# Linear Equation class 10 ncert solutions

## Introduction

If two linear equations have the two same variables, they are called a pair of linear equations in two variables. Following is the most general form of linear equations:

Here, a1, a2, b1, b2, c1 and c2 are real numbers such that;

A pair of linear equations can be represented and solved by the following methods:
1. Graphical method
2. Algebraic method

### Graphical Method:

For a given pair of linear equations in two variables, the graph is represented by two lines.
1. If the lines intersect at a point, that point gives the unique solution for the two equations. If there is a unique solution of the given pair of equations, the equations are called consistent.
2. If the lines coincide, there are indefinitely many solutions for the pair of linear equations. In this case, each point on the line is a solution. If there are infinitely many solutions of the given pair of linear equations, the equations are called dependent (consistent).
3. If the lines are parallel, there is no solution for the pair of linear equations. If there is no solution of the given pair of linear equations, the equations are called inconsistent.

#### Algebraic Method:

There are following methods for finding the solutions of the pair of linear equations:
1. Substitution method
2. Elimination method
3. Cross-multiplication method
If a pair of linear equations is given by  then following situations can arise.
Situation 1:

In this case, the pair of linear equations is consistent. This means there is unique solution for the given pair of linear equations. The graph of the linear equations would be two intersecting lines.
Situation 2:

In this case, the pair of linear equations is inconsistent. This means there is no solution for the given pair of linear equations. The graph of linear equations will be two parallel lines.
Situation 3:

In this case, the pair of linear equations is dependent and consistent. This means there are infinitely many solutions for the given pair of linear equations. The graph of linear equations will be coincident lines.

## NCERT Solution

### Exercise 3.1 (NCERT)

Question – 1 - Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically.
Solution: Let us assume Aftab’s current age = x and his daughter’s current age = y
Seven years ago:
Aftab’s age = x - 7 and daughter’s age = y - 7
As per question;
This equation gives following values for x and y
Three years from now:
Aftab’s age = x + 3 and daughter’s age = y + 3
As per question;
This equation gives following values for x and y:
The following graph is plotted for the given pair of linear equations:
Daughter’s age = 12 years
Aftab’s age = 42 years
Question – 2 - The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Solution: Let us assume that price of one bat = x and price of one ball = y. Following equations can be written as per the question:
This equation will give following values for x and y
This equation will give following values for x and y
The following graph is plotted for the given pair of linear equations.
Price of one bat = Rs. 1300
Price of one ball = zero
Question – 3 - The cost of 2 kg apples and 1 kg grapes was found to be Rs. 160. After a month, the cost of 4 kg apples and 2 kg grapes is Rs. 300. Represent this situation algebraically and graphically.
Answer: Let us assume that cost of 1 kg apple = x and cost of 1 kg grapes = y. Following equations can be written as per the question.
This equation gives following values for x and y
This equation will give following values for x and y:
The following graph is plotted for the given pair of linear equations.
Since we get parallel lines so there will be no solution for this pair of linear equations.

### Exercise 3.2

Question – 1 - Form a pair of linear equations in the following problems, and find their solutions graphically.
(a) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Solution: Let us assume that number of boys = x and number of girls = y. We get following equations as per question:
This equation will give following values for x and y;
This equation will give following values for x and y;
Following graph is plotted for the given pair of linear equations:
(b) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Solution: Let us assume that price of a pencil is x and that of a pen is y. We get following equations as per question:
This equation will give following values for x and y;
This equation will give following values for x and y;
Following graph is plotted for the given pair of linear equations.
Price of one pencil = Rs. 3
Price of one pen = Rs. 5
Question – 2- On comparing the ratios  find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.
Solution: In the given pair of linear equations;
It is clear that;
Hence the lines representing the given pair of linear equations intersect at a point.
Solution: In the given pair of linear equations;
It is clear that;
Hence the lines representing the given pair of linear equations will be coincident.
Solution: For the given pair of linear equations;
It is clear that;
Hence the lines representing the given pair of linear equations will be parallel.
Question – 3 - On comparing the ratios  find out whether the following pairs of linear equations are consistent or inconsistent.
Solution: For the given pair of linear equations;
It is clear that;
Hence, the given pair of linear equations is consistent.
Solution: For the given pair of linear equations;
It is clear that;
Hence the given pair of linear equations is inconsistent.
Solution: For the given pair of linear equations;
It is clear that;
Hence the given pair of linear equations is consistent.
Solution: For the given pair of linear equations;
It is clear that;
Hence the given pair of linear equations is dependent and consistent.

Question – 4 - Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
Solution: For the given pair of linear equations;
It is clear that;
Hence the given pair of linear equations is dependent and consistent. The lines for the equations will be coincident. This means there would be infinitely many solutions for the given pair of linear equations.
Solution: For the given pair of linear equations;
It is clear that;
Hence the given pair of linear equations is inconsistent.
Solution: For the given pair of linear equations;
It is clear that;
Hence the given pair of linear equations is consistent.
Following graph can be plotted with the given pair of linear equations;
Value of x = 2 and that of y = 2
Solution: For the given pair of linear equations;
It is clear that;
Hence the given pair of linear equations is inconsistent.
Question – 5 - Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution: Let us assume that width of garden = x and length = y. Then as per question;
Substitution the value of y from equation (1) in equation (2) we get;
Hence, length = 20 m and breadth = 16 m
Question – 6 - Given the linear equation  write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(a) Intersecting lines
Solution: For intersecting line, the linear equations should meet following condition:
For getting another equation to meet this criterion, multiply the coefficient of x with any number and multiply the coefficient of y with any other number. A possible equation can be as follows:
(b) Parallel lines
Solution: For parallel lines, the linear equations should meet following condition:
For getting another equation to meet this criterion, multiply the coefficients of x and y with the same number and multiply the constant term with any other number. A possible equation can be as follows:
(c) Coincident lines
Solution: For getting coincident lines, the equations should meet following condition;
For getting another equation to meet this criterion, multiply the whole equation with any number. A possible equation can be as follows:
Question – 7 - Draw the graphs of the equations Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution: Following graph can be plotted with the given pair of linear equations:
The coordinates of vertices of the triangle are; (-1, 0), (2, 3) and (4, 0)

### Exercise 3.3

Solving by substitution method:
• One of the equations is picked and in this, one variable is expressed in terms of another variable.
• The value of one variable (in terms of another variable) is substituted in the second equation to calculate the value of the remaining variable.
Question – 1 - Solve the following pair of linear equations by the substitution method.
Solution: Let us use the first equation to express one variable in terms of another variable;
Substituting the value of x in second equation we get;
Substituting the value of y in first equation, we get;
Solution: Let us use the first equation to express one variable in terms of another variable;
Substituting the value of s in second equation, we get;
Substituting the value of t in first equation, we get;
Solution: Let us use first equation to express one variable in terms of another variable.
The second equation is same as the first equation;
Dividing the equation by 3, we get;
Hence, the given pair of linear equations has infinitely many solutions.
Solution: Let us use the first equation to express one variable in terms of another variable;
Substituting the value of x in second equation, we get;
Substituting the value of y in first equation, we get;
Solution: Let us use first equation to express one variable in terms of another variable;
Substituting the value of x in second equation, we get;
This can be possible only when the value of y is zero. Substituting the value of y in first equation, we get;
Solution: Let us use first equation to express one variable in terms of another variable;
Substituting the value of y in second equation, we get;
Substituting the value of x in first equation, we get;
Question – 2 - Solve  and hence find the value of ‘m’ for which
Solution: Let us use first equation to express one variable in terms of another variable;
Substituting the value of x in second equation, we get;
Substituting the value of y in first equation, we get;
Now; we have to find the value of m in following equation;

Question – 2 - Form a pair of linear equations for the following problems and find their solution by substitution method.
(a) The difference between two numbers is 26 and one number is three times the other. Find them.
Solution: Let us assume the larger number is x and smaller number is y. Then we have following equations, as per question:
Substituting the value of x from second equation in the first equation, we get;
Substituting the value of y in second equation, we get;
(b) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution: Let us assume the larger angle is x and smaller angle is y. Then we have following equations;
Substituting the value of x from first equation in second equation, we get;
Substituting the value of y in first equation, we get;
(c) The coach of a cricket team buys 7 bats and 6 balls for Rs, 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Solution: Let us assume, price of a bat is x and that of a ball is y. Then we have following equations;
Let us use first equation to express one variable in terms of another variable;
Substituting the value of x in second equation, we get;
Substituting the value of y in first equation, we get;
Hence price of a bat=Rs.500 and price of a ball=Rs.50
(d) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for the journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution: Let us assume the fixed charge is x and charge for 1 km journey is y. Then we have following equations;
Let us use first equation to express one variable in terms of another variable;
Substituting the value of x in second equation, we get;
Substituting the value of y in first equation, we get;
(e) A fraction becomes  if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes  Find the fraction.
Solution: Let us assume, numerator is x and denominator is y. Then we have following equations;
Let us use first equation to express one variable in terms of another variable;
Substituting the value of y in second equation, we get;
Substituting the value of x in first equation, we get;
(f) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution: Let us assume Jacob’s current age is x and his son’s current age is y.
Five years hence, Jacob’s age = x + 5 and his son’s age = y + 5
As per question;
Five years ago, Jacob’s age = x – 5 and his son’s age = y – 5
As per question;
Substituting the value of x from first equation in this equation, we get;
Substituting the value of y in first equation, we get;
Hence, Jacob’s age = 40 years and son’s age = 10 years
Elimination Method:
In elimination method, one of the equations is multiplied by a suitable number so that the coefficient of one of the variables in both the equations becomes equal. After that, one equation is subtracted or added to another equation to eliminate one of the variables.

### Exercise 3.4

Question – 1 - Solve the following pair of linear equations by the elimination method and the substitution method.
Solution: Elimination Method:
Multiply the first equation by 3 as follows:
Now, add second equation to this equation;
Substituting the value of x in first equation, we get;
Substitution method;
Let us use first equation to express one variable in terms of another variable;
Substitute the value of x in second equation;
Substitute the value of y in first equation;
Solution: Elimination method:
Multiply second equation by 2;
Add this equation with first equation;
Substituting the value of x in first equation, we get;
Substitution Method:
Let us use second equation to express one variable in terms of another variable;
Substituting the value of x in first equation, we get;
Substituting the value of y in second equation, we get;
Solution: Elimination Method:
Multiply the first equation by 3;
Subtract the second equation from this equation;
Substituting the value of y in first equation, we get;
Substitution Method;
Let us use first equation to express one variable in terms of another variable;
Substituting the value of x in second equation, we get;
Substituting the value of y in first equation, we get;
Solution: Elimination method:
Equation 1:
Equation 2:
Subtracting second equation from first equation, we get;
Substituting the value of y in second equation, we get;
Substitution Method:
Let us use second equation to express one variable in terms of another variable;
Substituting the value of y in first equation, we get;
Substituting the value of x in second equation, we get;

Question – 2 - Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method.
(a) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes  if we only add 1 to the denominator. What is the fraction?
Solution: Let us assume that numerator is x and denominator is y. Then as per question;
The second condition can give following equation;
Adding first and second equations, we get;
Substituting the value of x in first equation, we get;
(b) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Solution: Let us assume Nuri’s age = x and Sonu’s age = y. Then we can get following equations;
Five years ago:
Nuri’s age = x – 5 and Sonu’s age = y – 5
Five years from now;
Nuri’s age = x + 5 and Sonu’s age = y + 5
Subtracting the second equation from first equation, we get;
Substituting the value of y in second equation, we get;
(c) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution: Let us assume, one of the digits is x and another digit is y.
The number can be given as;
And the number obtained after reversing the digits is;
As per question;
Multiplying the first equation by 11, we get;
Adding this to the second equation, we get;
Substituting the value of x in first equation, we get;
Hence, the required number = 18
(d) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find out how many notes of Rs. 50 and Rs. 100 she received.
Solution: Let us assume the number of Rs. 50 notes is x and that of Rs. 100 notes is y.
Multiplying the first equation by 50 we get;
Subtracting this equation from second equation, we get;
Substituting the value of y in first equation, we get;
Hence, number or Rs. 50 notes = 10 and number of Rs. 100 notes = 15
(e) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution: Let us assume fixed charge is x and per day charge after three days is y.
Amount paid by Saritha:
Amount paid by Susy:
Subtracting the second equation from first equation, we get;
Substituting the value of y in second equation, we get;
Hence, fixed charge for three days = Rs. 15 and per day charge after three days = Rs. 3

### Exercise 3.4 - Part

Question – 1 - Solve the following pair of linear equations by the elimination method and the substitution method.
Solution: Elimination Method:
Multiply the first equation by 3 as follows:
Now, add second equation to this equation;
Substituting the value of x in first equation, we get;
Substitution method;
Let us use first equation to express one variable in terms of another variable;
Substitute the value of x in second equation;
Substitute the value of y in first equation;
Solution: Elimination method:
Multiply second equation by 2;
Add this equation with first equation;
Substituting the value of x in first equation, we get;
Substitution Method:
Let us use second equation to express one variable in terms of another variable;
Substituting the value of x in first equation, we get;
Substituting the value of y in second equation, we get;
Solution: Elimination Method:
Multiply the first equation by 3;
Subtract the second equation from this equation;
Substituting the value of y in first equation, we get;
Substitution Method;
Let us use first equation to express one variable in terms of another variable;
Substituting the value of x in second equation, we get;
Substituting the value of y in first equation, we get;
Solution: Elimination method:
Equation 1:
Equation 2:
Subtracting second equation from first equation, we get;
Substituting the value of y in second equation, we get;
Substitution Method:
Let us use second equation to express one variable in terms of another variable;
Substituting the value of y in first equation, we get;
Substituting the value of x in second equation, we get;

### Exercise 3.5 (NCERT)

Question – 1 - Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
Solution:
It is clear that;
Hence, there will be no solution for the given pair of linear equations.
Solution:
It is clear that;
Hence, there will be unique solution for the given pair of linear equations.
From cross-multiplication method, we know;
Solution:
It is clear that;
Hence, there will be infinitely many solutions for the given pair of linear equations.
Solution:
It is clear that;
Hence, there will be unique solution for the given pair of linear equations.
From cross-multiplication method, we know;
Question – 2 - For which values of a and b does the following pair of linear equations have an infinite number of solutions?
Solution:
For infinite number of solutions, the equations should fulfill following criterion:
From equations (1) and (2), it is clear;
Substituting the value of b in equation (1), we get;
Question – 3 - For which value of k will the following pair of linear equations have no solution?
Solution:
For no solution, the equations should fulfill following criterion:
Question – 4 - Solve the following pair of linear equations by the substitution and cross-multiplication methods:
Solution: Substitution method:
Let us use the first equation to express one variable in terms of another variable;
Substituting the value of x in second equation, we get;
Substituting the value of y in first equation, we get;
Cross-multiplication Method:
From cross-multiplication method, we know;

Question – 5 - Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(a) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Solution: Let us assume that fixed charges is x and per day charges is y
Amount paid by student A:

Amount paid by student B:

It is clear that;

Hence, there will be unique solution for the given pair of linear equations.
Subtracting first equation from second equation, we get;

Substituting the value of y in first equation, we get;

Hence, fixed charges = Rs. 400 and per day charges = Rs. 30
(b) A fraction becomes  when 1 is subtracted from the numerator and it becomes  when 8 is added to its denominator. Find the fraction.
Solution: Let us assume numerator is x and denominator is y.
First condition:
Second condition:
It is clear that;
Hence, there will be unique solution for the given pair of linear equations.
Subtracting first equation from second equation, we get;
Substituting the value of x in second equation, we get;
(c) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Solution: Let us assume that the number of correct answers = x and number of incorrect answers = y
First condition:
Second condition:
It is clear that;
Hence, there will be unique solution for the given pair of linear equations.
Multiply the first equation and subtract second equation from the resultant:
Substituting the value of x in second equation, we get;
Hence, number of correct answers = 15 and number of incorrect answers = 10
(d) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Solution: Let us assume that the speed of car A is x and that of car B is y.
First condition: When both the cars are going in the same direction, their relative speed is;
Hence,
Second condition: When the cars are moving towards each other, their relative speed is;
Hence,
Adding the two equations, we get;
Substituting the value of x in first equation, we get;
Hence, speed of car A = 60 km/h and speed of car B = 40 km/h
(e) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution: Let us assume length is x and breadth is y.
Area of rectangle = xy
First condition: Length = x - 5 and breadth = y + 3
Area of rectangle:
Second condition: Length = x +3 and breadth = y + 2
Area of rectangle:
Multiplying the first equation by 2, we get;
Multiplying the second equation by 3, we get;
Subtracting these two equations, we get;
Substituting the value of y in any of the above equations, we get;
Hence, length = 17 unit and breadth = 9 unit

### Exercise 3.6 (NCERT)

Question – 1 - Solve the following pairs of equations by reducing them to a pair of linear equations:
Solution: First equation:
Second equation:
Multiplying the first equation by 4 and subtracting the resultant from second equation, we get;
Substituting the value of x in first equation, we get;
Solution: First equation:
Second equation:
From first and second equation, it is clear;
Substituting the value of  in any of the above equation, we get;
Substituting the value of  in second equation, we get;
Solution: First equation:
Second equation:
Multiply first equation by 4 and second equation by 3 and add the resultant equations;
Substituting the value of x in first equation, we get;

Solution: First equation;
Second equation:
Multiply the second equation by 2 and subtract first equation from resultant:
Substituting the value of x in first equation, we get;
Similarly, substituting the value of y in second equation, we get;
From above two equations, it is clear;
Solution: First equation:
Second equation:
Multiplying the first equation by 3 and subtracting the resultant from second equation, we get;
Substituting the value of x in first equation, we get;
Solution: Multiplying first equation by 5 and second equation by 6 and subtracting them, we get;
Substituting the value of y in first equation, we get;
This is possible only when x=1 or zero
Taking the non-zero value and substituting in second equation, we get;

Solution: First equation:
Second equation:
From first and second equations, it is clear;
Substituting the value of x in first equation, we get;

Solution: First equation:
Second equation:
From first and second equation, it is clear;
Substituting the value of x in first equation, we get;
Substituting the value of x in second equation, we get;

Question – 2 - Formulate the following problems as a pair of equations, and hence find their solutions.
(a) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Solution: Let us assume the speed of boat in still water is x and speed of current is y
Hence, downstream speed = x + y and upstream speed = x – y
First condition: Using time = distance/speed;
Second condition: Using time = distance/speed;
Adding first and second equation, we get;
Speed of boat = 6 km/h and speed of current = 4 km/h
(b) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work and also that taken by 1 man alone.
Solution: First condition: One day’s work done by 2 women and 5 men;
Second condition: One day’s work done by 3 women and 6 men;
From first and second equation, it is clear;
This means that 1 woman’s work is equal to 2 men’s work. Substituting the value of w in first equation, we get;
This means that 1 woman will take 18 days to finish the work.
Similarly,
This means that 1 man will take 36 days to finish the work.
(c) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution: Let us assume that speed of train is x and speed of bus is y
First condition: She travels 60 km by train and 240 km by bus;
Second condition: She travels 100 km by train and 200 km by bus;
Subtracting first equation from second, we get;
Substituting the value of x in first equation, we get;
Speed of train = 60 km/h and speed of bus = 80 km/h