RD SHARMA Solutions for Class 10 Maths Chapter 3 - Pairs of Linear Equations in Two Variables

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.1

Solution 1
Solution 2
Let the present age of Aftab and his daughter be x and y respectively.

Seven years ago,
Age of Aftab = x - 7
Age of his daughter = y - 7

According to the given condition,

 
Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3

According to the given condition,

 

Thus, the given conditions can be algebraically represented as:
x - 7y = -42
x - 3y = 6

 
Three solutions of this equation can be written in a table as follows:
x -7 0 7
y 5 6 7


 

Three solutions of this equation can be written in a table as follows:
x 6 3 0
y 0 -1 -2


The graphical representation is as follows:



Concept insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter's present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words 'seven years ago' means we have to subtract 7 from their present ages, and 'three years from now' or 'three years hence' means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.
Solution 3

Solution 4
Solution 5(i)
Solution 5(ii)
Solution 5(iii)
Solution 6(i)
Solution 6(ii)
Solution 6(iii)
Solution 7
Let the cost of 1 kg of apples and 1 kg grapes be Rsx and Rsy.
The given conditions can be algebraically represented as:


Three solutions of this equation can be written in a table as follows:


x
50 60 70
y 60 40 20




Three solutions of this equation can be written in a table as follows:


x
70 80 75
y 10
-10
0


The graphical representation is as follows:



Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale.

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.2

Solution 1

Solution 2

Solution 3




Solution 4

Solution 6




Solution 7

 

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Since, the graph of the two lines coincide, the given system of equations have infinitely many solutions.
Solution 13


Solution 14

Solution 15

Solution 16

Solution 17

Solution 18




Solution 19(i)




Solution 19(ii)




Solution 20




Solution 21(i)

Solution 21(ii)

Solution 22(i)




Solution 22(ii)




Solution 22(iii)




Solution 22(iv)

Solution 22(v)




Solution 22(vi)




Solution 23(i)





Solution 23(ii)





Solution 23(iii)





Solution 24

Solution 25

 
Solution 26




Solution 27



Solution 28(i)

Solution 28(ii)





Solution 28(iii)





Solution 28(iv)





Solution 29



Solution 30

Solution 31






Solution 32



Three solutions of this equation can be written in a table as follows:

x 0 1 2
y -5 0 5


x 0 1 2
y -3 0 3


The graphical representation of the two lines will be as follows:

 



It can be observed that the required triangle is ABC.
The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

Area space of space Triangle space increment ABC space equals space 1 half cross times BC cross times AO
equals 1 half cross times 2 cross times 1
equals 1 space sq. space unit

Concept insight: In order to find the coordinates of the vertices of the triangle so formed, find the points where the two lines intersects the y-axis and also where the two lines intersect each other. Here, note that the coordinates of the intersection of lines with y-axis is taken and not with x-axis, this is because the question says to find the triangle formed by the two lines and the y-axis.

Solution 33

(i) Let the number of girls and boys in the class be x and y respectively.
According to the given conditions, we have:
x + y = 10
x - y = 4
x + y = 10  x = 10 - y
Three solutions of this equation can be written in a table as follows:

x 4 5 6
y 6 5 4



x - y = 4 x = 4 + y
Three solutions of this equation can be written in a table as follows:

x 5 4 3
y 1 0 -1



The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (7, 3).
So, x = 7 and y = 3.

Thus, the number of girls and boys in the class are 7 and 3 respectively.


(ii)    Let the cost of one pencil and one pen be Rs x and Rs y respectively.

According to the given conditions, we have:
5x + 7y = 50
7x + 5y = 46

Three solutions of this equation can be written in a table as follows:

x 3 10 -4
y 5 0 10




Three solutions of this equation can be written in a table as follows:

x 8 3 -2
y -2 5 12



The graphical representation is as follows:


From the graph, it can be observed that the two lines intersect each other at the point (3, 5).
So, x = 3 and y = 5.

Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively.



(iii)

Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are:

y = 2x - 2 ...(1)

and y = 4x - 4 ...(2)

Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.

They are given in Table

x 2 0
y = 2x - 2 2 -2



x 0 1
y = 4x - 4 -4 0



Plot the points and draw the lines passing through them to represent the equations, as shown in fig.,

The two lines intersect at the point (1,0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

Concept insight: Read the question carefully and examine what are the unknowns. Represent the given conditions with the help of equations by taking the unknowns quantities as variables. Also carefully state the variables as whole solution is based on it. On the graph paper, mark the points accurately and neatly using a sharp pencil. Also, take at least three points satisfying the two equations in order to obtain the correct straight line of the equation. Since joining any two points gives a straight line and if one of the points is computed incorrect will give a wrong line and taking third point will give a correct line. The point where the two straight lines will intersect will give the values of the two variables, i.e., the solution of the two linear equations. State the solution point.

Solution 34(i)





Solution 34(ii)





Solution 35

Solution 36

(i) For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0, to be intersecting, we must have

So, the other linear equation can be 5x + 6y - 16 = 0
 
(ii)  For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0, to be parallel, we must have

So, the other linear equation can be 6x + 9y + 24 = 0,

(iii)  For the two lines a1x + b1x + c1 = 0 and a2x + b2x + c2 = 0 to be coincident, we must have

So, the other linear equation can be 8x + 12y - 32 = 0,


Concept insight: In order to answer such type of problems, just remember the conditions for two lines to be intersecting, parallel, and coincident. This problem will have multiple answers as their can be many equations satisfying the required conditions.

Solution 37(i)

Solution 37(ii)

Solution 38

The lines AB and CD intersect at point R(1, 4). Hence, the solution of the given pair of linear equations is x = 1, y = 4.

 

From R, draw RM X-axis and RN Y-axis.

Then, from graph, we have

RM = 4 units, RN = 1 unit, AP = 4 units, BQ = 4 units

 

Solution 39

From the graph, the vertices of the triangle AOP formed by the given lines are A(4, 4), O(0, 0) and P(6, 2).

Solution 40

The graph of x = 3 is a straight line parallel to Y-axis at a distance of 3 units to the right of Y-axis.

 

The graph of x = 5 is a straight line parallel to Y-axis at a distance of 5 units to the right of Y-axis.

 

Solution 41

The graph of x = -2 is a straight line parallel to Y-axis at a distance of 2 units to the left of Y-axis.

 

The graph of y = 3 is a straight line parallel to X-axis at a distance of 3 units above X-axis.

Solution 42

  

Solution 5

Given equations are:

x - y + 1 = 0 … (i)

3x + 2y - 12 = 0 … (ii)

From (i) we get, x = y - 1

When x = 0, y = 1

When x = -1, y = 0

When x = 1, y = 2

We have the following table:

x

0

-1

1

y

1

0

2

From (ii) we get,   

When x = 0, y = 6

When x = 4, y = 0

When x = 2, y = 3

We have the following table:

x

0

4

2

y

6

0

3

Graph of the given equations is:

As the two lines intersect at (2, 3). 

Hence, x = 2, y = 3 is the solution of the given equations.

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.3

Solution 1
Solution 2
Solution 3

Solution 4

Solution 5
Solution 6
Solution 7

Solution 8
Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14
Solution 15

Solution 16
Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

fraction numerator 10 over denominator straight x plus straight y end fraction plus fraction numerator 2 over denominator straight x minus straight y end fraction equals 4
fraction numerator 15 over denominator straight x plus straight y end fraction minus fraction numerator 9 over denominator straight x minus straight y end fraction equals negative 2

Let space fraction numerator 1 over denominator straight x plus straight y end fraction equals straight p space and space fraction numerator 1 over denominator straight x minus straight y end fraction equals straight q
10 straight p plus 2 straight q minus 4 equals 0....... left parenthesis straight i right parenthesis
15 straight p minus 9 straight q plus 2 equals 0........ left parenthesis ii right parenthesis

Using space cross minus multiplication space method comma space we space obtain colon
fraction numerator straight p over denominator 4 minus 36 end fraction equals fraction numerator straight q over denominator negative 60 minus 20 end fraction equals fraction numerator 1 over denominator negative 90 minus 30 end fraction
fraction numerator straight p over denominator negative 32 end fraction equals fraction numerator straight q over denominator negative 80 end fraction equals fraction numerator 1 over denominator negative 120 end fraction
fraction numerator straight p over denominator negative 4 end fraction equals fraction numerator straight q over denominator negative 10 end fraction equals fraction numerator 1 over denominator negative 15 end fraction
straight p equals 4 over 15 comma straight q equals 10 over 15 equals 2 over 3

Substituting space the space values space of space straight p space and space straight q comma
straight x plus straight y equals 15 over 4....... left parenthesis straight i right parenthesis space
straight x minus straight y equals 3 over 2...... left parenthesis ii right parenthesis
left parenthesis straight i right parenthesis plus left parenthesis ii right parenthesis rightwards double arrow
2 straight x equals 21 over 4
straight x equals 21 over 8 comma straight y equals 9 over 8

Solution 41

Solution 42

Solution 43

152 straight x minus 378 straight y equals negative 74....... left parenthesis straight i right parenthesis
minus 378 straight x plus 152 straight y equals negative 604....... left parenthesis ii right parenthesis

Adding space the space equations space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space obtain colon
minus 226 straight x minus 226 straight y equals negative 678
rightwards double arrow straight x plus straight y equals 3........... left parenthesis 3 right parenthesis

Subtracting space the space equation space left parenthesis 2 right parenthesis space from space equation space left parenthesis 1 right parenthesis comma space we space obtain colon
530 straight x minus 530 straight y equals 530
rightwards double arrow straight x minus straight y equals 1........... left parenthesis 4 right parenthesis

Adding space equations space left parenthesis 3 right parenthesis space and space left parenthesis 4 right parenthesis comma space we space obtain colon
straight x equals 2
Substituting space straight x equals 2 space in space equation space left parenthesis 3 right parenthesis comma space we space get
straight y equals 1

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.5

Solution 29

The given system of equations will have infinite number of solutions if

  

Solution 1
Solution 2
Solution 3

Solution 4
Solution 5
Solution 6

Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17

Solution 18

Solution 19

Solution 20

Solution 21
Solution 22
Solution 23
Solution 24
Solution 25

Solution 26
Solution 27
Solution 28

Solution 30

Solution 31
Solution 32
Solution 33

Solution 34

Solution 35
Solution 36 (i)
Solution 36 (ii)
Solution 36 (iii)
Solution 36 (iv)

Solution 36 (v)

2 straight x plus 3 straight y minus 7 equals 0
left parenthesis straight a minus straight b right parenthesis straight x plus left parenthesis straight a minus straight b right parenthesis straight y minus left parenthesis 3 straight a plus straight b minus 2 right parenthesis equals 0
Here comma
straight a subscript 1 equals 2 comma space straight b subscript 1 equals 3 comma straight c subscript 1 equals negative 7
straight a subscript 2 equals left parenthesis straight a minus straight b right parenthesis comma straight b subscript 2 equals left parenthesis straight a plus straight b right parenthesis comma straight c subscript 2 equals negative left parenthesis 3 straight a plus straight b minus 2 right parenthesis

straight a subscript 1 over straight a subscript 2 equals fraction numerator 2 over denominator left parenthesis straight a minus straight b right parenthesis end fraction comma space straight b subscript 1 over straight b subscript 2 equals fraction numerator 3 over denominator left parenthesis straight a plus straight b right parenthesis end fraction comma space straight c subscript 1 over straight c subscript 2 equals fraction numerator negative 7 over denominator negative left parenthesis 3 straight a plus straight b minus 2 right parenthesis end fraction equals fraction numerator 7 over denominator left parenthesis 3 straight a plus straight b minus 2 right parenthesis end fraction
For space the space equations space to space have space infinitely space many space solutions comma space we space have
straight a subscript 1 over straight a subscript 2 equals space straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2
fraction numerator 2 over denominator left parenthesis straight a minus straight b right parenthesis end fraction equals fraction numerator 7 over denominator left parenthesis 3 straight a plus straight b minus 2 right parenthesis end fraction
6 straight a plus 2 straight b minus 4 equals 7 straight a minus 7 straight b
straight a minus 9 straight b equals negative 4........... left parenthesis straight i right parenthesis

fraction numerator 2 over denominator left parenthesis straight a minus straight b right parenthesis end fraction equals fraction numerator 3 over denominator left parenthesis straight a plus straight b right parenthesis end fraction
2 straight a plus 2 straight b equals 3 straight a minus 3 straight b
straight a minus 5 straight b equals 0............ left parenthesis ii right parenthesis
Subtracting space left parenthesis straight i right parenthesis space from space left parenthesis ii right parenthesis comma space we space get
4 straight b equals 4
straight b equals 1
Substituting space the space value space of space straight b space in space equation space left parenthesis 2 right parenthesis comma space we space obtain colon
straight a space minus space 5 cross times 1 equals 0
straight a equals 5
Thus space the space values space of space straight a space and space straight b space are space 5 space and space 1 space respectively.

Solution 36 (vi)

Solution 36 (vii)

Solution 36(viii)

Solution 36(ix)

Solution 37(i)

Solution 37(ii)

Solution 37(iii)

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.8

Solution 9

Let the fraction be

According to the given conditions, we have

Subtracting (ii) from (i), we get x = 7

Substituting the value of x in (ii), we get

y = 15

Solution 1
Solution 2

Solution 3
Solution 4
Solution 5

Solution 6
Solution 7
Solution 8
Solution 9Old

Solution 10

Solution 11

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.4

Solution 1
Solution 2
Solution 3
Solution 4

Solution 5
Solution 6
Solution 7
Solution 8
Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20

 

Solution 21
Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.6

Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7

Solution 8

Solution 9

Let space the space cost space of space straight a space bat space be space straight x space and space straight y space respectively.
According space to space the space given space information comma
7 straight x plus 6 straight y equals 3800....... left parenthesis 1 right parenthesis
3 straight x plus 5 straight y equals 1750....... left parenthesis 2 right parenthesis
From space left parenthesis 1 right parenthesis comma space we space obtain comma
straight y equals fraction numerator 3800 minus 7 straight x over denominator 6 end fraction........ left parenthesis 3 right parenthesis
Substituting space this space value space in space equation space left parenthesis 2 right parenthesis comma space we space obtain
3 straight x plus 5 open parentheses fraction numerator 3800 minus 7 straight x over denominator 6 end fraction close parentheses equals 1750
3 straight x plus fraction numerator 19000 minus 35 straight x over denominator 6 end fraction equals 1750
3 straight x minus fraction numerator 35 straight x over denominator 6 end fraction equals 1750 minus 19000 over 6
fraction numerator 18 straight x minus 35 straight x over denominator 6 end fraction equals fraction numerator 10500 minus 19000 over denominator 6 end fraction
fraction numerator 17 straight x over denominator 6 end fraction equals 8500 over 6
straight x equals 500........ left parenthesis 4 right parenthesis
Substituting space this space equation left parenthesis 3 right parenthesis comma space we space obtain comma
straight y equals fraction numerator 3800 minus 7 cross times 500 over denominator 6 end fraction
equals 300 over 6 equals 50
Hence comma space the space cost space of space straight a space bat space is space Rs space 500 space and space that space of space straight a space ball space is space Rs space 50.

bold Concept bold space bold Insight colon space Cost space of space bats space and space balls space need space to space be space found space so space the space cost space of space straight a space ball
and space bat space will space be space taken space as space the space variables. space Apply space the space conditions space of space total space cost space of space
bats space and space balls space algebraic space equations space will space be space obtained. space The space pair space of space equations space
can space then space be space solved space by space suitable space substitution.

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.7

Solution 1
Solution 2
Solution 3
Solution 4
Solution 5

Solution 6

Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13

Solution 14
Solution 15
Solution 16

Solution 17

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.9

Solution 1
Solution 2
Solution 3

Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9

Solution 10

Solution 11

The difference between the ages of Ani and Biju is given as 3 years. So, either Biju is 3 years older than Ani or Ani is 3 years older than Biju.

Let the age of Ani and Biju be x years and y years respectively.
Age of Dharam = 2 × x = 2x years

Case I: Ani is older than Biju by 3 years
x - y = 3        ... (1)

begin mathsize 12px style 2 straight x minus straight y over 2 equals 30 end style
4x - y = 60     ....(2)

Subtracting (1) from (2), we obtain:
 
3x = 60 - 3 = 57

begin mathsize 12px style straight x equals 57 over 3 equals 19 end style
Age of Ani = 19 years
Age of Biju = 19 - 3 = 16 years

Case II: Biju is older than Ani by 3 years
y - x = 3        ... (3)

begin mathsize 12px style 2 straight x minus straight y over 2 equals 30 end style

4x - y = 60        ... (4)

Adding (3) and (4), we obtain:
3x = 63
x = 21

Age of Ani = 21 years
Age of Biju = 21 + 3 = 24 years

Concept Insight: In this problem, ages of Ani and Biju are the unknown quantities. So, we represent them by variables x and y. Now, note that here it is given that the ages of Ani and Biju differ by 3 years. So, it is not mentioned that which one is older. So, the most important point in this question is to consider both cases  Ani is older than Biju and  Biju is older than Ani. For second condition the relation  on the ages of Dharam and Cathy can be implemented . Pair of linear equations can be solved using a suitable algebraic method.

Solution 12

Solution 13

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.10

Solution 1
Solution 2
Solution 3

Solution 4
Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

 

 

Solution 11

Solution 12
Solution 13

Solution 14

Solution 15
Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.
Or, d = xt        ... (1)

According to the question,
 



By using equation (1), we obtain:
3x - 10t = 30        ... (3)

Adding equations (2) and (3), we obtain:
x = 50
Substituting the value of x in equation (2), we obtain:
(-2) x (50) + 10t = 20
-100 + 10t = 20
10t = 120
t = 12
From equation (1), we obtain:
d = xt = 50 x 12 = 600

Thus, the distance covered by the train is 600 km.

Concept insight: To solve this problem, it is very important to remember the relation . Now, all these three quantities are unknown. So, we will represent these

by three different
variables. By using the given conditions, a pair of equations will be obtained. Mind one thing that the equations obtained will not be linear. But they can be reduced to linear form by using the fact that . Then two linear equations can be formed which can

be solved easily by elimination method.

Solution 16

Solution 17

Solution 18

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.11

Solution 1

Solution 2
Solution 3
Solution 4
Solution 5
Solution 6

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

A + C = 180
4y + 20 - 4x = 180
-4x + 4y = 160
x - y = -40            ... (1)
Also, B + D = 180
3y - 5 - 7x + 5 = 180
-7x + 3y = 180        ... (2)
Multiplying equation (1) by 3, we obtain:

3x - 3y = -120        ... (3)
Adding equations (2) and (3), we obtain:
-4x = 60
x = -15
Substituting the value of x in equation (1), we obtain:
-15 - y = -40
y = -15 + 40 = 25

A = 4y + 20 = 4(25) + 20 = 120o
B = 3y - 5 = 3(25) - 5 = 70o
C = -4x = -4(-15) = 60o
D = -7x + 5 = -7(-15) + 5 = 110o

Concept insight: The most important idea to solve this problem is by using the fact that the sum of the measures of opposite angles in a cyclic quadrilateral is 180o. By using this relation, two linear equations can be obtained which can be solved easily by eliminating a suitable variable.

Solution 7

Solution 8

Solution 9
Solution 10

Let space the space number space of space right space answers space and space wrong space answers space be space straight x space and space straight y space respectively.
According space to space the space question comma
3 straight x minus straight y equals 40......... left parenthesis 1 right parenthesis
4 straight x minus 2 straight y equals 50
rightwards double arrow 2 straight x minus straight y equals 25......... left parenthesis 2 right parenthesis
Subtracting space equation space left parenthesis 2 right parenthesis space from space equation space left parenthesis 1 right parenthesis comma space we space obtian colon
straight x equals 15
Substituting space the space value space of space straight x space in space equation space left parenthesis 2 right parenthesis comma space we space obtain colon
30 minus straight y equals 25
straight y equals 5

Thus comma space the space number space of space right space answers space and space the space number space of space wrong space answers space is space 15
and space 5 space respectively. space Therefore space the space total space number space of space questions space is space 20.

Concept space insight colon space In space this space problem comma space the space number space of space write space answers space and space the space
number space of space wrong space answers space answered space by space Yash space are space the space unknown space variable space straight y
has space the space same space coefficient space in space both space the space equations comma space so space it space will space be space easier space to space find space the
solution space by space eliminating space straight y space from space both space the space equations.

Solution 11

Solution 12

Solution 13

Solution 14
Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,
x + 100 = 2(y - 100)
x + 100 = 2y - 200
x - 2y = -300        ... (1)

6(x - 10) = (y + 10)
6x - 60 = y + 10
6x - y = 70            ... (2)

Multiplying equation (2) by 2, we obtain:
12x - 2y = 140        ... (3)
Subtracting equation (1) from equation (3), we obtain:
11x = 140 + 300
11x = 440
x = 40
Putting the value of x in equation (1), we obtain:
40 - 2y = -300
40 + 300 = 2y
2y = 340
y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.

Concept insight: This problem talks about the amount of capital with two friends. So, we will represent them by variables x and y respectively. Now, using the given conditions, a pair of linear equations can be formed which can then be solved easily using elimination method.

Solution 23

Solution 24

Solution 25

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise 3.114

Solution 1

begin mathsize 11px style table attributes columnalign left end attributes row cell We text  know , if  a end text subscript text 1 end text end subscript straight x plus straight b subscript 2 straight y plus straight c subscript 1 equals 0 text    ---- end text box enclose 1 end cell row cell text                    a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text    ---- end text box enclose 2 end cell row cell text for unique solution end text end cell row cell straight a subscript 1 over straight a subscript 2 not equal to straight b subscript 1 over straight b subscript 2 text        ---- end text box enclose 3 end cell row cell text Given equations are end text end cell row cell text kx - y = 2  ---- end text box enclose 4 end cell row cell text 6x - 2y =3 ---- end text box enclose 5 end cell row cell text from  end text box enclose 1 text  &  end text box enclose 4 end cell row cell text a end text subscript text 1 end text end subscript text =k    b end text subscript 1 equals negative 1 end cell row cell from text   end text box enclose 2 text  &  end text box enclose 5 end cell row cell text a end text subscript text 2 end text end subscript text =6    b end text subscript text 2 end text end subscript equals negative 2 end cell row cell from text   end text box enclose 3 end cell row cell straight k over 6 not equal to fraction numerator negative 1 over denominator negative 2 end fraction end cell row cell rightwards double arrow box enclose straight k not equal to 3 end enclose end cell end table end style

So, the correct option is (b).

Solution 2

begin mathsize 11px style table attributes columnalign left end attributes row cell we text  know, if  a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text     --- end text box enclose 1 end cell row cell text                   a end text subscript text 2 end text end subscript straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text    --- end text box enclose 2 text   end text end cell row cell text for infinite solution  end text end cell row cell text                   end text text a end text subscript text 1 end text end subscript over text a end text subscript text 2 end text end subscript equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2 text        --- end text box enclose 3 end cell row cell Given text  equation are end text end cell row cell text                   2x+3y=5     --- end text box enclose 4 end cell row cell text                   4x+ky=10   --- end text box enclose 5 text    end text end cell row cell text from equations   end text box enclose 1 text  &  end text box enclose 4 text  a end text subscript text 1 end text end subscript equals 2 text      end text straight b subscript 1 equals 3 text      end text straight c subscript 1 equals negative 5 end cell row cell text from equations  end text box enclose 2 text  &  end text box enclose 5 text  a end text subscript text 2 end text end subscript equals 4 text      end text straight b subscript 2 equals straight k text      end text straight c subscript 2 equals negative 10 end cell row cell text from equation  end text left parenthesis 3 right parenthesis end cell row cell 2 over 4 text  =  end text 3 over straight k text  =  end text fraction numerator negative 5 over denominator negative 10 end fraction end cell row cell rightwards double arrow 3 over straight k equals 1 half end cell row cell rightwards double arrow box enclose straight k equals 6 end enclose end cell end table
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 3

begin mathsize 11px style table attributes columnalign left end attributes row cell We text  know if   a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text     ---- end text box enclose 1 end cell row cell text                   a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text     ---- end text box enclose 2 end cell row cell text for no solution  end text end cell row cell text                   end text text a end text subscript text 1 end text end subscript over text a end text subscript 2 equals text b end text subscript text 1 end text end subscript over text b end text subscript 2 not equal to straight c subscript 1 over straight c subscript 2 text         ---- end text box enclose 3 end cell row cell Given text  equations are end text end cell row cell text                  x+2y-3=0        ---- end text box enclose 4 end cell row cell text                  5x+ky+7=0     ---- end text box enclose 5 end cell row cell text from  end text box enclose 1 text  &  end text box enclose 4 text     a end text subscript text 1 end text end subscript text = 1      b end text subscript text 2 end text end subscript text =2    c end text subscript text 1 end text end subscript equals negative 3 end cell row cell text from  end text box enclose 2 text  &  end text box enclose 5 text     a end text subscript 2 text = 5      b end text subscript text 2 end text end subscript text =k    c end text subscript 2 equals 7 end cell row cell text from  end text box enclose 3 text   end text 1 fifth equals 2 over straight k not equal to fraction numerator negative 3 over denominator 7 end fraction end cell row cell rightwards double arrow box enclose straight k equals 10 end enclose end cell row cell text  So, the correct option is (a). end text end cell end table end style

Solution 4

begin mathsize 11px style table attributes columnalign left end attributes row cell text if the equations are   a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text     ---- end text box enclose 1 end cell row cell text                a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text     ---- end text box enclose 2 end cell row blank row cell text for infinite solutions   end text end cell row cell text                   end text text a end text subscript text 1 end text end subscript over text a end text subscript 2 equals text b end text subscript text 1 end text end subscript over text b end text subscript 2 text         ---- end text box enclose 3 end cell row blank row cell Given text  equations are end text end cell row cell text                  3x + 5y = 0        ---- end text box enclose 4 end cell row cell text                  kx + 10y = 0     ---- end text box enclose 5 end cell row blank row cell on text   end text comparing text    end text box enclose 1 text ,  end text box enclose 2 text ,  end text box enclose 4 text ,  end text box enclose 5 end cell row cell text   end text rightwards double arrow text     a end text subscript text 1 end text end subscript text  = 3      b end text subscript text 1 end text end subscript text = 5     end text end cell row cell text           a end text subscript 2 equals straight k text       b end text subscript 2 text = 10 end text end cell row blank row cell text from  end text box enclose 3 text      end text 3 over straight k equals 5 over 10 end cell row cell rightwards double arrow box enclose straight k equals 6 end enclose text   end text end cell end table
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Solution 5

begin mathsize 11px style table attributes columnalign left end attributes row cell for text   end text infinite text  solution end text end cell row cell straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 equals straight c subscript 1 over straight c subscript 2 end cell row cell rightwards double arrow text   end text fraction numerator 2 over denominator straight a plus straight b end fraction text   end text equals text   end text fraction numerator 3 over denominator 2 straight a minus straight b end fraction text   end text equals text   end text 7 over 21 end cell row cell rightwards double arrow text   end text fraction numerator 2 over denominator straight a plus straight b end fraction text   end text equals text   end text 7 over 21 text      &     end text fraction numerator 3 over denominator 2 straight a minus straight b end fraction text   end text equals text   end text 7 over 21 end cell row cell rightwards double arrow straight a text   end text plus text   end text straight b equals 6 text    -- end text box enclose 1 text            end text rightwards double arrow 9 text   end text equals text  2 end text straight a minus straight b text    -- end text box enclose 2 end cell row cell text on adding  end text box enclose 1 text   &   end text box enclose 2 end cell row cell rightwards double arrow 3 straight a equals 15 end cell row cell rightwards double arrow box enclose straight a equals 5 end enclose text     &    end text box enclose straight b equals 1 end enclose text      end text end cell row cell So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end cell end table end style

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise 3.115

Solution 6

begin mathsize 11px style table attributes columnalign left end attributes row cell if text        end text straight a subscript 1 straight x text  + b end text subscript text 1 end text end subscript text y + c end text subscript text 1 end text end subscript text =0   --- end text box enclose 1 end cell row cell text          end text straight a subscript 2 straight x text  + b end text subscript 2 text y + c end text subscript 2 text =0   --- end text box enclose 2 end cell row blank row cell text for inconsistent solution end text end cell row cell text         end text straight a subscript 1 over straight a subscript 2 text  =   end text straight b subscript 1 over straight b subscript 2 text     --- end text box enclose 3 end cell row cell text Given equations are end text end cell row cell text            3x + y =1   --- end text box enclose 4 text   end text end cell row cell text             end text left parenthesis 2 straight k minus 1 right parenthesis text x +  end text left parenthesis straight k minus 1 right parenthesis text y = 2k+1   --- end text box enclose 5 end cell row cell text from  end text left parenthesis 1 right parenthesis text   &  end text left parenthesis 4 right parenthesis end cell row cell rightwards double arrow text  a end text subscript text 1 end text end subscript text  = 3       b end text subscript text 1 end text end subscript equals 1 text      c end text subscript text 1 end text end subscript equals negative 1 end cell row cell from text    end text left parenthesis 2 right parenthesis text   &  end text left parenthesis 5 right parenthesis end cell row cell text        end text straight a subscript 2 equals 2 straight k minus 1 text      b end text subscript text 2 end text end subscript equals straight k minus 1 text     c end text subscript 2 text = 2k+1 end text end cell row cell text from  end text box enclose 3 end cell row cell text                 end text fraction numerator 3 over denominator 2 straight k minus 1 end fraction equals fraction numerator 1 over denominator straight k minus 1 end fraction end cell row cell text          end text rightwards double arrow text   3k - 3 = 2k-1 end text end cell row cell text          end text rightwards double arrow text     end text box enclose straight k equals 2 end enclose end cell row cell text  So, the correct option is (d). end text end cell end table end style

Solution 7

begin mathsize 11px style table attributes columnalign left end attributes row cell for text  unique solution end text end cell row cell text             end text straight a over straight l not equal to straight b over straight m end cell row cell text          end text rightwards double arrow text am  end text not equal to text  bl end text end cell row cell text which is the given condition  end text end cell row cell text Hence the given equation end text end cell row cell text            am end text not equal to text bl end text end cell row cell text is the condition to a unique solution. end text end cell end table end style

So, the correct option is (a).

Solution 8

begin mathsize 11px style table attributes columnalign left end attributes row cell if text   a end text subscript text 1 end text end subscript straight x plus text  b end text subscript text 1 end text end subscript straight y plus straight c subscript 1 equals 0 text     --- end text box enclose 1 end cell row cell text     a end text subscript text 2 end text end subscript straight x plus text  b end text subscript 2 straight y plus straight c subscript 2 equals 0 text     --- end text box enclose 2 end cell row cell text    for infinite many soluations end text end cell row cell text           end text straight a subscript 1 over text a end text subscript text 2 end text end subscript equals text b end text subscript text 1 end text end subscript over text b end text subscript 2 equals straight c subscript 1 over straight c subscript 2 text      --- end text box enclose 3 end cell row cell Given text  equations are end text end cell row cell text          2x + 3y = 7      --- end text box enclose 4 end cell row cell text          2ax + (a+b)y = 28     --- end text box enclose 5 end cell row cell from text    end text box enclose 1 text   &   end text box enclose 4 end cell row cell text          a end text subscript text 1 end text end subscript text  =  end text 2 text       b end text subscript 1 text  = 3     end text straight c subscript 1 equals text  -7  end text end cell row cell from text    end text box enclose 2 text   &   end text box enclose 5 end cell row cell text          a end text subscript 2 text  =  end text 2 straight a text       b end text subscript 2 text  = a+b    end text straight c subscript 2 equals text  -28 end text end cell row cell text from  end text box enclose 3 text    end text end cell row cell rightwards double arrow text      end text fraction numerator 2 over denominator 2 straight a end fraction text  =  end text fraction numerator 3 over denominator straight a plus straight b end fraction text  =  end text fraction numerator negative 7 over denominator negative 28 end fraction text    end text end cell row cell rightwards double arrow text      end text 1 over straight a text  =  end text fraction numerator 3 over denominator straight a plus straight b end fraction text    end text rightwards double arrow text    end text box enclose 2 straight a equals straight b end enclose end cell end table end style

So, the correct option is (b).

Solution 9

begin mathsize 11px style table attributes columnalign left end attributes row cell for text  equation a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text      --- end text box enclose 1 end cell row cell text            a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text      --- end text box enclose 2 end cell row cell for space no text  solution end text end cell row cell text             end text text a end text subscript text 1 end text end subscript over straight b subscript 2 text  =  end text text a end text subscript text 1 end text end subscript over straight b subscript 2 not equal to text   end text text c end text subscript text 1 end text end subscript over straight c subscript 2 text    --- end text box enclose 3 end cell row cell given text  equations are end text end cell row cell text                    x+2y=5     --- end text box enclose 4 end cell row cell text            3x+ky+15=0     --- end text box enclose 5 end cell row cell from text   end text box enclose 1 text   &   end text box enclose 4 end cell row cell text        a end text subscript text 1 end text end subscript equals 1 text      b end text subscript text 1 end text end subscript text =2      c end text subscript text 1 end text end subscript text  =-5 end text end cell row cell text   end text from text   end text box enclose 2 text   &   end text box enclose 5 end cell row cell text        a end text subscript 2 equals 3 text      b end text subscript 2 text =k      c end text subscript 2 text  =15 end text end cell row cell from text   end text box enclose 3 text    end text end cell row cell text         end text 1 third equals 2 over straight k not equal to fraction numerator negative 5 over denominator 15 end fraction end cell row cell rightwards double arrow box enclose straight k equals 6 end enclose end cell end table end style

So, the correct option is (a).

Solution 10

begin mathsize 11px style table attributes columnalign left end attributes row cell for text  conicident lines end text end cell row cell text        end text fraction numerator 2 over denominator straight a plus straight b end fraction text   end text equals text   end text fraction numerator negative 3 over denominator negative left parenthesis straight a plus straight b minus 3 right parenthesis end fraction text  =  end text fraction numerator 7 over denominator 4 straight a plus straight b end fraction end cell row cell rightwards double arrow fraction numerator 2 over denominator straight a plus straight b end fraction equals fraction numerator 7 over denominator 4 straight a plus straight b end fraction end cell row cell rightwards double arrow 8 straight a plus 2 straight b equals 7 straight a plus 7 straight b end cell row cell rightwards double arrow text   end text box enclose straight a equals 5 straight b end enclose end cell end table
rightwards double arrow straight a space minus space 5 straight b equals 0 end style

So, the correct option is (c).

Solution 11

Consistent solution means either linear equations have unique solutions or infinite solutions.

⇒ In case of unique solution; lines are intersecting

⇒ If solutions are infinite, lines are coincident.

So, lines are either intersecting or coincident

So, the correct option is (d).

Solution 12

begin mathsize 11px style Intercept space on space straight x space minus space axis space equals space straight a
Intercept space on space straight y minus space axis space equals space straight b

table attributes columnalign left end attributes row cell space space space area text   end text of text   end text triangle equals 1 half cross times straight a cross times straight b end cell row cell text                               end text equals 1 half ab end cell end table end style

So, the correct option is (c).

Solution 13

begin mathsize 11px style table attributes columnalign left end attributes row cell area text  of triangle =  end text 1 half cross times 6 cross times 6 end cell row cell text                             = 18 sq. units    end text end cell end table end style

So, the correct option is (b).

Solution 14

begin mathsize 11px style table attributes columnalign left end attributes row cell If text  the equations,       a end text subscript text 1 end text end subscript straight x plus straight b subscript 1 straight y plus straight c subscript 1 equals 0 text       --- end text box enclose 1 end cell row cell text                                 a end text subscript 2 straight x plus straight b subscript 2 straight y plus straight c subscript 2 equals 0 text       --- end text box enclose 2 end cell row cell For space infinitely text   end text many text  solution    end text text a end text subscript blank to the power of text 1 end text end exponent end subscript over straight a subscript 2 equals text b end text subscript blank to the power of text 1 end text end exponent end subscript over straight b subscript 2 equals text c end text subscript blank to the power of text 1 end text end exponent end subscript over straight c subscript 2 text  --- end text box enclose 3 end cell row cell Comparing text  the given equations end text to the power of blank text  to  end text box enclose 1 text   &   end text box enclose 2 end cell row cell We text  get, end text end cell row cell text           a end text subscript text 1 end text end subscript equals 2 text          b end text subscript text 1 end text end subscript equals 3 text          c end text subscript text 1 end text end subscript equals negative 5 end cell row cell text           a end text subscript 2 equals 4 text          b end text subscript 2 equals straight k text          c end text subscript text 1 end text end subscript equals negative 10 end cell row cell rightwards double arrow text     end text from text   end text box enclose 3 end cell row cell text                    end text 2 over 4 equals 3 over straight k equals fraction numerator negative 5 over denominator negative 10 end fraction end cell row cell text               end text rightwards double arrow text   end text 3 over straight k equals 1 half end cell row cell text              end text rightwards double arrow text     end text box enclose straight k equals 6 end enclose end cell end table end style

So, the correct option is (d).

Solution 15

begin mathsize 11px style table attributes columnalign left end attributes row cell If text  equations end text to the power of blank text           a end text subscript text 1 end text end subscript straight x text  + b end text subscript text 1 end text end subscript text y + c end text subscript text 1 end text end subscript text  = 0         --- end text box enclose 1 end cell row cell text                      and   a end text subscript 2 straight x text  + b end text subscript 2 text y + c end text subscript text 2 end text end subscript text  = 0         --- end text box enclose 2 end cell row cell have text  no solution, then end text end cell row cell text                  end text straight a subscript 1 over straight a subscript 2 equals straight b subscript 1 over straight b subscript 2 not equal to straight c subscript 1 over straight c subscript 2 text       --- end text box enclose 3 end cell row cell On text  comparing with given equation end text to the power of blank text   to   end text box enclose 1 text    &   end text box enclose 2 end cell row cell we text   end text get end cell row cell text          a end text subscript text 1 end text end subscript equals straight k text        b end text subscript text 1 end text end subscript equals text  -5      c end text subscript text 1  end text end subscript equals text  -2 end text end cell row cell text          a end text subscript 2 equals 6 text        b end text subscript 2 equals text  2      c end text subscript 2 equals text  -7 end text end cell row cell text From equation  end text box enclose 3 end cell row cell text             end text straight k over 6 equals text   end text fraction numerator negative 5 over denominator 2 end fraction text   end text not equal to text   end text 2 over 7 end cell row cell text       end text rightwards double arrow text    end text box enclose straight k equals negative 15 end enclose end cell end table end style

So, the correct option is (d).

Solution 16

begin mathsize 11px style table attributes columnalign left end attributes row cell area text  of triangle  =  end text 1 half cross times 1 cross times 1 end cell row cell text                              =  end text 1 half text  sq. unit end text end cell end table end style

So, the correct option is (a).

Chapter 3 - Pairs of Linear Equations in Two Variables Exercise 3.116

Solution 17

begin mathsize 11px style From space figure space shown comma space the space required space triangle space is space as space shown space above.
BD space equals space 5 space Units space space space and space space space space space PO space equals space 3 space Units
table attributes columnalign left end attributes row cell area text   end text of space triangle text   end text equals 1 half cross times 5 cross times 3 end cell row cell text                               =  end text 15 over 2 equals text  7.5 sq. Units end text end cell end table end style

So, the correct option is (b).

Solution 18

Solution 19

Since x = a and y = b is the solution of given system of equations x - y = 2 and x + y = 4, we have

a - b = 2 ….(i)

a + b = 4 ….(ii)

Adding (i) and (ii), we have

2a = 6 a = 3

b = 4 - 3 = 1

Hence, correct option is (a).

Solution 20

Solution 21