0 ...(i)\\\text{and again}\\\frac{x}{y+8}=\frac{1}{4}\\4x = y + 8\\4x – y – 8 = 0 ...(ii)\\\text{Solve equations (i) and (ii)}\\3x – y – 3 = 0\\4x – y – 8 = 0\\\frac{-++}{– x + 5 = 0}\text{Subtracting}\\– x = – 5\\x = 5\\\text{Put the value of x in equation (i)}\\3 × 5 – y – 3 = 0\\– y = – 12\\y = 12\\\text{Therefore,}\\\text{Numerator x = 5}\\\text{denominator y = 12}\\\text{Then}\\\text{Fraction}=\frac{5}{12}$$

(iii) Let be number of right answers x and wrong answers y.
From the given information.
3x – y = 40
3x – y – 40 = 0 ...(i)
Again
4x – 2y = 50
2(2x – y – 25) = 0
2x – y – 25 = 0 ...(ii)
By equations (i) and (ii)
3x – y – 40 = 0
2x – y – 25 = 0
$$\frac{-++}{x – 15 = 0}\text{By subtracting}\\x = 15\\\text{Put the value of x in equation (i)}\\3 × 15 – y = 40\\– y = – 5\\y = 5\\\text{Number of questions = x + y}\\= 15 + 5\\= 20$$

(iv) Suppose that
Speed of I car = x km/h
Speed of II car = y km/h
If the cars travel in the same direction.
$$x – y =\frac{100}{5}\\x – y = 20...(i)\\\text{If the cars travel towards each other}\\x + y =\frac{100}{1}\\x + y = 100 ...(ii)\\\text{By equations (i) and (ii)}\\x – y = 20\\x + y = 100\\\text{Adding}\\2x = 120\\x = 60 km/h\\\text{Put the value of x in equation (i)}\\60 – y = 20\\y = 40 km/h\\\text{Therefore,}\\\text{Speed of I car = 60 km/h}\\\text{Speed of II car= 40 km/h}$$

(v) Suppose that x and y are respectively the length and breadth of the rectangle.
Area of the rectangle = xy
From the given information
(x – 5)(y + 3) = (xy – 9)
xy + 3x – 5y – 15 = xy – 9
3x – 5y = – 9 + 15 + xy – xy
3x – 5y = 6 ...(i)
Again
(x + 3)(y + 2) = (xy + 67)
xy + 2x + 3y + 6 = xy + 67
2x + 3y = xy + 67 – xy – 6
2x + 3y = 61 ...(ii)
By equations (i) and (ii)
3x –5y = 6
2x + 3y = 61
Multiply equation (i) by 2 and equation (ii) by 3, then
6x – 10y = 12
6x + 9y = 183
$$\frac{---}{– 19y = – 171}\text{Subtracting}$$
y = 9
Put the value of y in equation (i)
3x – 5 × 9 = 6
3x = 6 + 45
3x = 51
x = 17
Therefore,
Length of rectangle (x) = 17 units
Breadth of rectangle (y) = 9 units

Exercise 3.6

1. Solve the following pairs of equations by reducing them to a pair of linear equations :
$$\textbf{(i)}\frac{1}{2x}+\frac{1}{3y}= 2 \text{and}\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}\\\textbf{(ii)}\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}= 2 \text{and}\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}= – 1\\\textbf{(iii)}\frac{4}{x}+3y=14\text{and}\frac{3}{x}-4y=23\\\textbf{(iv)}\frac{5}{(x-1)}+\frac{1}{(y-1)}= 2 \\\text{and}\frac{6}{(x-1)}-\frac{3}{(y-2)}=1\\\textbf{(v)}\frac{7x-2y}{xy}= 5 \text{and}\frac{8x+7y}{xy}=15\\\textbf{(vi)} 6x + 3y = 6xy and 2x + 4y = 5xy\\textbf{vii}\frac{10}{(x+y)}+\frac{1}{(3x-y)}=4\text{and}\\\frac{15}{(x+y)}-\frac{5}{(3x-y)}=-2\\\textbf{viii}\frac{1}{(3x+y)}+\frac{1}{(3x-y)}=\\\frac{3}{4}\text{and}\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8}$$

$$\text{Sol. (i)}\frac{1}{2x}+\frac{1}{3y}=2\\\frac{1}{3x}+\frac{1}{2y}=\frac{13}{6}\\\text{Let}\frac{1}{x}=u\text{and}\frac{1}{y}= v\\\text{Then,}\frac{u}{2}+\frac{v}{3}=2\\3u + 2v = 12\\3u + 2v – 12 = 0 ...(i)\\\text{Again}\\\frac{u}{3}+\frac{v}{2}=\frac{13}{6}\\\frac{2u+3v}{6}=\frac{13}{6}\\2u + 3v – 13 = 0 ...(ii)\\\text{By equations (i) and (ii)}\\3u + 2v – 12 = 0\\2u + 3v – 13 = 0\\\text{Multiply equation (i) by 2}\text{ and equation (ii) by 3, then}\\6u + 4v – 24 = 0\\6u + 9v – 39 = 0\\\frac{--+}{-5v+15=0}\\\text{Subtracting}\\– 5v = – 15\\v = 3\\\text{Put the value of v in equation (i)}\\u = 2\\\text{but}\\\frac{1}{x}=u,\frac{1}{y}=v,\\\\\frac{1}{x}=2,\frac{1}{y}=3,\\\text{then}\\x=\frac{1}{2},y=\frac{1}{3}$$

$$\textbf{(ii)}\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\\\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}= – 1\\\text{Let us put}\frac{1}{\sqrt{x}}= \text{u and}\frac{1}{\sqrt{y}}= v.\text{ Then the given equations}\\2(\frac{1}{\sqrt{x}})+3(\frac{1}{\sqrt{y}})=2\\4(\frac{1}{\sqrt{x}})-9(\frac{1}{\sqrt{y}})=-1\\\text{Can be written as}\\2u + 3v = 2 ...(i)\\4u – 9v = – 1 ...(ii)\\\text{Equations (i) and (ii) from a pair of linear equations in the general form.}\\\text{Solve these equation}\\2u + 3v = 2\\4u – 9v = – 1\\\text{Multiply equation (i) by 4 and equation (ii) by 2, then}\\8u + 12v = 8\\8u – 18v = – 2\\\frac{-++}{30v = 10}\text{Subtracting}\\v=\frac{1}{3}\\\text{Put the value of v in equation (i)}\\u=\frac{1}{2}\\\text{Now, substituting}\frac{1}{\sqrt{x}}\\\text{for u we have}\\\frac{1}{\sqrt{x}}=\frac{1}{2}\\\text{Squaring both sides, we get}\\x = 4\\\text{Similarly, substituting}\frac{1}{\sqrt{y}}\text{for v.}\\\frac{1}{\sqrt{y}}=\frac{1}{3}\\\text{Squaring both sides, we get}\\\frac{1}{y}=\frac{1}{9}\\y = 9\\\text{Hence, x = 4, y = 9 is the required solution of the given pair of equations.}$$

$$\textbf{(iii)}\frac{4}{x}+3y=14\\\frac{3}{x}-4y=23\\\text{Let us put}\frac{1}{x}=u\\\text{Then the given equations}\\4(\frac{1}{x})+3y=14\\3(\frac{1}{x})-4y=23\\\text{Can be written as}\\4u + 3y = 14 ...(i)\\3u – 4y = 23 ...(ii)\\\text{Multiply equation (i) by 3 and equation (ii) by 4, we get}\\12u + 9y = 42\\12u – 16y = 92\\\frac{-+-}{25y = – 50}\text{Subtracting}\\25y = – 50\\y = – 2\\\text{Put the value of y in equation (i)}\\4u – 6 = 14\\4u = 20\\u = 5\\\text{Now, substituting}\frac{1}{x}\text{for u.}\\\text{We have}\\\frac{1}{x},\text{then}x=\frac{1}{5}\\\text{Hence,}x=\frac{1}{5},\text{y = – 2 is the required solution of the given pair of equations.}$$

$$\textbf{(iv)}\frac{4}{(x-1)}+\frac{4}{(y-2)}=2 ...(i)\\\frac{6}{(x-1)}+\frac{3}{(y-2)}=1 ...(ii)\\\text{Let us put}\frac{1}{x-1}= u \text{and}\frac{1}{y-2}= v.\\\text{in equations (i) and (ii) Then}\\5u + v = 2 ...(iii)\\6u – 3v = 1 ...(iv)\\15u + 3v = 6\\6u – 3v = 1\\\text{Adding}\\21u = 7\\u=\frac{1}{3}\\\text{Put the value of u in equation (iii)}\\5×\frac{1}{3}+v= 2\\v = 2-\frac{5}{3}\\v=\frac{1}{3}\\\text{Now, substituting}\frac{1}{x-1}\text{for u, we get}\\\frac{1}{x-1}=\frac{1}{3}\\x – 1 = 3\\x = 4\\\text{Similarly, substituting}\frac{1}{y-2}\\\text{for v, we get}\\\frac{1}{y-2}=\frac{1}{3}\\3 = y – 2\\y = 5\\\text{Hence, x = 4, y = 5 is the required solution of the given pair of equations}$$

$$\textbf{(v)}\frac{7x-2y}{xy}=5...(i)\\\frac{8x+7y}{xy}=15...(ii)\\\text{Rearranging equations (i) and (ii)}\\\frac{7x}{xy}-\frac{2y}{xy}=5\\\frac{8x}{xy}-\frac{7y}{xy}=15\\\text{we get}\\\frac{7}{y}-\frac{2}{x}=5\\\frac{8}{y}+\frac{7}{x}=15\\\text{Let us put}\frac{1}{x}= u \text{and}\frac{1}{y}=v\\\text{Then, we get}\\7v – 2u = 5 ...(iii)\\8v + 7u = 15 ...(iv)\\\text{Multiply equation (i) by 8 and equation (ii) by 7, then}\\56v – 16u = 40\\56v + 49u = 105\\\frac{---}{– 65u = – 65}\text{Subtracting}\\u = 1\\\text{Put the value of u in equation (iii)}\\7v – 2 = 5\\v = 1\\\text{Now, substituting}\frac{1}{x}\text{for u and}\frac{1}{y}\text{for v.}\\\frac{1}{x}=1,\frac{1}{y}=1\\x = 1, y = 1\\\text{Hence, x = 1, y = 1 is the required solution of the given pair of equations.}$$

$$\textbf{(vi)}6x + 3y = 6xy ...(i)\\2x + 4y = 5xy ...(ii)\\\text{Rearranging equations (i) and (ii), we get}\\\frac{6x}{xy}+\frac{3y}{xy}=6\\\text{and}\\\frac{2x}{xy}+\frac{4y}{xy}=5\\\text{Then}\\\frac{6}{y}+\frac{3}{x}=6..(iii)\\\frac{2}{y}+\frac{4}{x}=6...(iv)\\\text{Let us put}\frac{1}{x}= u \text{and}\frac{1}{y}= \text{v. Then we get}\\6v + 3u = 6 ...(v)\\2v + 4u = 5 ...(vi)\\\text{Solve the equations (v) and (vi) by cross multiplication method.}\\\frac{v}{15-24}=\frac{u}{12-30}=\frac{-1}{24-6}\\\frac{v}{-9}=\frac{u}{-18}=frac{1}{-18}\\\text{Take}\\\frac{v}{-9}=-frac{-1}{18}\\v=\frac{1}{2}\\\text{Again take}\\\frac{u}{-18}=\frac{-1}{18}\\u = 1\\\text{Now, substituting}\frac{1}{x}\text{for u and}\frac{1}{y}\text{for v.}\\\frac{1}{x}=1,\frac{1}{y}=\frac{1}{2}\\\text{Then}\\x = 1, y = 2\\\text{Hence, x = 1, y = 2 is the required solution of the given pair of equations.}$$

$$\textbf{(vii)}\frac{10}{x+y}+\frac{2}{x-y}=4\\\frac{15}{x+y}+\frac{5}{x-y}=-2\\\text{Let us put}\frac{1}{x+y}= u \text{and}\frac{1}{x-y}=v \text{in equations (i) and (ii), we get}\\10u + 2v = 4\\15u – 5v = – 2\\\text{Rearranging}\\10u + 2v – 4 = 0 ...(iii)\\15u – 5v + 2 = 0 ...(iv)\\\text{Solve equations (iii) and (iv) by cross multiplication method.}\\\frac{u}{4-20}=\frac{v}{-60-20}=\frac{1}{-50-30}\\\frac{u}{-16}=\frac{v}{-80}=\frac{1}{-80}\\\text{Take}\frac{u}{-16}=\frac{v}{-80}\\u=\frac{1}{5}\\\text{Again take}\\\frac{v}{-80}=\frac{1}{-80}\\v = 1\\\text{Now, substituting}\frac{1}{x+y}\text{for u, we have}\frac{1}{x+y}=\frac{1}{5}\\x + y = 5 ...(v)\\\text{Similarly, substituting}\frac{1}{x-y}\text{for v, we get}\\\frac{1}{x-y}=1\\x – y = 1 ...(vi)\\\text{By equations (v) and (vi)}\\x + y = 5\\x – y = 1\\\text{Adding both equations, we get}\\2x = 6\\x = 3\\\text{By equations (v) – (vi)}\\2y = 4\\y = 2\\\text{Hence, x = 3, y = 2 is the required solution of the given pair of equations.}$$

$$\textbf{(viii)}\frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4}...(i)\\\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=\frac{-1}{8}...(ii)\\\text{Let us put}\frac{1}{3x+y}= u \text{and}\frac{1}{3x-y}= v\text{ in equations (i) and (ii)}\\u+v=\frac{3}{4}\\⇒ 4u + 4v = 3 ...(iii)\\\frac{u}{2}-\frac{v}{2}=\frac{-1}{8}\\⇒\frac{u-v}{2}=\frac{-1}{8}\\⇒ 4u – 4v = – 1 ...(iv)\\\text{Add equations (iii) and (iv)}\\8u = 2\\u=\frac{2}{8}\\u=\frac{1}{4}\\\text{Subtracting equation (iii) from equation (iv)}\\8v = 4\\v=\frac{1}{2}\\\text{Now, substituting}\frac{1}{3x+y}\text{for u, we have}\\\frac{1}{3x+y}=\frac{1}{4}\\Then 3x + y = 4 ...(v)\\\text{Similarly, substituting}\frac{1}{3x-y}\text{for v, we get}\\\frac{1}{3x-y}=\frac{1}{2}\\\text{Then 3x – y = 2 ...(vi)}\\\text{Equations (v) and (vi) from a pair of linear equations in the general form. Use elimination method solve these equations we get}\\x = 1, y = 1\\\text{Hence, x = 1, y = 1 is the required solution of the given pair of equations.}$$

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.

(iii) 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.

Sol. (i) Let the speed of the Ritu in still water be
x km/h and speed of the stream be y km/hr. Then the speed of the boat in downstream
= (x + y) km/h.
The speed of the boat in upstream
= (x –y) km/h
In the first case, Ritu goes 20 km downstream in 2 hours.
$$\text{Time} =\frac{Distance}{Speed}\\2 =\frac{20}{x+y}\\\text{Then}\\x + y = 10 ...(i)\\\text{In the second case, Ritu goes 4 km upstream in 2 hours.}\\2=\frac{4}{x-y}\\x – y = 2 ...(ii)\\\text{Adding equations (i) and (ii), we get}\\2x = 12\\x = 6\\\text{Subtracting equation (ii) from (i)}\\2y = 8\\y = 4\\\text{Hence, the speed of the Ritu in still water is 6 km/h and the speed of stream is 4 km/hr.}$$

(ii) Let be x and y are the number of days taken by 1 woman and 1 man to finish the embroidery work.

$$\text{Work done by women in one day}=\frac{1}{x}\\\text{Work done by men in one day }=\frac{1}{y}\\\text{As per question given}\\\frac{2}{x}+\frac{5}{y}=\frac{1}{4}...(i)\\\frac{3}{x}+\frac{6}{y}=\frac{1}{3}...(i)\\\text{Now, put}\frac{1}{x}= u \text{and}\frac{1}{y}= v \text{in equation (i)}\\2u + 5v =\frac{1}{4}\\8u + 20v = 1 ...(iii)\\\text{Similarly, put}\frac{1}{x}= u \text{and}\frac{1}{y}= v\text{ in equation (ii)}\\3u + 6v =\frac{1}{3}\\9u + 18v = 1 ...(iv)\\\text{Solve equations (iii) and (iv) by cross multiplication method.}\\\frac{u}{20-18}=\frac{v}{9-8}=\frac{-1}{144-180}\\\text{Then}\\\frac{u}{2}=\frac{v}{2}=\frac{-1}{-36}\\\frac{u}{2}=\frac{1}{36}\text{and}\frac{v}{1}=\frac{1}{36}\\u =\frac{1}{18}\text{and}v=\frac{1}{36}\\\text{You know that}u=\frac{1}{x}\text{and}v =\frac{1} {y}\\\text{Substitute the value of u and v to get}\\\frac{1}{18}=\frac{1}{x}\text{and}\frac{1}{36}=\frac{1}{y}\\x = 18 \text{and} y = 36\\\text{Hence,}\\\text{Number of days taken 1 woman alone to finish the work (x) = 18 days Number of days take 1 man alone to finish the work (y) = 36 days}$$

(iii) Suppose that
Speed of the train = x km/h
Speed of the bus = y km/h
According to question (I Case)

$$\frac{60}{x}+\frac{240}{y}=4\\4\biggl(\frac{15}{x}+\frac{60}{y}\biggr)=4\\\frac{15}{x}+\frac{60}{y}=1...(i)\\\text{According to question (II Case)}\\\frac{100}{x}+\frac{200}{y}=\frac{25} {6}\\25\biggl(\frac{4}{x}+\frac{8}{y}\biggr)=\frac{25}{6}\\\frac{4}{x}+\frac{8}{y}=\frac{1}{6}\\\frac{24}{x}+\frac{48}{y}=1...(ii)\\\text{Now, put}\frac{1}{x}= u\text{ and}\frac{1}{y}= v\text{in equations}\\\text{(i) and (ii) then, we get}\\15u + 60v = 1 ...(iii)\\24u + 48v = 1 ...(iv)\\\text{Solve equations (iii) and (iv) by cross multiplication method}\\\frac{u}{60-48}=\frac{v}{24-15}=\frac{-1}{720-1140}\\\frac{u}{12}=\frac{v}{9}=\frac{-1}{-720}\\\text{Take}\\\frac{u}{12}=\frac{1}{720}\text{and}\\\frac{v}{9}=\frac{1}{720}\\u=\frac{1}{60}\text{and}v=\frac{1}{180}\\\text{We know that}\\u=\frac{1}{x}\text{and}v=\frac{1}{y}\\\text{Now, put the value of u and v}\\\frac{1}{60}=\frac{1}{x}\text{and}\\\frac{1}{80}=\frac{1}{y}\\x = 60 \text{and} y = 80\\\text{Hence,}\\\text{Speed of train (x) = 60 km/h.}\\\text{Speed of bus (y) = 80 km/h.}$$

Exercise 3.7 (Optional)

1. The ages of two friends Ani and Biju differ by 3 years. Ani's 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.

Sol. Let us consider
Age of Ani = A
Age of Biju = B
Case I : If Ani > Biju.
Then according to question
A – B = 3 ...(i)
Age of Dharam = 2A
Age of Cathy =B/2
2A-B/2=30
4A – B = 60 ...(ii)
Solve the equations (i) and (ii) by Elimination method.
A – B = 3
4A – B = 60
$$\frac{-+-}{– 3A = – 57}\text{Subtracting}$$
A = 19 years
Put the v alue of A in equation (i)
B = 16 years
In this case,
Ani's age = 19 years
Biju's age = 16 years
Case II : If Biju > Ani
B – A = 3 ...(iii)
But no change in equation (ii)
4A –B = 60 ...(ii)
Solve equation (ii) and (iii)
3A = 63
A = 21
B = 24
In II Case
Ani's age = 21 years
Biju's age = 16 years

2. One says, ''Give me a hundred, friend! I shall then become twice as rich as you''. The other replies, ''If you give me ten, I shall be six times as rich as you''. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
[Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)]

Sol. Let us The money with the first person = ₹ x
Money with the second person = ₹ y
As per question given (one says)
(x + 100) = 2(y – 100)
x + 100 = 2y – 200
x – 2y + 300 = 0 ...(i)
Other Replies
6(x – 10) = (y + 10)
6x – 60 = y + 10
6x – y – 70 = 0 ...(ii)
Rewrite the equations and solv e the cross-multiplication method.
x – 2y + 300 = 0
6x – y – 70 = 0
Then
$$\frac{x}{140+300}=\frac{y}{1800+70}=\frac{1}{-1+12}\\\frac{x}{440}=\frac{y}{1870}=\frac{1}{11}\\\text{Take}\\\frac{x}{440}=\frac{1}{11}\text{and}\frac{y}{1870}=\frac{1}{11}\\x = ₹ 40\space \text{and} \space y = ₹ 170\\\text{Hence}\\\text{First person's money = ₹ 40}\\\text{Second person's money = ₹ 170}$$

3. A train covered a certain distance at a uniform speed. If the train would have been 10 kmh–1 faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 kmh–1; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Sol. Let us distance covered by train is d km.
Train's uniform speed = x km/h
Scheduled time = t hours
$$Time =\frac{Distance}{Speed}\\t =\frac{d}{x}...(i)\\\text{According to question}\\\frac{d}{x+10}=t-2\\\text{Put the value of t}\\\frac{d}{x+10}=\frac{d}{x}-2\\d[\frac{1}{x+1}-\frac{1}{x}]=-2\\d[\frac{x-x-10}{x(x+10)}]=-2\\\frac{-10d}{x^2+10x}=-2\\d=\frac{x^2+10x}{5}...(ii)\\\text{Again according to question}\\\frac{d}{x-10}= t + 3\\\text{Put the value t in equation (i), we get}\\\frac{d}{x-10}= \frac{d}{x}+3
$$Time =\frac{Distance}{Speed}\\t =\frac{d}{x}...(i)\\\text{According to question}\\\frac{d}{x+10}=t-2\\\text{Put the value of t}\\\frac{d}{x+10}=\frac{d}{x}-2\\d[\frac{1}{x+1}-\frac{1}{x}]=-2\\d[\frac{x-x-10}{x(x+10)}]=-2\\\frac{-10d}{x^2+10x}=-2\\d=\frac{x^2+10x}{5}...(ii)\\\text{Again according to question}\\\frac{d}{x-10}= t + 3\\\text{Put the value t in equation (i), we get}\\\frac{d}{x-10}= \frac{d}{x}+3\\d[\frac{1}{x-10}-\frac{1}{x}]=3\\d[\frac{x-x+10}{x(x-10)}]=3\\d=\frac{3x^2-30x}{10}...(iii)\\\text{Comparing equations (ii) and (iii)}\\\frac{x^2+10x}{5}=\frac{3x^2-30x}{10}\\2x^2 + 20x = 3x^2 – 30x\\x^2 – 50x = 0\\x(x – 50) = 0\\x = 0 (Impossible)\\x – 50 = 0\\x = 50 km/h\\\text{Put the value of x in equation (ii)}\\d=\frac{50^2+10×50}{5}\\d=\frac{2500+500}{5}\\d=\frac{3000}{5}\\d = 600 km\\\text{Hence, distance covered by the train = 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.

Sol. Suppose that
Number of students in one Row = x
Number of Row = y
Then
Number of students in class = xy ...(i)
According to question (I case)
Number of students in one Row = (x + 3)
Number of Row = y – 1
Then
Number of students in class
= (x + 3)(y – 1) ...(ii)
By equations (i) and (ii)
xy = (x + 3)(y – 1)
x – 3y + 3 = 0 ...(iii)
Accoding to question (II Case)
Number of students in one Row = (x – 3)
Number of Row = (y + 2)
Then
Number of students in class
= (x – 3)(y + 2) ...(iv)
By equations (i) and (iv)
xy = (x – 3)(y + 2)
xy = xy + 2x – 3y – 6
2x – 3y – 6 = 0 ...(v)
Rewrite equations (iii), (v) and solve by elimination method.
x – 3y + 3 = 0
2x – 3y – 6 = 0
$$\frac{-++}{– x + 9 = 0}\text{Adding}$$
x = 9
Put the value of x in equation (iii)
y = 4
Hence
Number of students in class = xy
= 9 × 4
= 36 students

5. In a DABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.

Sol. Given that
∠C = 3∠B = 2(∠A + ∠B)
T ake
∠C = 3∠B ...(i)
Again take
3∠B = 2(∠A + ∠B)
∠B = 2∠A ...(ii)
Put the value of ∠B in equation (i)
∠C = 6∠A ...(iii)
We know that
Sum of internal angles in a D is 180°.
Then
∠A + ∠B + ∠C = 180° ...(iv)
Put the value ∠B and ∠C in equation (iv)
∠A + 2∠A + 6∠A = 180°
9∠A = 180°
∠A = 20°
Put the value of ∠A in equation (ii) and (iii)
∠B = 40°
∠C = 120°
Hence, the angles are 20°, 40° and 120°.

6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the Y-axis.

Sol. 5x – y = 5 ...(i)
3x – y = 3 ...(ii)
By equation (i)

x	-1	0	1
y	-10	-5	0

Put the value of x in equation (i)
5 × (– 1) – y = 5
– y = 10
y = – 10
Put x = 0
y = – 5
Put x = 1
5 – y = 5
y = 0
By equation (ii)

x	-1	0	1
y	-6	-3	0

Put x = – 1 in equation (ii)
3 × (– 1) – y = 3
y = – 6
Put x = 0
y = – 3
Put x = 1
3 – y = 3
y = 0

7. Solve the following pair of linear equations :

(i) px + qy = p – q
qx – py = p + q
(ii) ax + by = c
bx + ay = 1 + c
$$\text{(iii)}\frac{x}{a}-\frac{x}{b}=0$$
ax + by = a² + b²
(iv) (a – b)x + (a + b)y = a² – 2ab – b²
(a + b)(x + y) = a² + b²
(v) 152x – 378y = – 74
– 378x + 152y = – 604

Sol. (i) px + qy = p – q ...(i)
qx – py = p + q ...(ii)
Solve by cross-multiplication method.
$$\frac{x}{pq+q^2-(-p^2+pq)}=\frac{y}{pq-q^2-(p^2+pq)}=\frac{-1}{-p^2-q^2}\\\frac{x}{pq+q^2+p^2-pq)}=\frac{y}{pq-q^2-p^2-pq}=\frac{1}{p^2-q^2}\\\frac{x}{p^2+q^2}=\frac{y}{-(p^2+q^2)}=\frac{1}{p^2+q^2}\\\frac{x}{p^2+q^2}=\frac{1}{p^2+q^2}\\\text{and}\frac{y}{-(p^2+q^2)}=\frac{1}{p^2+q^2}\\x = 1 \text{and} y = – 1.$$

(ii) ax + by = c ...(i)
bx + ay = 1 + c ...(ii)
Solve by cross-multiplication method.
$$\frac{x}{b+bc+ac}=\frac{y}{bc-a-ac}=\frac{-1}{a^2-b^2}\\\text{From I and II, we have}\\\frac{x}{b+bc+ac}=\frac{-1}{a^2-b^2}\\x=\frac{−(b+bc-ac)}{a^2-b^2}\\x=\frac{ac-bc-b}{a^2-b^2}\\x=\frac{c(a-b)-b}{a^2-b^2}\\\text{From II and III, we have}\\\frac{y}{bc-a-ac}=\frac{-1}{a^2-b^2}\\y=\frac{a+ac-bc}{a^2-b^2}\\y=\frac{c(a-b)+a}{a^2-b^2}$$

$$\textbf{(iii)}\frac{x}{a}-\frac{y}{b}=0...(i)\\ax + by = a2 + b2 ...(ii)\\\text{Rearranged equation (i)}\\bx – ay = 0 ...(iii)\\\text{Rewrite equations (ii) and (iii) and solve by cross-multiplication method.}\\ax + by = a2 + b2\\bx – ay = 0\\\text{Then}\\\frac{x}{0-a^3-ab^2}=\frac{y}{a^2b+b^3-0}=\frac{-1}{-a^2-b^2}\\\text{Now, take}\\\frac{x}{-a^3-ab^2}=\frac{-1}{-a^2-b^2}\\x =\frac{-a(a^2+b^2)}{a^2+b^2}\\x = – a\\\text{Now, take}\\\frac{y}{a^2b+b^3}=\frac{-1}{-a^2-b^2}\\y=\frac{b(a^2+b^2)}{a^2+b^2}\\y = b$$

(iv) (a – b)x + (a + b)y = a² – 2ab – b²
(a + b)(x + y) = a² + b²
Rearranged given equations
(a – b)x + (a + b)y = a² – 2ab – b² ...(i)
(a + b)x + (a + b)y = a² + b² ...(ii)
$$\frac{----}{(a – b)x – (a + b)x = – 2ab – 2b^2}\text{Subtract}$$
x[a – b – a – b] = – 2b(a + b)
– 2bx = – 2b(a + b)
x = (a + b)
Put the value of x in equation (i)
(a – b)(a + b) + (a + b)y = a² – 2ab – b²
a² – b² + (a + b)y = a² – 2ab – b²
(a + b)y = a² – 2ab – b² – a² + b²

$$\frac{−2ab}{a +b}\\\text{Hence}\\x = (a + b), y =\frac{−2ab}{a +b}\\\text{is required solution of given pair of equations.}$$

(v) 152x – 378y = – 74 ...(i)
– 378x + 152y = – 604 ...(ii)
$$\text{Solve by cross-multiplication method.}\\\frac{x}{228312+11248}=\frac{y}{27972+91808}\\=\frac{-1}{23104-142884}\\\frac{x}{239560}=\frac{y}{119780}\\=\frac{-1}{-119780}\\\text{Now take}\\\frac{x}{239560}=\frac{-1}{-119780}\text{and}\frac{y}{119780}=\frac{1}{119780}\\x = 2 and y = 1$$

Hence,
Required D formed DABC. Its vertices are A(1, 0),
B(0, – 3) and C(0, – 5).

8. ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

Sol. We know that
Sum of opposite angles in cyclic quadrilateral is 180°.
Then
∠A + ∠C = 180°
4y + 20 – 4x = 180°
– 4x + 4y = 160°
– x + y = 40° ...(i)
Again
∠B + ∠D = 180°
3y – 5 – 7x + 5 = 180°
– 7x + 3y = 180° ...(ii)
Rewrite and solve equations (i) and (ii)
– x + y = 40
– 7x + 3y = 180
Multiply equation (i) by 3, then
– 3x + 3y = 120
– 7x + 3y = 180
$$\frac{+--}{4x = – 60}\text{By subtracting}$$
4x = – 60
x = – 15
Put value of x in equation (i)
15 + y = 40°
y = 25°
Hence,
∠A = 4y + 20 = 120°
∠B = 3y – 5 = 70°
∠C = – 4x = 60°
∠D = – 7x + 5 = 110°