NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 - Pair of Linear Equations in two variables

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 - Pair of Linear Equations in two variables

Updated on 30 May 2025, 02:40 PM IST

In the previous exercise, it was explained how to show the linear equation using the graphical method, but this method is convenient when the solutions' coordinates are non-integral. There is another method for solving the equation is the arithmetic method for solving linear equations has been explained. In this exercise arithmetic method for solving equations is used to solve the questions. There are many arithmetic methods for solving the equation, namely the Graphical method, substitution method, cross Multiplication method, elimination method and determinant method. In this exercise substitution method has been explained to solve the equation.

This Story also Contains

  1. Assess NCERT Solutions for Class 10 Chapter 3 Exercise: 3.2
  2. Topics Covered in Chapter 3 Pair of Linear Equations in Two Variables: Exercise 3.2
  3. NCERT Solutions of Class 10 Subject Wise
  4. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 - Pair of Linear Equations in two variables
NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2

This exercise deals with the substitution method for solving algebraic equations. Each solutions are explained in detail and crafted in easy-to-understand language. These NCERT solutions are essential for every student to understand the concept. Practice these questions and answers to command the concepts, boost confidence and in-depth understanding of concepts. These solutions are according to the NCERT Books and cover all the methods according to the latest syllabus.

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Assess NCERT Solutions for Class 10 Chapter 3 Exercise: 3.2

Q1(i) Solve the following pair of linear equations by the substitution method.

$x+y=14$

$x−y=4$

Answer:

Given two equations,

$x+y=14$.......(1)

$x−y=4$........(2)

Now, from (1), we have

$y=14−x$........(3)

Substituting this in (2), we get

$x−(14−x)=4$

$\Rightarrow x−14+x=4$

$\Rightarrow 2x=4+14=18$

$\Rightarrow x=9$

Substituting this value of x in (3)

$\Rightarrow y=14−x=14−9=5$

Hence, the solution of the given equations is x = 9 and y = 5.

Q1(ii) Solve the following pair of linear equations by the substitution method

$s−t=3$

$\frac{s}{3}+\frac{t}{2}=6$

Answer:

Given two equations,

$s−t=3$..........(1)

$\frac{s}{3}+\frac{t}{2}=6$....... (2)

Now, from (1), we have

$s=t+3$........(3)

Substituting this in (2), we get

$\frac{t+3}{3}+\frac{t}{2}=6$

$\Rightarrow 2t+6+3t=36$

$\Rightarrow 5t+6=36$

$\Rightarrow 5t=30$

$\Rightarrow t=6$

Substituting this value of t in (3)

$\Rightarrow s=t+3=6+3=9$

Hence, the solution of the given equations is s = 9 and t = 6.

Q1(iii) Solve the following pair of linear equations by the substitution method.

$3x−y=39$

$9x−3y=9$

Answer:

Given two equations,

$3x−y=3$......(1)

$9x−3y=9$.....(2)

Now, from (1), we have

$y=3x−3$........(3)

Substituting this in (2), we get

$9x−3(3x−3)=9$

$\Rightarrow 9x−9x+9=9$

$\Rightarrow 9=9$

This is always true, and hence this pair of equations has infinite solutions.

As we have

$y=3x−3$,

One of many possible solutions is x = 1, and y = 0.

Q1(iv) Solve the following pair of linear equations by the substitution method.

$0.2x+0.3y=1.3$

$0.4x+0.5y=2.3$

Answer:

Given two equations,

$0.2x+0.3y=1.3$

$0.4x+0.5y=2.3$

Now, from (1), we have

$y=\frac{1.3−0.2x}{0.3}$........(3)

Substituting this in (2), we get

$0.4x+0.5\frac{1.3−0.2x}{0.3}=2.3$

$\Rightarrow 0.12x+0.65−0.1x=0.69$

$\Rightarrow 0.02x=0.69−0.65=0.04$

$\Rightarrow x=2$

Substituting this value of x in (3)

$\Rightarrow y=\frac{1.3−0.2x}{0.3} =\frac{1.3−0.2 × 2}{0.3} =3$

Hence, the solution of the given equations is

x = 2 and y = 3.

Q1(v) Solve the following pair of linear equations by the substitution method.

$\sqrt 2x+ \sqrt 3y=0$

$\sqrt 3x−\sqrt 8y=0$

Answer:

Given two equations,

$\sqrt 2x+ \sqrt 3y=0$

$\sqrt 3x−\sqrt 8y=0$

Now, from (1), we have

$x=\frac{−\sqrt 3y}{\sqrt 2}$........(3)

Substituting this in (2), we get

$\sqrt 3 (\frac{−\sqrt 3y}{\sqrt 2}) - \sqrt 8y=0$

$\Rightarrow \frac{−\sqrt 3y}{\sqrt 2} - 2\sqrt 2y = 0$

$\Rightarrow y(\frac{−\sqrt 3}{\sqrt 2} - 2\sqrt 2) = 0$

$\Rightarrow y=0$

Substituting this value of y in (3)

$\Rightarrow x = 0$

Hence, the solution of the given equations is,

x = 0, and y = 0 .

Q1(vi) Solve the following pair of linear equations by the substitution method.

$\frac{3x}{2}−\frac{5y}{3} =−2$

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Answer:

Given,

$\frac{3x}{2}−\frac{5y}{3} =−2$ ....... (1)

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$ ........ (2)

From (1) we have,

$x=\frac{(-12 + 10y)}{9}$........(3)

Putting this in (2), we get,

$\frac{\frac{(-12 + 10y)}{9}}{3}+ \frac{y}{2} = \frac{13}{6}$

$\frac{(-12 + 10y)}{27}+ \frac{y}{2} = \frac{13}{6}$

$\frac{(-24 + 20y + 27y)}{54} = \frac{13}{6}$

$47y=141$

$y=3$

Putting this value in (3), we get,

$x=\frac{(-12 + 10× 3)}{9}$

$x =2$

Hence, x = 2 and y = 3.

Q2 Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

Answer:

Given two equations,

$2x+3y=11$......(1)

$2x−4y=−24$.......(2)

Now, from (1), we have

$y=\frac{11−2x}{3}$........(3)

Substituting this in (2), we get

$2x−4(\frac{11−2x}{3})=−24$

$\Rightarrow 6x−44+8x=−72$

$\Rightarrow 14x=44−72$

$\Rightarrow 14x=−28$

$\Rightarrow x=−2$

Substituting this value of x in (3)

$\Rightarrow y=\frac{11−2x}{3}=\frac{11−2×(−2)}{3}=\frac{15}{3}=5$

Hence, the solution of the given equations is,

x = −2, and y = 5.

Now,

As it satisfies $y=mx+3$,

$\Rightarrow 5=m(−2)+3$

$\Rightarrow 2m=3−5$

$\Rightarrow 2m=−2$

$\Rightarrow m=−1$

Hence, the value of m is -1.

Q3(i) Form the pair of linear equations for the following problem and find their solution by the substitution method.

The difference between the two numbers is 26, and one number is three times the other. Find them.

Answer:

Let two numbers be x and y, and the bigger number is y.

Now, according to the question,

$y−x=26$......(1)

And

$y=3x$......(2)

Now, substituting the value of y from (2) in (1), we get,

$3x−x=26$

$\Rightarrow 2x=26$

$\Rightarrow x=13$

Substituting this in (2)

$\Rightarrow y=3x=3(13)=39$

Hence, the two numbers are 13 and 39.

Q3(ii) Form the pair of linear equations for the following problem and find their solution by the substitution method

The larger of the two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer:

Let the larger angle be x and the smaller angle be y

Now, as we know, the sum of supplementary angles is 180. so,

$x+y=180^0$.......(1)

Also given in the question,

$x−y=18^0$.......(2)

Now, from (2) we have,

$y=x−18^0$.......(3)

Substituting this value in (1)

$x+x−18^0=180^0$

$\Rightarrow 2x=180^0+18^0$

$\Rightarrow 2x=198^0$

$\Rightarrow x=99^0$

Now, substituting this value of x in (3), we get

$\Rightarrow y=x−18^0=99^0−18^0=81^0$

Hence, the two supplementary angles are

$99^0$ and $81^0$.

Q3 Form the pair of linear equations for the following problems and find their solution by the substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Answer:

Let the cost of 1 bat is x and the cost of 1 ball is y.

Now, according to the question,

$7x+6y=3800$......(1)

$3x+5y=1750$......(2)

Now, from (1) we have

$y=\frac{(3800−7x)}{6}$........(3)

Substituting this value of y in (2)

$3x+5\frac{(3800−7x)}{6}=1750$

$\Rightarrow 18x+19000−35x=1750×6$

$\Rightarrow −17x=10500−19000$

$\Rightarrow −17x=−8500$

$\Rightarrow x=\frac{8500}{17}$

$\Rightarrow x=500$

Now, Substituting this value of x in (3)

$y=\frac{(3800−7x)}{6}$

$y =\frac{3800−7×500}{6}$

$y =\frac{3800−3500}{6}=50$

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

Q3. Form the pair of linear equations for the following problems and find their solution by the substitution method. (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105, and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Answer:

Let the fixed charge is x and the per km charge is y.

Now, according to the question

$x+10y=105$.......(1)

And,

$x+15y=155$.......(2)

Now, from (1) we have,

$x=105−10y$........(3)

Substituting this value of x in (2), we have

$105−10y+15y=155$

$\Rightarrow 5y=155−105$

$\Rightarrow 5y=50$

$\Rightarrow y=10$

Now, substituting this value in (3)

$x=105−10y$

$=105−10(10)$

$=105−100=5$

Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.

Now, Fair for 25 km :

$\Rightarrow x+25y=5+25(10) $

$=5+250=255$

Hence, fair for 25km is 255 Rs.

Q3.Form the pair of linear equations for the following problems and find their solution by the substitution method. (v) A fraction becomes 911, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 56. Find the fraction.

Answer:

Let the numerator of the fraction be x, and the denominator of the fraction be y

Now, according to the question,

$\frac{x+2}{y+2}=\frac{9}{11}$

$\Rightarrow 11(x+2)=9(y+2)$

$\Rightarrow 11x+22=9y+18$

$\Rightarrow 11x−9y=−4$.........(1)

Also,

$\frac{x+3}{y+3}=\frac{5}{6}$

$\Rightarrow 6(x+3)=5(y+3)$

$\Rightarrow 6x+18=5y+15$

$\Rightarrow 6x−5y=−3$...........(2)

Now, from (1) we have

$y=\frac{11x+4}{9}$.............(3)

Substituting this value of y in (2)

$6x−5(\frac{11x+4}{9})=−3$

$\Rightarrow 54x−55x−20=−27$

$\Rightarrow −x=20−27$

$\Rightarrow x=7$

Substituting this value of x in (3)

$y={11x+4}{9}$

$=\frac{11(7)+4}{9}=\frac{81}{9}=9$

Hence, the required fraction is

$x =\frac{7}{9}$

Q3 Form the pair of linear equations for the following problems and find their solution by the substitution method. (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer:

Let x be the age of Jacob and y be the age of Jacob's son.

Now, according to the question

$x+5=3(y+5)$

$\Rightarrow x+5=3y+15$

$\Rightarrow x−3y=10$........(1)

Also,

$x−5=7(y−5)$

$\Rightarrow x−5=7y−35$

$\Rightarrow x−7y=−30$.........(2)

Now,

From (1) we have,

$x=10+3y$...........(3)

Substituting this value of x in (2)

$10+3y−7y=−30$

$\Rightarrow −4y=−30−10$

$\Rightarrow 4y=40$

$\Rightarrow y=10$

Substituting this value of y in (3),

$x=10+3y=10+3(10)=10+30=40$

Hence, the present age of Jacob is 40 years and the present age of Jacob's son is 10 years.

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Topics Covered in Chapter 3 Pair of Linear Equations in Two Variables: Exercise 3.2

  1. Formation of Linear Equation: In this, students need to understand the problem and convert it into a linear equation to solve the problem.
  2. Algebraic Methods: This exercise deals with the various techniques for determining the solutions for linear equations.
  3. Substitution Method: In this method, the equation is solved for one variable and the value of this variable is substituted into the other equation to solve the question.
  4. Verification of Solutions: In this method, we can verify whether the solution is correct or not by replacing the value of the variables in the given question equation.

Also see-

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

Q: Define substitution method according to NCERT solutions for Class 10 Maths chapter 3 exercise 3.2?
A:

The substitution method is defined as substituting the value of one variable in another equation to obtain the value of another variable

Q: In the NCERT solutions for Class 10 Maths chapter 3 exercise 3.2, how many questions are there?
A:

In the NCERT solutions for Class 10 Maths chapter 3 exercise 3.2, there are three questions.

Q: Find the value of x in 6x-y=10 if y=2 .
A:

Putting y=2 in 6x-y=10 

6x-2=10 

6x=10+2

Thus x =12/6=2

Q: Mothers age is six times her daughter’s age. Six years after the age of mother will be four times the age of her daughter. Find the present age of the mother and her daughter.
A:

Let the present age of mother be x years and the present age of daughter be y years. 

Given that, x=6y 

Also, x+6=4(y+6) 

Now substituting x=6y in x+6=4(y+6)  

6y+6=4(y+6)

6y-4y+6-24=0 

2y-18=0 

Thus y=9 

The daughter’s age is 9 years then her mother’s age is 54 years. 

Q: Solve x-3y=4 and 2x+y=1 by substitution method.
A:

Given, x-3y=4 ⇒ x=4+3y 

Now substituting x=4+3y in 2x+y=1 

2(4+3y)+y=1

8+6y+y=1 

7y=1-8 ⇒ y=-1 

Substitute y=-1 in x-3y=4 

x-3(-1)=4 

x=4-3 ⇒ x=1 

Thus (x, y)=(1, -1) 

Q: Mention the different methods to solve the system of equations linear equations in two variables.
A:

The methods used to find the solution of a pair of linear equations are 

• Graphical method 

• Algebraic method 

  • Substitution Method

  • Elimination Method

  • Cross- multiplication Method

Q: What is the general form of linear equation?
A:

The General form is ax+by+c=0, where a, b, c are real numbers.

Q: Whether the substitution method can be used to solve the system of equations in three variables?
A:

Yes, we can use the substitution method to solve the system of equations in three variables.

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