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

Increase your scores in CBSE Class 10 by scoring full marks in Mathematics. TopperLearning answers every query and provides solutions to every need of a 10th grader. Preparing CBSE Class 10 Maths becomes easier with Maths NCERT Solution for CBSE Class 10. Every student must try to attempt all the questions from the NCERT textbook. These NCERT Solutions for CBSE Class 10 Maths will help you with detailed and stepwise answers. NCERT Solutions for CBSE Class 10 Maths not only include the solutions but also assist you with conceptual insight to gain perfect clarity. Students will be able to grasp the concepts better with these handy NCERT solutions. This will help them to develop their interest in CBSE Class 10 Mathematics and also increase their confidence. These solutions are drafted by an experienced team of academic experts following CBSE's proper pattern and guidelines.

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

Algebra gives an astounding sense of confidence to students. Since it based on calculations and solutions, it is popular as a scoring aspect of CBSE Class 10 Maths. Mathematics Chapter Pairs of Linear equations in two Variables helps in achieving this goal for the students. As per the CBSE guidelines, Maths Chapter Pairs of Linear equations in two Variables is one of the highest weightage carrying chapters in CBSE Class 10 Maths. Students are acquainted with this concept learnt in previous grades. In CBSE Class 10 Maths, students will encounter different and interesting techniques to solve equations that have two variables. Students will also understand the geometrical application in Pairs of Linear equations in two Variables. Given below is a list of topics focussed on Maths Chapter Pairs of Linear equations in two Variables:

●      Pair of Linear Equations in Two Variables: Standard Equation and its terminologies

●      Geometrical application of the linear equations

●      Graphical Method of Solution of a Pair of Linear Equations

●      Algebraic Methods of Solving a Pair of Linear Equations

●      Substitution Method

●      Elimination Method

●      Cross-Multiplication Method

●      Equations Reducible to a Pair of Linear Equations in Two Variables

Go through every exercise from this chapter and refer to Pairs of Linear equations in two Variables NCERT Solutions

CBSE Class 10 Mathematics Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.1

Discover the easiest way of finding a solution to the given linear equations with two variables using geometrical applications. Locate the point on the graph and assess your solution. Attempt every question and refer to CBSE Class 10 Maths textbook Solutions to assess your answers. This exercise often has its reference in the multiple-choice questions. Attempt every kind of question and prepare yourself for the board examination with CBSE Class 10 Maths study material. TopperLearning’s CBSE Class 10 Maths study material helps students score more marks.

CBSE Class 10 Mathematics Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.2

Taking reference from the previous exercise, use the graphical method to find the solution of the linear equation with two variables. Students will require a graph paper for this exercise. Stepwise deductions and processes leading to the solutions are explained in CBSE Class 10 Maths video lessons. Practise with CBSE Class 10 Maths sample papers, CBSE Class 10 Maths most important questions and CBSE Class 10 Maths past year papers and gain precision in this method. CBSE Class 10 Maths doubt solver will solve the doubts that you face.

CBSE Class 10 Mathematics Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.3

Use the Substitution Method to solve the linear equations and tally your solutions with Pairs of Linear equations in two Variables NCERT Solutions. Substitute the value of either variable from one of the equations and substitute it in the other. All the answers including calculations and stepwise methods can be referred to by the students in Pairs of Linear equations in two Variables NCERT Solutions.

CBSE Class 10 Mathematics Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.4

Consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Figure out how to use this method with CBSE Class 10 Maths video lessons. Solve the linear equations using the method of elimination in Mathematics Chapter Pairs of Linear equations in two Variables and tally your answers with CBSE Class 10 Maths textbook Solutions. This method is useful for practical applications too. Be careful of the calculations and practise more to gain precision. Assessing your knowledge also helps in gaining a better perspective at any method hence use CBSE Class 10 Maths study material and increase your scores. Attempt all the important questions using CBSE Class 10 Maths most important questions

CBSE Class 10 Mathematics Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.5

CBSE Class 10 Mathematics introduces one more algebraic method to solve a pair of linear equations which for many reasons is a very useful method of solving the equations. It is the cross multiplication method. It process by cross multiplying the coefficients of both the variables of the given two equations. It reduces the complexity and allows students to find the solutions of the equations in a stepwise manner with basic calculations as elaborated in CBSE Class 10 Maths video lessons. CBSE Class 10 Maths textbook Solutions helps you with all the solutions and calculations drafted in a comprehensible manner by our team of subject experts.

CBSE Class 10 Mathematics Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.6

Students are expected to be able to solve linear equations with two variables choosing any of the methods learnt in previous exercises. In this section, we will discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. This is the advanced level application of Algebra that can be understood with stepwise progress using CBSE Class 10 Maths study material. Attempt these questions and observe all the steps with calculations as mentioned in Pairs of Linear equations in two Variables NCERT Solutions.

Questions from this exercise are popular to be asked in the 3-4 mark category.

CBSE Class 10 Mathematics Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.7

Pairs of Linear equations in two Variables NCERT Solutions also comprises solutions to optional exercises from this chapter. Attempt some higher-order questions and word problems most simply to gain better conceptual clarity with CBSE Class 10 Maths textbook Solutions. Get all your concerns about these questions resolved with CBSE Class 10 Maths Undoubt. With this exercise, students are advised to focus on the method and steps in solving the problems.

Revision during the exams should be crisp and should take less time. Considering this fact, TopperLearning brings out CBSE Class 10 Maths revision notes for quick and apt revision.

Use these and ace your exams.

Page / Exercise

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

Solution 1
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 representatino 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 2

Let the cost of a bat and a ball be Rs x and Rs y respectively.

The given conditions can be algebraically represented as:
3x + 6y = 3900 x + 2y = 1300   .... (i)

x + 3y = 1300   .... (ii)

From (i), we get

x = 1300 - 2y

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

 x 300 100 -100 y 500 600 700

From (ii), we get

x = 1300 - 3y

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

 x 100 400 -200 y 400 300 500

The graphical representation is as follows:

Concept insight: Cost of bats and balls needs to be found so the cost of a ball and bat will be taken as the variables. Applying the conditions of total cost of bats and balls algebraic equations will be obtained. Then, in order to represent the obtained equations graphically take atleast three ordered pairs on both the equations in order to avoid any computational errors.

Solution 3
Let the cost of 1 kg of apples and 1 kg grapes be Rs. x and Rs. y.
The given condtions 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 like 1 cm = 20.

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

Solution 1
(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 5 4 6 y 5 6 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.

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 2

(i)  5x - 4y + 8 = 0
7x + 6y - 9 = 0
Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,     we get:

Since , the given pair of equations intersect at exactly one point.

(ii)  9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get:

Since , the given pair of equation are coincident.

(iii)   6x - 3y + 10 = 0 , 2x - y + 9 = 0

Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get:

Since , the given pair of equation are parallel to each other.

Concept insight: In order to answer such questions, remember the condition for the pair of linear equations to be intersecting, parallel or coincident. Also, while writing the coefficients, don't forget to take the signs.

Solution 3

Since , the given pair of equation has only one solution.

Thus, the pair of linear equations is consistent.

Thus, the pair of linear equations is inconsistent.

Thus, the pair of linear equations is consistent.

Thus, the pair of linear equations is consistent.

Thus, the pair of linear equations is consistent.

Concept Insight: If a pair of linear equations has one or more than one solution then they are said to be consistent and if they have no solution then they are said to be inconsistent. So, to identify the consistency of a given pair of equations, apply the conditions involving the coefficients of the given pair of equations.
In case, two consistent linear equations are plotted, they will either intersect or overlap each other.

Solution 4

Thus, the pair of linear equations is consistent.
Now, x + y = 5 x = 5 - y
Three solutions of this equation can be written in a table as follows:
 x 4 3 2 y 1 2 3

Three solutions of this equation can be written in a table as follows:
 x 4 3 2 y 1 2 3

Thus, the graphical representation is as follows:

From the graph, it can be observed that the two lines coincide. Thus, the given pair of equations has infinite solutions.
Let x = k, then y = 5 - k. So, the ordered pair (k, 5 - k) , where k is a constant, will be the solution of the given pair of linear equations.

Thus, the pair of linear equations is inconsistent.

Thus, the pair of linear equations is consistent.

Now, 2x + y - 6 = 0  y = 6 - 2x
Three solutions of this equation can be written in a table as follows:
 x 0 1 2 y 6 4 2

Three solutions of this equation can be written in a table as follows:
 x 1 2 3 y 0 2 4

Thus, the graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (2, 2). Thus, the solution of the given pair of equations is (2, 2).

Thus, the pair of linear equations is inconsistent.

Concept insight: If a pair of linear equations has one or more than one solutions then they are said to be consistent and if they have no solution then they are said to be inconsistent. The graph of each equation can be plotted by taking at least three ordered pairs which are the solutions of the equations. The point where both the lines intersect will be the solution of the given pair of equations. Remember two overlapping lines intersect each other at infinitely many points. State the solution explicitly.
Solution 5
Let the width and length of the rectangular garden be x and y respectively.

According to the given conditions,
y - x = 4
y + x = 36

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

 x 0 8 12 y 4 12 16

y + x = 36

Three solutions of this equation can be written in a table as follows:
 x 0 36 16 y 36 0 20

Thus, the graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (16, 20). So, x = 16 and y = 20.
Thus, the length and width of the rectangular garden is 20 m and 16 m respectively.

Concept insight: Here dimensions of the rectangular garden needs to be found. Since opposite sides of the rectangle are equal so length and breadth can be taken as variables. Applying conditions given in the problem two linear equations in the 2 variables can be obtained. Now, in order to represent the obtained equations graphically, take the values of variables as whole numbers only, because then it will be easier to represent the values on the graph. The point where the two equations intersect will give the required dimensions. State the dimensions length and breadth from the values of the variables.
Solution 6
(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 7
x - y + 1 = 0  x = y - 1
Three solutions of this equation can be written in a table as follows:

 x 0 1 2 y 1 2 3

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

 x 4 2 0 y 0 3 6

Now, these equations can be drawn on a graph. The triangle formed by the two lines and the x-axis can be shown by the shaded part as:

From the graph, it can be observed that the coordinates of the vertices of the triangle so formed are (2, 3), (-1, 0), and (4, 0).

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 x-axis and also where the two lines intersect each other. Note here that the coordinates of the intersection of lines with x-axis is taken and not with y-axis, this is because the question says to find the triangle formed by the two lines and the x-axis.

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

Solution 1
(i) x + y = 14      ... (i)
x - y = 4        ... (ii)
From (i), we obtain:
x = 14 - y      ... (iii)
Substituting this value in equation (ii), we obtain:

Substituting the value of y in equation (iii), we obtain:

From (i), we obtain:
s = t + 3                   ...(iii)

Substituting this value in equation (ii), we obtain:

Substituting the value of t in equation (iii), we obtain:
s = 9

(iii)  3x - y = 3        ... (i)
9x - 3y = 9        ... (ii)
From (i), we obtain
y = 3x - 3        ... (iii)
Substituting this value in equation (ii), we obtain:

9 = 9
This is always true.

Thus, the given pair of equations has infinitely many solutions and the relation between these variables can be given by
y  = 3x - 3
So, one of the possible solutions can x = 3, y = 6.

Substituting the value of y in equation (iii), we obtain:
x = 0

Concept insight: In order to solve the given pairs of equations, we need to substitute the value of any one of the variable from any one of the equation. But make sure to substitute the value of that variable which simplifies calculations. For example, in part (iv) it is most convenient to substitute the value of x from the first equation to the second equation, as the division by 0.2 is more easier than the division by 0.3, 0.4 and 0.5.
Solution 2

Thus, the value of m is -1.

Concept insight: Firstly the solution of the given pair of linear equations can be found out by substituting the value of one variable, say x, from one equation into the other equation. Then after finding out the values of x and y, substitute them in the equation y = m x + 3 in order to find the value of m.
Solution 3
(i) Let one number be x and the other number be y such that y > x.

According to the question:

Substituting this in equation (1), we obtain
y = 39
Hence, the numbers are 13 and 39.

Concept insight: In this problem, two relations between two numbers are given. So, the two numbers have to be found out here. So the two numbers will be represented by variables x and y explicitly state the greater variable.

A pair of equations can be obtained from the given conditions. The pair of equations can then be solved by suitable substitution.

(ii)  Let the larger angle be x and smaller angle be y.
We know that the sum of the measures of angles of a supplementary pair is always 180º.
According to the given information,

From (1), we obtain
x = 180º - y     (3)
Substituting this in equation (2), we obtain

Putting this in equation (3), we obtain
x = 180º - 81º
= 99º
Hence, the angles are 99º and 81º.

Concept insight: This problem talks about the measure of two supplementary angles. So, the two angles will be written as variables. The pair of equations can be formed using the fact that the sum of two supplementary angles is 180° and using the condition given in the problem. The pair of equations can then be solved by suitable substitution.

(iii)  Let the cost of a bat and a ball be x and y respectively.
According to the given information,

Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.

Concept insight: Cost of bats and balls needs to be found so the cost of a ball and bat will be taken as the variables. Applying the conditions of total cost of bats and balls algebraic equations will be obtained. The pair of equations can then be solved by suitable substitution.

Concept insight: In this problem, we are required to find out the fixed charge and the charge per km. So, we will represent these two by using different variables. Now, two linear equations can be written by using the conditions given in the problem. The pair of equations can then be solved by suitable substitution.

(v)  Let the fraction be
According to the given information,

Concept insight: This problem talks about a fraction which is not known to us. So numerator and denominators will be taken to be variables x and y respectively and y will be strictly non zero. Then, a pair of linear equations can be formed from the given conditions. The pair of equations can then be solved by suitable substitution.

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

Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.

Concept insight: Here, Jacob's and his son's present age are not known. So, we will write both these in terms of variables. The problem talks about their ages five years ago and five years hence. Here, five years ago means we have to subtract 5 from their present ages, and five years hence means we have to add 3 to their present ages. So, using the given conditions, a pair of linear equations can be formed. The pair of equations can then be solved by suitable substitution.

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

Solution 1

(ii)      Elimination method:
3x + 4y = 10      ...(1)
2x - 2y = 2         ...(2)
Multiplying equation (2) by 2, we obtain:
4x - 4y = 4         ...(3)
Adding equation (1) and (3), we obtain
7x = 14
x = 2
Substituting the value of x in equation (1), we obtain:
6 + 4y = 10
4y = 4
y = 1
Hence, x = 2, y = 1

Substitution method:
From equation (2), we obtain:
x = 1 + y            ...(4)
Putting this value in equation (1), we obtain:
3(1 + y) + 4y = 10.
7y = 7
y = 1
Substituting the value of y in equation (4), we obtain:

(iv)      Elimination Method:

Substitution method:
From equation (2), we obtain:
y = 3x - 9        ... (3)
Putting this value in equation (1), we obtain:
3x + 4(3x - 9) = -6
15x = 30
x = 2
Substituting the value of x in equation (3), we obtain:
y = 6 - 9 = -3

Concept insight: In order to solve the given pairs of equations by elimination method, it is required to eliminate one variable to get a linear equation in one variable. For elimination, we multiply one equation (or sometimes even both equations) by suitable constant to make the coefficient of one variable same in both the equations. The trick is to eliminate that variable which involves lesser computations.

In order to solve the given pair of equations by substitution method, we need to substitute the value of any one of the variable from any one of the equation. The trick here is to make sure you substitute the value of that variable which simplifies your calculations.
Note that the solution must be same in both the cases.

Solution 2

(i)    Let the fraction be

According to the question,

Subtracting equation (1) from equation (2), we obtain:
x = 3
Substituting this value of x in equation (1), we obtain:

Concept insight: This problem talks about a fraction. The numerator and denominator of the fraction are not known so we represent these as variables x and y respectively where variable y must be non zero. Then, a pair of linear equations can be formed from the given conditions. The pair of equations can then be solved by eliminating a suitable variable.

(ii)  Let present age of Nuri and Sonu be x and y respectively.

According to the question,

Subtracting equation (1) from equation (2), we obtain:
y = 20
Substituting the value of y in equation (1), we obtain:

Thus, the age of Nuri and Sonu are 50 years and 20 years respectively.

Concept insight: Here, Nuri's and Sonu's present age are not known. So, we will write both these in terms of variables. Then, using the given conditions, a pair of linear equations can be formed. The pair of equations can then be solved by eliminating a suitable variable.

(iii)  Let the units digit and tens digit of the number be x and y respectively.

Number = 10y + x
Number after reversing the digits = 10x + y

According to the question,
x + y = 9          ... (1)
9(10y + x) = 2(10x + y)
88y - 11x = 0
- x + 8y =0          ... (2)
Adding equations (1) and (2), we obtain:
9y = 9
y = 1
Substituting the value of y in equation (1), we obtain:
x = 8
Thus, the number is 10y + x = 10 x 1 + 8 = 18

Concept insight: This problem talks about a two digit number. Here, remember that a two digit number xy can be expanded as 10x + y. Then, using the two given conditions, a pair of linear equations can be formed which can be solved by eliminating one of the variables.

(iv)  Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.
According to the question,

Multiplying equation (1) by 50, we obtain:
50x + 50 y = 1250           .... (3)
Subtracting equation (3) from equation (2), we obtain:

50y = 750
y = 15

Substituting the value of y in equation (1), we obtain:
x = 10
Hence, Meena received 10 notes of Rs 50 and 15 notes of Rs 100.

Concept insight: This problem talks about two types of notes, Rs 50 notes and Rs 100 notes. And the number of both these notes with Meena is not known. So, we denote the number of Rs 50 notes and Rs 100 notes by variables x and y respectively. Now two linear equations can be formed by the given conditions which can be solved by eliminating one of the variables.

(v)    Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
According to the question,

Subtracting equation (2) from equation (1), we obtain:

2y = 6
y = 3

Substituting the value of y in equation (1), we obtain:

x + 12 = 27
x = 15

Hence, the fixed charge is Rs 15 and the charge per day is Rs 3.

Concept insight: Here, the fixed charges for the first three days and per day charges are not known so, they will be represented  using two different variables. The two equations can then be obtained by using the given conditions which can be solved by eliminating one of the variables.

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

Solution 1

Thus, the given pair of equations has no solution.

Thus, the given pair of equations has unique solution.

By cross-multiplication method,

Thus, the given pair of equations has infinite solutions.
The solutions can be obtained by assuming the value of x to be k, where k is any

constant. So, the ordered pairs , where k is a constant, are the solutions of

the given pair of equations.

Thus, the given pair of equations has unique solution.

By cross-multiplication,

Concept Insight: In order to answer such questions, remember the condition for the pair of linear equations to have unique, infinitely many or no solution relating the coefficients. Also, remember the formula used to solve by cross multiplication method. While applying the formula, be careful about the signs of the coefficients. And above all don't forget to first write the linear equations in standard form which is ax + by + c = 0.

Solution 2

Concept insight: In order to solve such problems, firstly write the linear equations in standard form which is ax + by + c = 0.To find the value of the unknowns, the key idea is to remember the conditions for a given pair of equations to have infinite solutions and no solution.  In case of infinite solutions rule out the values which does not satisfy all the ratios. Also, be careful about the signs of the coefficients.
Solution 3

Substituting this value of y in equation (1), we obtain:

Cross-multiplication method:

Concept insight: In order to solve the given pair of equations by cross multiplication method, remember the formula to be used and convert the system of equations to standard form. While applying the formula, be careful about the signs of the coefficients.

In order to solve the given pair of equations by substitution method, substitute the value of any one of the variable from any one of the equation. Make sure you substitute the value of that variable which simplifies your calculations.
Solution will be same in both cases

Solution 4

(i)  Let the fixed charge of the food and the charge for food per day be x and y respectively.
According to the question,
x + 20y = 1000     ...(1)
x + 26y = 1180   ...(2)

x = 1180 - 26y

Substituting the value of x in equation (1) , we obtain
6y = 180
y = 30

Substituting the value of y in equation (1), we obtain:
x+ 20 x 30 = 1000
x = 1000 - 600
x = 400

Thus, the fixed charge of the food and the charge per day are Rs 400 and Rs 30 respectively.

Concept insight: Here, the fixed charge of the food and charge for food per day are the unknown quantities. So they are taken as variables x and y. The two equations can then be obtained by using the given conditions. You will observe that the variable x has the same coefficient in both the equations, so it will be easier to find the solution by eliminating x from both the equations. Also, one can solve the system by other methods.

(ii)  Let the fraction be According to the question,

According to the question,

Subtracting equation (1) from equation (2), we obtain:
x = 5
Putting the value of x in equation (1), we obtain:
15 - y = 3
y = 12

Concept insight: Since the problem asks for a fraction. The numerator and denominator of the fraction need to be represented by variables. A pair of linear equations can be obtained from the given conditions. Observe that the variable y has the same coefficient in both the equations, so it will be easier to find the solution by eliminating y from both the equations.

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

According to the question,

Subtracting equation (2) from equation (1), we obtain:
x = 15
Substituting the value of x in equation (2), we obtain:
30 - y = 25
y = 5

Thus, the number of right answers and the number of wrong answers is 15 and 5 respectively.
Therefore, the total number of questions in the test is 20.

Concept insight: In this problem, the number of right answers and the number of wrong answers answered by Yash are the unknown quantities. So, they must be represented by two variables. A pair of linear  equations can be obtained by applying the given condition. Variable y has the same coefficient in both the equations, so it will be easier to find the solution by eliminating y from both the equations.

(iv)  Let the speed of first car and second car be u km/h and v km/h respectively.

Speed of both cars while they are travelling in same direction = (u - v) km/h
Speed of both cars while they are travelling in opposite directions i.e., when they are travelling towards each other = (u + v) km/h

Distance travelled = Speed x Time

According to the question,

Substituting the value of u in equation (2), we obtain:
v = 40

Hence, speed of the first car is 60 km/h and speed of the second car is 40 km/h.

Concept insight: In this problem, the speeds of the two cars are not known so we will represent them by variables. Remember that when the cars are travel in the same direction, then the speed will be equal to the difference of their speeds and when they travel in opposite direction, then the speed will be equal to the sum of their speeds. Don't miss to write the unit.

(v)  Let length and breadth of rectangle be x unit and y unit respectively.

Area = xy

According to the question,

Thus, the length and breadth of the rectangle are 17 units and 9 units respectively.

Concept insight: Here, the length and the breadth of the rectangle will be represented by variables. Then, the pair of equations will be written from the given conditions. Solution can be easily computed  using the cross multiplication method.

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

Solution 1

Using cross-multiplication method, we obtain:

Concept insight: Here, the two given equations are not linear as the variables x and y both are in the denominator. By substituting  given equation reduces to

linear equation in variables p and q. Note here that we are assuming the variables x and y to be non-zero, because only then the given equations will make sense. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Multiplying equation (1) by 3, we obtain:
6p + 9q = 6       (3)
Adding equation (2) and (3), we obtain:

Concept insight: Here, the two given equations are not linear as it involves  x-1/2  and y-1/2  If we substitute  in the equations, then they will reduce to

linear form in variables p and q. Note here that we are assuming the variables x and y to be non negative real numbers. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the variable x is in the denominator. If we substitute   in the equations, then they will reduce to linear form in

variables p and y. Note here that we are assuming the variable x to be non-zero. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the value of p to find x.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. If we substitute   in the

equations, then they will reduce to linear form in variables p and q. Note here that we are assuming x  1 and y 2. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. So, here our first step will be to break the expression on the LHS. Thereafter, if we will substitute  in the equations, then they will

reduce to linear form in variables p and q. Note here that we are assuming x 0 and y 0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS of both the equations are not linear. So, first step will be to divide both LHS and RHS by xy. Thereafter, substitute  in the equations, then they will reduce to linear

form in variables p and q. Note here that we are assuming x 0 and y 0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to find x and y.

Concept insight: Here, the two given equations are not linear as the expressions on the LHS are not linear. If we substitute  in the equations, then they will

reduce to linear form in variables p and q. Note here that we are assuming x + y 0 and x - y 0, otherwise  the equations will not make sense. The linear equations can then be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q and two linear equations x + y = 5 and x - y = 1 will be obtained. And these equations can be solved easily by elimination method.

The given equations reduce to:

Concept insight: Here, the two given equations are not linear as the expressions on the LHS are not linear. If we substitute  in the equations, then they will

reduce to linear form in variables p and q. Note here that we are assuming 3x + y 0 and 3x - y 0. The linear equations can be solved by using any suitable algebraic method. Also, do not forget here that we need to find the values of x and y. So, put the values of p and q to obtain two linear equations
3x + y = 4 and 3x - y = 2. These equations can then be solved easily by elimination method.
Solution 2

(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

Speed of Ritu while rowing downstream = (x + y) km /h

According to the question,

Adding equations (1) and (2), we obtain:

2x = 12

x = 6

Putting the value of x in equation (1), we obtain:

y = 4

Thus, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

Concept insight: Two unknown quantities are speed in still water and the speed of the current, which can be represented by variables x and y respectively. Then using the given conditions, two linear equations can be formed. Remember the speed while rowing upstream will be x + y and while rowing downstream, it will be x – y. Now, the two equations can be solved easily by the substitution method.

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

Work done by a woman in 1 day =

Work done by a man in 1 day =

According to the question,

(iii)     Let the speed of train and bus be u km/h and v km/h respectively.

Concept insight: In this problem it is mentioned that a girl Roohi travels to her home partly by train and partly by bus. So, we represent the speed of the train and bus by variables u and v respectively. Here, to answer this problem, remember the fact that speed is the ratio of distance and time. Using this and the given conditions, formulate pair of linear equations. Observe that the equations obtained are not linear. So, first step will be to convert them into linear form by suitable substitution and then solve it using an appropriate algebraic method.

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

Solution 1
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)

4x - y    = 60    ... (2)

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

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 2
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 3
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 4

Let the number of rows be x and number of students in a row be y.
Total number of students in the class = Number of rows x Number of students in a row
= xy

According to the question,
Total number of students = (x - 1) (y + 3)
xy = (x - 1) (y + 3)

= xy - y + 3x - 3
3x - y - 3 = 0
3x - y = 3        ... (1)

Total number of students = (x + 2) (y - 3)
xy = xy + 2y - 3x - 6
3x - 2y = -6    ... (2)

Subtracting equation (2) from (1), we obtain:
y = 9

Substituting the value of y in equation (1), we obtain:
3x - 9 = 3
3x = 9 + 3 = 12
x = 4

Number of rows = x = 4
Number of students in a row = y = 9

Total number of students in a class = xy = 4 x 9 = 36

Concept insight: This problem talks about some number of students who are made to stand in a row. So, in order to know the total number of students, it is must  to know the number of students standing in each row and the total number of rows. So, these two quantities are represented by variables x and y. Use the conditions given in the question to obtain a pair of equations. The pair of equations can then be solved easily by eliminating a suitable variable.

Solution 5
C = 3B = 2(A + B)
3B = 2(A + B)
B = 2A
2A - B = 0 ... (1)

We know that the sum of the measures of all angles of a triangle is 180°.
A + B + C = 180°
A + B + 3B = 180°
A + 4B = 180° ... (2)

Multiplying equation (1) by 4, we obtain:
8A - 4B = 0 ... (3)

Adding equations (2) and (3), we obtain:

9A = 180° A = 20°

From equation (2), we obtain:
20° + 4B = 180° 4B = 160°
B = 40°
C = 3B = 3 x 40° = 120°
Thus, the measure of A, B and C are 20°, 40°, and 120° respectively.

Concept insight: Here the problem is about three angles of a triangle. By simplifying the given relation, a linear equation is obtained. Using the angle sum property of triangles, i.e., the sum of all the three angles of a triangle is 180° another relation is obtained. So, from these two relations, a pair of linear equations will be obtained which can be easily solved by elimination method.
Solution 6

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).

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 7
(i)
px + qy = p - q        ... (1)
qx - py = p + q        ... (2)
Multiplying equation (1) by p and equation (2) by q, we obtain:
p2x + pqy = p2 - pq        ... (3)
q2x - pqy = pq + q2        ... (4)
Adding equations (3) and (4), we obtain:
p2x + q2x = p2 + q2
(p2 + q2)x = p2 + q2

Substituting the value of x in equation (1), we obtain:
p(1) + qy = p - q
qy = -q
y = -1

(ii)
ax + by = c             ... (1)
bx + ay = 1 + c        ... (2)
Multiplying equation (1) by a and equation (2) by b, we obtain:
a2x + aby = ac         ... (3)
b2x + aby = b + bc    ... (4)
Subtracting equation (4) from equation (3),
(a2 - b2)x = ac - bc - b

Substituting the value of x in equation (1), we obtain:

bx - ay  = 0                         ... (1)
ax + by = a2 + b2                 ... (2)
Multiplying equation (1) and (2) by b and a respectively, we obtain:
b2x - aby = 0                       ... (3)
a2x + aby = a3 + ab2            ... (4)
Adding equations (3) and (4), we obtain:
b2x + a2x = a3 + ab2
x(b2 + a2) = a(a2 + b2)
x = a
Substituting the value of x in equation (1), we obtain:
b(a) - ay = 0
ab - ay = 0
ay = ab
y = b

(iv)
(a - b)x + (a + b)y = a2 - 2ab - b2        ... (1)
(a + b)(x + y) = a2 + b2

Subtracting equation (2) from (1), we obtain:
(a - b)x - (a + b)x = (a2 - 2ab - b2) - (a2 + b2)
(a - b - a - b)x = -2ab - 2b2
-2bx = -2b(a + b)
x = a + b
Substituting the value of x in equation (1), we obtain:
(a - b)(a + b) + (a + b)y = a2 - 2ab - b2
a2 - b2 + (a + b)y = a2 - 2ab - b2
(a + b)y = -2ab

Concept insight: In all the above parts, the given pair of equations can be easily solved by eliminating a suitable variable. Although one can use any of the other methods but the elimination method is the best suited over here.

(v)
152x - 378y = -74                    ... (1)
-378x + 152y = -604                 ... (2)
Adding the equations (1) and (2), we obtain:
-226x - 226y = -678
Subtracting the equation (2) from equation (1), we obtain:
530x - 530y = 530
Adding equations (3) and (4), we obtain:
2x = 4
x = 2
Substituting the value of x in equation (3), we obtain:
y = 1

Concept insight: Here, since the coefficients of x and y in the two equations are interchanging this types of problems can be easily  solved by once adding the given equations and then  subtracting the given equations. By doing this  an equation of the form x - y = a and x + y = b, where a and b are constants will be obtained. And these equations can now be solved easily by elimination method.
Solution 8
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 = 120°
B = 3y - 5 = 3(25) - 5 = 70°
C = -4x = -4(-15) = 60°
D = -7x + 5 = -7(-15) + 5 = 110°

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 180°. By using this relation, two linear equations can be obtained which can be solved easily by eliminating a suitable variable.

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