# NCERT Solutions Class 10 for Maths Chapter 3 Linear Equations in Two Variables

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# Class 10 Maths Chapter 3 – Linear Equations in Two Variables

## Exercise 3.1 (Page 44)

Question 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.

Let the present age of Aftab be ‘x’.
And, the present age of his daughter be ‘y’.

Now, we can write, seven years ago,
Age of Aftab = x-7
Age of his daughter = y-7

According to the question,
x−7 = 7(y−7)
⇒x−7 = 7y−49
⇒x−7y = −42 ………………………(i)

Also, three years from now or after three years,
Age of Aftab will become = x+3.
Age of his daughter will become = y+3

According to the situation given,
x+3 = 3(y+3)
⇒x+3 = 3y+9
⇒x−3y = 6       …………..…………………(ii)

Subtracting equation (i) from equation (ii) we have
(x−3y)−(x−7y) = 6−(−42)
⇒−3y+7y = 6+42
⇒4y = 48
⇒y = 12

The algebraic equation is represented by
x−7y = −42
x−3y = 6
For, x−7y = −42 or x = −42+7y

The solution table is

For, x−3y = 6 or x = 6+3y

The solution table is

The graphical representation is:

Question 2:
The coach of a cricket team buys 3 bats and 6 balls for Rs.3900. Later, she buys another bat and 3 more balls of the same kind for Rs.1300. Represent this situation algebraically and geometrically.

Let cost of one bat = Rs x
Cost of one ball = Rs y
3 bats and 6 balls for Rs 3900 So that
3+ 6y = 3900 … (i)
Dividing equation by 3, we get
+ 2y = 1300
Subtracting 2y both side we get
x = 1300 – 2y
Putting y = -1300, 0 and 1300 we get
x = 1300 – 2 (-1300) = 1300 + 2600 = 3900
= 1300 -2(0) = 1300 – 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = – 1300

Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x + 2= 1300 … (ii)
Subtracting 2y both side we get
= 1300 – 2y
Putting y = – 1300, 0 and 1300 we get
x = 1300 – 2 (-1300) = 1300 + 2600 = 3900
= 1300 – 2 (0) = 1300 – 0 = 1300
= 1300 – 2(1300) = 1300 – 2600 = -1300

Algebraic representation
3+ 6y = 3900 … (i)
+ 2= 1300 … (ii)
Graphical representation,

Question 3:
The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically and geometrically.

Let cost each kg of apples = Rs x
Cost of each kg of grapes = Rs y
Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160
So that
= 160 … (i)
2x = 160 – y
x = (160 – y)/2
Let y = 0 , 80 and 160,  we get
x = (160 – ( 0 )/2 = 80
x = (160- 80 )/2 = 40
x = (160 – 2 × 80)/2 = 0

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x + 2= 300 … (ii)
Dividing by 2 we get
2x + y = 150
Subtracting 2x both side, we get
= 150 – 2x
Putting x = 0 , 50 , 100 we get
= 150 – 2 × 0 = 150
= 150 – 2 ×  50 = 50
= 150 – 2 × (100) = -50

Algebraic representation,
2y = 160 … (i)
4x + 2y = 300 … (ii)

Graphical representation,

## Exercise 3.2 (Page 49)

Question 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 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

(i)Let there are x number of girls and y number of boys. As per the given question, the algebraic expression can be represented as follows.

x +y = 10
x– y = 4

Now, for x+y = 10 or x = 10−y, the solutions are;

For x – y = 4 or x = 4 + y, the solutions are;

The graphical representation is as follows;

(ii) Let 1 pencil costs Rs.x and 1 pen costs Rs.y.
According to the question, the algebraic expression cab be represented as;

5x + 7y = 50
7x + 5y = 46

For, 5x + 7y = 50 or x = (50-7y)/5, the solutions are

For 7x + 5y = 46 or x = (46-5y)/7, the solutions are;

Hence, the graphical representation is as follows;|

Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.

Question 2:
On comparing the ratios a1/a2 , b1/b2 , 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

(i) Given expressions;

5x−4y+8 = 0
7x+6y−9 = 0

Comparing these equations with a1x+b1y+c1 = 0
And a2x+b2y+c2 = 0
We get,

a1 = 5, b1 = -4, c1 = 8
a2 = 7, b2 = 6, c2 = -9

(a1/a2) = 5/7
(b1/b2) = -4/6 = -2/3
(c1/c2) = 8/-9

Since, (a1/a2) ≠ (b1/b2)
So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(ii) Given expressions;

9x + 3y + 12 = 0
18x + 6y + 24 = 0

Comparing these equations with a1x+b1y+c1 = 0
And a2x+b2y+c2 = 0

We get,

a1 = 9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24

(a1/a2) = 9/18 = 1/2
(b1/b2) = 3/6 = 1/2
(c1/c2) = 12/24 = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2)
So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) Given Expressions;

6x – 3y + 10 = 0
2x – y + 9 = 0

Comparing these equations with a1x+b1y+c1 = 0
And a2x+b2y+c2 = 0

We get,

a1 = 6, b1 = -3, c1 = 10
a2 = 2, b2 = -1, c2 = 9

(a1/a2) = 6/2 = 3/1
(b1/b2) = -3/-1 = 3/1
(c1/c2) = 10/9

Since (a1/a2) = (b1/b2) ≠ (c1/c2)
So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point and there is no possible solution for the given pair of equations.

Question 3:
On comparing the ratio, (a1/a2) , (b1/b2) , (c1/c2) find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii)(3/2)x+(5/3)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v)(4/3)x+2y = 8 ; 2x + 3y = 12

(i) 3x + 2y = 5 ; 2x – 3y = 7
a1/a2 = 3/2
b1/b2 = -2/3 and
c1/c2 = 5/7
Hence, a1/a2 ≠ b1/b2
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii) 2x – 3y = 8 ; 4x – 6y = 9
a1/a2 = 2/4 = 1/2
b1/b2 = -3/-6 = 1/2 and
c1/c2 = 8/9
Hence, a1/a2 = b1/bc1/c2
Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 3/2x + 5/3y = 7 ; 9– 10y = 14
a1/a2 = 3/2/9 = 1/6
b1/b2 = 5/3/-10 = -1/6 and
c1/c2 = 7/14 = 1/2
Hence, a1/a2 ≠ b1/b2
Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv) 5x – 3y = 11 ; – 10x + 6y = –22
a1/a2 = 5/-10 = -1/2
b1/b2 = -3/6 = -1/2 and
c1/c2 = 11/-22 = -1/2
Hence, a1/a2 = b1/b2 =c1/c2
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

(v) 4/3x + 2y =8 ; 2x + 3y = 12
a1/a2 = 4/3/2 = 2/3
b1/b2 = /3 and
c1/c2 = 8/12 = 2/3
Hence, a1/a2 = b1/b2 =c1/c2
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

Question 4:
Which of the following pairs of linear equations are consistent/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

(i)Given, x + y = 5 and 2x + 2y = 10

(a1/a2) = 1/2
(b1/b2) = 1/2
(c1/c2) = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2)
∴The equations are coincident and they have infinite number of possible solutions.

So, the equations are consistent.
For, x + y = 5 or x = 5 – y

For 2x + 2y = 10 or x = (10-2y)/2

So, the equations are represented in graphs as follows

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(ii) x – y = 8, 3x – 3y = 16
a1/a2 = 1/3
b1/b2 = -1/-3 = 1/3 and
c1/c2 = 8/16 = 1/2
Hence, a1/a2 = b1/bc1/c2

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a1/a2 = 2/4 = 1/2
b1/b2 = -1/2 and
c1/c2 = -6/-4 = 3/2
Hence, a1/a2 ≠ b1/b2
Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.2x + y – 6 = 0
y = 6 – 2x

And, 4x – 2y -4 = 0
y = 4x – 4/2

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a1/a2 = 2/4 = 1/2
b1/b2 = -2/-4 = 1/2 and
c1/c2 = 2/5
Hence, a1/a2 = b1/bc1/c2
Therefore, these linear equations are parallel to each other and thus, have no possible solution. Hence, the pair of linear equations is inconsistent.

Question 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.

:Let us consider.
The width of the garden is x and length is y.
Now, according to the question, we can express the given condition as;

y – x = 4
and
y + x = 36

Now, taking y – x = 4 or y = x + 4

For y + x = 36, y = 36 – x

The graphical representation of both the equation is as follows;

From the graph you can see, the lines intersects each other at a point(16, 20). Hence, the width of the garden is 16 and length is 20.

Question 6:
Given the 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

(i) Intersecting lines:
For this condition,
a1/a2 ≠ b1/b2
The second line such that it is intersecting the given line is
2x + 4y – 6 = 0 as
a1/a2 = 2/2 = 1
b1/b2 = 3/4 and
a1/a2 ≠ b1/b2

(ii) Parallel lines
For this condition,a1/a2 = b1/bc1/c2
Hence, the second line can be
4x + 6y – 8 = 0 as
a1/a2 = 2/4 = 1/2
b1/b2 = 3/6 = 1/2 and
c1/c2 = -8/-8 = 1
and a1/a2 = b1/bc1/c2

(iii) Coincident lines
For coincident lines,
a1/a2 = b1/b2 =c1/c2
Hence, the second line can be
6x + 9y – 24 = 0 as
a1/a2 = 2/6 = 1/3
b1/b2 = 3/9 = 1/3 and
c1/c2 = -8/-24 = 1/3
and a1/a2 = b1/b2 =c1/c2

Question 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 the x-axis, and shade the triangular region.

x – y + 1 = 0
x = y – 1

3x + 2y – 12 = 0x = 12 – 2y/3

Graphical representation

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at ( – 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( – 1, 0), and (4, 0).

## Exercise 3.3 (Page 53)

Question 1:
Solve the following pair of linear equations by the substitution method

(i) x + y = 14
x – y = 4

(ii) s – t = 3
(s/3) + (t/2) = 6

(iii) 3x – y = 3
9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3

(v) √2 x+√3 y = 0
√3 x-√8 y = 0

(vi) (3x/2) – (5y/3) = -2
(x/3) + (y/2) = (13/6)

(i) Given,
x + y = 14 and x – y = 4 are the two equations.

From 1st equation, we get,
x = 14 – y

Now, substitute the value of x in second equation to get,

(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5

By the value of y, we can now find the exact value of x;
∵ x = 14 – y
∴ x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5.

(ii) Given,
s – t = 3 and (s/3) + (t/2) = 6 are the two equations.

From 1st equation, we get,
s = 3 + t ____(1)
Now, substitute the value of s in second equation to get,

(3+t)/3 + (t/2) = 6
⇒ (2(3+t) + 3t )/6 = 6
⇒ (6+2t+3t)/6 = 6
⇒ (6+5t) = 36
⇒5t = 30
⇒t = 6

Now, substitute the value of t in equation (1)
s = 3 + 6 = 9
Therefore, s = 9 and t = 6.

(iii) Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.

From 1st equation, we get,
x = (3+y)/3

Now, substitute the value of x in the given second equation to get,
9(3+y)/3 – 3y = 9
⇒9 +3y -3y = 9
⇒ 9 = 9

Therefore, y has infinite values and since, x = (3+y) /3, so x also has infinite values.

(iv)Given,
0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3are the two equations.

From 1st equation, we get,
x = (1.3- 0.3y)/0.2 _____(1)

Now, substitute the value of x in the given second equation to get,
0.4(1.3-0.3y)/0.2 + 0.5y = 2.3
⇒2(1.3 – 0.3y) + 0.5y = 2.3
⇒ 2.6 – 0.6y + 0.5y = 2.3
⇒ 2.6 – 0.1 y = 2.3
⇒ 0.1 y = 0.3
⇒ y = 3

Now, substitute the value of y in equation (1), we get,
x = (1.3-0.3(3))/0.2 = (1.3-0.9)/0.2 = 0.4/0.2 = 2
Therefore, x = 2 and y = 3.

(v) Given,
√2 x + √3 y = 0 and √3 x – √8 y = 0

are the two equations.
From 1st equation, we get,

x = – (√3/√2)y ______(1)
Putting the value of x in the given second equation to get,

√3(-√3/√2)y – √8y = 0 ⇒ (-3/√2)y- √8 y = 0
⇒ y = 0

Now, substitute the value of y in equation (1), we get,
x = 0

Therefore, x = 0 and y = 0.

(vi)Given,
(3x/2)-(5y/3) = -2 and (x/3) + (y/2) = 13/6 are the two equations.
From 1st equation, we get,

(3/2)x = -2 + (5y/3)
⇒ x = 2(-6+5y)/9 = (-12+10y)/9 ………………………(1)

Putting the value of x in the given second equation to get,

((-12+10y)/9)/3 + y/2 = 13/6
⇒y/2 = 13/6 –( (-12+10y)/27 ) + y/2 = 13/6
Now, substitute the value of y in equation (1), we get,

(3x/2) – 5(3)/3 = -2
⇒ (3x/2) – 5 = -2
⇒ x = 2

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

2x + 3y = 11…………………………..(I)
2x – 4y = -24………………………… (II)

From equation (II), we get
x = (11-3y)/2 ………………….(III)

Substituting the value of x in equation (II), we get
2(11-3y)/2 – 4y = 24
11 – 3y – 4y = -24
-7y = -35
y = 5……………………………………..(IV)

Putting the value of y in equation (III), we get

x = (11-3×5)/2 = -4/2 = -2
Hence, x = -2, y = 5
Also,

y = mx + 3
5 = -2m +3
-2m = 2
m = -1

Question 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 Rs.3800. Later, she buys 3 bats and 5 balls for Rs.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 Rs 105 and for a journey of 15 km, the charge paid is Rs 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?

(i) Let the two numbers be x and y respectively, such that y > x.

According to the question,

y = 3x ……………… (1)
y – x = 26 …………..(2)

Substituting the value of (1) into (2), we get

3x – x = 26
x = 13 ……………. (3)

Substituting (3) in (1), we get y = 39
Hence, the numbers are 13 and 39.

(ii) Let the larger angle by xo and smaller angle be yo.

We know that the sum of two supplementary pair of angles is always 180o.
According to the question,

x + y = 180o……………. (1)
x – y = 18……………..(2)

From (1), we get x = 180o – y …………. (3)
Substituting (3) in (2), we get

180– y – y =18o
162o = 2y
y = 81o ………….. (4)

Using the value of y in (3), we get
x = 180o – 81o
= 99o
Hence, the angles are 99o and 81o.

(iii) Let the cost a bat be x and cost of a ball be y.

According to the question,

7x + 6y = 3800 ………………. (I)
3x + 5y = 1750 ………………. (II)

From (I), we get
y = (3800-7x)/6………………..(III)

Substituting (III) in (II). we get,
3x+5(3800-7x)/6 =1750
⇒3x+ 9500/3 – 35x/6 = 1750
⇒3x- 35x/6 = 1750 – 9500/3
⇒(18x-35x)/6 = (5250 – 9500)/3
⇒-17x/6 = -4250/3
⇒-17x = -8500
x = 500 ……………………….. (IV)

Substituting the value of x in (III), we get
y = (3800-7 ×500)/6 = 300/6 = 50

(iv) Let the fixed charge be Rs x and per km charge be Rs y.
According to the question,

x + 10y = 105 …………….. (1)
x + 15y = 155 …………….. (2)

From (1), we get x = 105 – 10y ………………. (3)
Substituting the value of x in (2), we get

105 – 10y + 15y = 155
5y = 50
y = 10 …………….. (4)

Putting the value of y in (3), we get
x = 105 – 10 × 10 = 5
Hence, fixed charge is Rs 5 and per km charge = Rs 10

(v) Let the fraction be x/y.
According to the question,

(x+2) /(y+2) = 9/11
11x + 22 = 9y + 18
11x – 9y = -4 …………….. (1)
(x+3) /(y+3) = 5/6
6x + 18 = 5y +15
6x – 5y = -3 ………………. (2)

From (1), we get x = (-4+9y)/11 …………….. (3)
Substituting the value of x in (2), we get

6(-4+9y)/11 -5y = -3
-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (4)

Substituting the value of y in (3), we get
x = (-4+9×9 )/11 = 7
Hence the fraction is 7/9.

(vi) Let the age of Jacob and his son be x and y respectively.
According to the question,

(x + 5) = 3(y + 5)
x – 3y = 10 …………………………………….. (1)
(x – 5) = 7(y – 5)
x – 7y = -30 ………………………………………. (2)

From (1), we get x = 3y + 10 ……………………. (3)
Substituting the value of x in (2), we get

3y + 10 – 7y = -30
-4y = -40
y = 10 ………………… (4)

Substituting the value of y in (3), we get
x = 3 x 10 + 10 = 40
Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.

## Exercise 3.4 (Page 56)

Question 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) x/2+ 2y/3 = -1 and x-y/3 = 3

(i) x + y = 5 and 2x – 3y = 4
By the method of elimination.

x + y = 5 ……………………………….. (i)
2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get

2x + 2y = 10 ……………………………(iii)
When the equation (ii) is subtracted from (iii) we get,

5y = 6
y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get,

x=5−6/5 = 19/5
∴x = 19/5 , y = 6/5

By the method of substitution.
From the equation (i), we get:

x = 5 – y………………………………….. (v)
When the value is put in equation (ii) we get,

2(5 – y) – 3y = 4
-5y = -6
y = 6/5

When the values are substituted in equation (v), we get:
x =5− 6/5 = 19/5
∴x = 19/5 ,y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2
By the method of elimination.

3x + 4y = 10……………………….(i)
2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get:
4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get:
7x = 14
x = 2 ……………………………….(iv)

Substituting equation (iv) in (i) we get,
6 + 4y = 10
4y = 4
y = 1

Hence, x = 2 and y = 1

By the method of Substitution
From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i) we get,

3(1 + y) + 4y = 10
7y = 7
y = 1

When y = 1 is substituted in equation (v) we get,\
A = 1 + 1 = 2

Therefore, A = 2 and B = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By the method of elimination:

3x – 5y – 4 = 0 ………………………………… (i)
9x = 2y + 7
9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied we get,
9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii) we get,

13y = -5
y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i) we get,
3x +25/13 −4=0
3x = 27/13
x =9/13
∴x = 9/13 and y = -5/13

By the method of Substitution:
From the equation (i) we get,

x = (5y+4)/3 …………………………………………… (v)
Putting the value (v) in equation (ii) we get,

9(5y+4)/3 −2y −7=0
13y = -5
y = -5/13

Substituting this value in equation (v) we get,

x = (5(-5/13)+4)/3
x = 9/13
∴x = 9/13, y = -5/13

(iv) x/2 + 2y/3 = -1 and x-y/3 = 3
By the method of Elimination.

3x + 4y = -6 …………………………. (i)
x-y/3 = 3
3x – y = 9 ……………………………. (ii)

When the equation (ii) is subtracted from equation (i) we get,

-5y = -15
y = 3 ………………………………….(iii)

When the equation (iii) is substituted in (i) we get,
3x – 12 = -6
3x = 6
x = 2
Hence, x = 2 , y = -3

By the method of Substitution:

From the equation (ii) we get,
x = (y+9)/3…………………………………(v)
Putting the value obtained from equation (v) in equation (i) we get,

3(y+9)/3 +4y =−6
5y = -15
y = -3

When y = -3 is substituted in equation (v) we get,
x = (-3+9)/3 = 2
Therefore, x = 2 and y = -3

Question 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 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 Rs.2000. She asked the cashier to give her Rs.50 and Rs.100 notes only. Meena got 25 notes in all. Find how many notes of Rs.50 and Rs.100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

(i) Let the fraction be x/y
According to the question,x + 1/y – 1 = 1
⇒ – = -2 … (i)x/y+1 = 1/2
⇒ 2x – = 1 … (ii)
Subtracting equation (i) from equation (ii), we get
x = 3 … (iii)
Putting this value in equation (i), we get
3 – y = -2
y = -5
y = 5
Hence, the fraction is 3/5

(ii) Let present age of Nuri = x
and present age of Sonu = y
According to the given information,question,(x – 5) = 3(y – 5)
x – 3y = -10 … (i)
(x + 10y) = 2(y + 10)
x – 2y = 10 … (ii)
Subtracting equation (i) from equation (ii), we get
y = 20 … (iii)
Putting this value in equation (i), we get
x – 60 = -10
= 50
Hence, age of Nuri = 50 years and age of Sonu = 20 years

(iii) Let the unit digit and tens digits of the number be and respectively.
Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
x + y = 9 … (i)
9(10x) = 2(10x + y)
88y – 11x = 0
– x + 8y =0 … (ii)
Adding equation (i) and (ii), we get
9y = 9
y = 1 … (iii)
Putting the value in equation (i), we get
x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.

(iv) Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.
According to the question,
= 25 … (i)
50x + 100y = 2000 … (ii)
Multiplying equation (i) by 50, we get
50x + 50y = 1250 … (iii)
Subtracting equation (iii) from equation (ii), we get
50y = 750
y = 15
Putting this value in equation (i), we have x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v) Let the fixed charge for first three days and each day charge thereafter be Rs and Rs respectively.
According to the question,
+ 4y = 27 … (i)
+ 2y = 21 … (ii)
Subtracting equation (ii) from equation (i), we get
2y = 6
y = 3 … (iii)
Putting in equation (i), we get
x + 12 =27
x = 15
Hence, fixed charge = Rs 15 and Charge per day = Rs 3.

## Exercise 3.5 (Page 62)

Question 1:
Which of the following pairs of linear equations has unique solution, 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

(i) Given, x – 3y – 3 =0 and 3x – 9y -2 =0
a1/a2=1/3 , b1/b2= -3/-9 =1/3, c1/c2=-3/-2 = 3/2
(a1/a2) = (b1/b2) ≠ (c1/c2)

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

(ii) Given, 2x + y = 5 and 3x +2y = 8
a1/a2 = 2/3 , b1/b2 = 1/2 , c1/c2 = -5/-8
(a1/a2) ≠ (b1/b2)

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:
x/(b1c2-c1b2) = y/(c1a2 – c2a=) = 1/(a1b2-a2b1)
x/(-8-(-10)) = y/(15+16) = 1/(4-3)
x/2 = y/1 = 1
∴ x = 2 and y =1

(iii) 3x – 5y = 20
6x – 10y = 40
a1/a2 = 3/6 = 1/2
b1/b2 = -5/-10 = 1/2 and
c1/c2 = -20/-40 = 1/2
a1/a2 =b1/b=c1/c2
Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) Given, x – 3y – 7 = 0 and 3x – 3y – 15 = 0
(a1/a2) = 1/3
(b1/b2) = -3/-3 = 1
(c1/c2) = -7/-15
a1/a2 ≠ b1/b2

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.
By cross multiplication,

x/(45-21) = y/(-21+15) = 1/(-3+9)
x/24 = y/ -6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
∴ x = 4 and y = 1.

Question 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

(i) 3y + 2x -7 =0
(a + b)y + (a-b)y – (3a + b -2) = 0
a1/a2 = 2/(a-b) ,               b1/b2 = 3/(a+b) ,               c1/c2 = -7/-(3a + b -2)

For infinitely many solutions,
a1/a2 = b1/b2 = c1/c2
Thus 2/(a-b) = 7/(3a+b– 2)

6a + 2b – 4 = 7a – 7b
a – 9b = -4  ……………………………….(i)
2/(a-b) = 3/(a+b)
2a + 2b = 3a – 3b
a – 5b = 0 ……………………………….….(ii)

Subtracting (i) from (ii), we get

4b = 4
b =1

Substituting this eq. in (ii), we get
a -5 x 1= 0
a = 5

Thus at a = 5 and b = 1 the given equations will have infinite solutions.

(ii) 3x + y -1 = 0
(2k -1)x  +  (k-1)y – 2k -1 = 0

a1/a2 = 3/(2k -1) ,           b1/b2 = 1/(k-1), c1/c2 = -1/(-2k -1) = 1/( 2k +1)
For no solutions

a1/a2 = b1/b2 ≠ c1/c2
3/(2k-1) = 1/(k -1) ≠ 1/(2k +1)
3/(2k –1) = 1/(k -1)
3k -3 = 2k -1
k =2
Therefore, for k = 2 the given pair of linear equations will have no solution.

Question 3:
Solve the following pair of linear equations by the substitution and cross-multiplication methods:

8x + 5y = 9
3x + 2y = 4

8x +5y = 9 … (i)
3x +2y = 4 … (ii)
From equation (ii), we get
x = 4-2y/3 … (iii)
Putting this value in equation (i), we get
8(4-2y/3) + 5y = 9
32 – 16y +15y = 27
y = -5
y = 5 … (iv)
Putting this value in equation (ii), we get
3x + 10 = 4
x = -2
Hence, x = -2, = 5

By cross multiplication again, we get
8x + 5y -9 = 03x + 2y – 4 = 0x/-20-(-18) = y/-27-(-32) = 1/16-15
x/-2 = y/5 = 1/1
x/-2 = 1 and y/5 = 1
x = -2 and y = 5

Question 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 Rs.1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs.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 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.

(i) Let be the fixed charge of the food and be the charge for food per day.

According to the question,x + 20y = 1000 … (i)x + 26y = 1180 … (ii)
Subtracting equation (i) from equation (ii), we get
6y = 180y = 180/6 = 30
Putting this value in equation (i), we get
x + 20 × 30 = 1000
x = 1000 – 600
x = 400
Hence, fixed charge = Rs 400 and charge per day = Rs 30

(ii) Let the fraction be x/y
According to the question,x-1/y = 1/3
⇒ 3x – y = 3… (i)
x/y+8 = 1/4
⇒ 4x – y = 8 … (ii)
Subtracting equation (i) from equation (ii), we getx = 5 … (iii)Putting this value in equation (i), we get15 – = 3= 12
Hence, the fraction is 5/12.

(iii) Let the number of right answers and wrong answers be and respectively.

According to the question,3x – = 40 … (i)4x – 2y = 50⇒ 2x – = 25 … (ii)Subtracting equation (ii) from equation (i), we get
= 15 … (iii)
Putting this value in equation (ii), we get
30 – = 25= 5Therefore, number of right answers = 15

And number of wrong answers = 5
Total number of questions = 20

(iv) Let the speed of 1st car and 2nd car be u km/h and v km/h.
Respective speed of both cars while they are travelling in same direction = (u -v) km/hRespective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (v) km/h
According to the question,5(– v) = 100⇒ u – v = 20 … (i)1(u + v) = 100 … (ii)Adding both the equations, we get2u = 120u = 60 km/h … (iii)Putting this value in equation (ii), we obtain
v = 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) Let length and breadth of rectangle be unit and y unit respectively.
Area = xy
According to the question,
(x – 5) (y + 3) = xy – 9
⇒ 3x – 5y – 6 = 0 … (i)
(x + 3) (y + 2) = xy + 67
⇒ 2x – 3y – 61 = 0 … (ii)
By cross multiplication, we get
x/305-(-18) = y/-12-(-183) = 1/9-(-10)
x/323 = y/171 = 1/19
= 17, y = 9
Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

## Exercise 3.6 (Page 67)

Question 1:
Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6

(ii) 2/√x + 3/√y = 2
4/√x + 9/√y = -1

(iii) 4/x + 3y = 14
3/x -4y = 23

(iv) 5/(x-1) + 1/(y-2) = 2
6/(x-1) – 3/(y-2) = 1

(v) (7x-2y)/ xy = 5
(8x + 7y)/xy = 15

(vi) 6x + 3y = 6xy
2x + 4y = 5xy

(vii) 10/(x+y) + 2/(x-y) = 4
15/(x+y) – 5/(x-y) = -2

(viii) 1/(3x+y) + 1/(3x-y) = 3/4
1/2(3x+y) – 1/2(3x-y) = -1/8

(i) 1/2x + 1/3y = 2
1/3x + 1/2= 13/6
Let 1/p and 1/q, then the equations changes as below:
p/2 + q/3 = 2
⇒ 3p + 2q -12 = 0 … (i)
p/3 + q/2 = 13/6
⇒ 2p + 3q -13 = 0 … (ii)

By cross-multiplication method, we getp/-26-(-36) = q/-24-(-39) = 1/9-4
p/10 = q/15 = 1/5
p/10 = 1/5 and q/15 = 1/5
= 2 and = 3

1/= 2 and 1/= 3
Hence, = 1/2 and = 1/3

(ii) 2/√x +3/√y = 2
4/√x – 9/√y = -1
Let 1/√p and 1/√y = q, then the equations changes as below:
2p + 3q = 2 … (i)
4p – 9q = -1 … (ii)
Multiplying equation (i) by 3, we get
6p + 9q = 6 … (iii)
Adding equation (ii) and (iii), we get
10p = 5p = 1/2 … (iv)Putting in equation (i), we get2 × 1/2 + 3q = 2
3q = 1
q = 1/3

= 1/√x = 1/2
x = 2
x = 4
and
= 1/√y = 1/3
y = 3
y = 9
Hence, x = 4, y = 9

(iii) 4/x + 3y = 14
3/x – 4y = 23
Putting 1/x = p in the given equations, we get
4p + 3y = 14
⇒ 4p + 3y – 14 = 0
3p – 4y = 23
⇒ 3p – 4y -23 = 0
By cross-multiplication, we get
p/-69-56 = y/-42-(-92) = 1/-16-9
⇒ –p/125 = y/50 = -1/25
Now,
p/125 = -1/25 and y/50 = -1/25

⇒ p = 5 and y = -2
Also, p = 1/x = 5
⇒ x = 1/5
So, x = 1/5 and y = -2 is the solution.

(iv) 5/x-1 + 1/y-2 = 2
6/x-1 – 3/y-2 = 1
Putting 1/x-1 = p and 1/y-2 = q in the given equations, we obtain
5p + q = 2 … (i)
6p – 3q = 1 … (ii)
Now, by multiplying equation (i) by 3 we get
15p + 3q = 6 … (iii)
Now, adding equation (ii) and (iii)
21p = 7
⇒ p = 1/3
Putting this value in equation (ii) we get,
6×1/3 – 3q =1

⇒ 2-3q = 1
⇒ -3q = 1-2
⇒ -3q = -1
⇒ q = 1/3
Now,
p = 1/x-1 = 1/3
⇒1/x-1 = 1/3
⇒ 3 = – 1
⇒ x = 4
Also,
q = 1/y-2 = 1/3
⇒ 1/y-2 = 1/3
⇒ 3 = y-2
⇒ y = 5

Hence, x = 4 and y = 5 is the solution.

(v) 7x-2y/xy = 5
⇒ 7x/xy – 2y/xy = 5
⇒ 7/y – 2/x = 5 … (i)
8x+7y/xy = 15
⇒ 8x/xy + 7y/xy = 15
⇒ 8/y + 7/x = 15 … (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
7q – 2p = 5 … (iii)
8q + 7p = 15 … (iv)
Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,
49q – 14p = 35 … (v)
16q + 14p = 30 … (vi)
Now, adding equation (v) and (vi) we get,

49q – 14p + 16q + 14p = 35 + 30
⇒ 65q = 65
⇒ q = 1
Putting the value of q in equation (iv)
8 + 7p = 15
⇒ 7p = 7
⇒ p = 1
Now,
p = 1/x = 1
⇒ 1/x = 1
⇒ x = 1
also, q = 1 = 1/y
⇒ 1/y = 1

⇒ = 1
Hence, =1 and y = 1 is the solution.

(vi) 6x + 3y = 6xy
⇒ 6x/xy + 3y/xy = 6
⇒ 6/y + 3/x = 6 … (i)
2x + 4y = 5xy
⇒ 2x/xy + 4y/xy = 5
⇒ 2/y + 4/x = 5 … (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
6q + 3p – 6 = 0
2q + 4p – 5 = 0
By cross multiplication method, we get
p/-30-(-12) = q/-24-(-15) = 1/6-24
p/-18 = q/-9 = 1/-18
p/-18 = 1/-18 and q/-9 = 1/-18

= 1 and q = 1/2
= 1/x = 1 and q = 1/y = 1/2
= 1, y = 2
Hence, = 1 and = 2

(vii) 10/x+y + 2/xy = 4
15/x+y – 5/xy = -2
Putting 1/x+y = p and 1/xy = q in the given equations, we get:
10p + 2q = 4
⇒ 10p + 2q – 4 = 0 … (i)
15p – 5q = -2
⇒ 15p – 5q + 2 = 0 … (ii)
Using cross multiplication, we get
p/4-20 = q/-60-(-20) = 1/-50-30
p/-16 = q/-80 = 1/-80
p/-16 = 1/-80 and q/-80 = 1/-80
p = 1/5 and q = 1
p = 1/x+y = 1/5 and q = 1/xy = 1
x + y = 5 … (iii)

and x – y = 1 … (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 …. (v)
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2

(viii) 1/3x+y + 1/3xy = 3/4
1/2(3xy) – 1/2(3xy) = -1/8
Putting 1/3x+y = p and 1/3xy = q in the given equations, we get
p + q = 3/4 … (i)
p/2 – q/2 = -1/8
– = -1/4 … (ii)
Adding (i) and (ii), we get
2p = 3/4 – 1/4
2p = 1/2
= 1/4
Putting the value in equation (ii), we get
1/4 – q = -1/4
q = 1/4 + 1/4 = 1/2

p = 1/3x+y = 1/4
3x + y = 4 … (iii)
q = 1/3xy = 1/2
3x – y = 2 … (iv)
Adding equations (iii) and (iv), we get
6x = 6
x = 1 … (v)
Putting the value in equation (iii), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1

Question 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.

(i) Let the speed of Ritu in still water and the speed of stream be x km/h
and y km/h respectively.
Speed of Ritu while rowing
Upstream = (x – y) km/h
Downstream = (x + y) km/hAccording to question,
2(x + y) = 20⇒ x + y = 10 … (i)
2(x – y) = 4
⇒ x – y = 2 … (ii)
Adding equation (i) and (ii), we getPutting this equation in (i), we gety = 4Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii) Let the number of days taken by a woman and a man be x and y respectively.
Therefore, work done by a woman in 1 day = 1/xAccording to the question,4(2/x + 5/y) = 12/x + 5/y = 1/43(3/x + 6/y) = 13/x + 6/y = 1/3Putting 1/x = p and 1/q in these equations, we get2p + 5q = 1/4By cross multiplication, we getp/-20-(-18) = q/-9-(-18) = 1/144-180p/-2 = q/-1 = 1/-36p/-2 = -1/36 and q/-1 = 1/-36

= 1/18 and q = 1/36p = 1/x = 1/18 and q = 1/y = 1/36= 18 and = 36Hence, number of days taken by a woman = 18 and number of days taken by a man = 36

(iii) Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,60/u + 240/= 4 … (i)
100/u + 200/= 25/6 … (ii)
Putting 1/u = p and 1/v = q in the equations, we get
60p + 240q = 4 … (iii)
100p + 200q = 25/6
600p + 1200q = 25 … (iv)
Multiplying equation (iii) by 10, we get
600p + 2400q = 40 …. (v)
Subtracting equation (iv) from (v), we get1200q = 15
q = 15/200 = 1/80 … (vi)
Putting equation (iii), we get
60p + 3 = 4
60p = 1
p = 1/60
p = 1/u = 1/60 and q = 1/v = 1/80
u = 60 and v = 80
Hence, speed of train = 60 km/h and speed of bus = 80 km/h