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NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Ex 3.2

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NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2 - FREE PDF Download

Linear Equation is a vital topic in Mathematics. No matter how advanced studies you go for, the linear equation concept will always help you solve critical problems by finding values of unknown variables. The study material of Class 10 Maths Chapter 3 Exercise 3.2 Solutions is a wonderful resource that helps you to understand this chapter in an easy way. It not only explains the basics in a step-by-step manner but also provides a wide array of problems and solutions. By practising these problems, you can surely get a good grasp of the chapter. Vedantu has come up with easy to access Maths NCERT Solutions for Exercise 3.2 Class 10 Chapter 3 pdf. Maths Class 10 NCERT Solutions Chapter 3 is available free! 

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Glance on NCERT Solutions Maths Chapter 3 Exercise 3.2 Class 10 | Vedantu

  • NCERT Solutions for Class 10th Maths Chapter 3 Exercise 3.2, focuses on solving pairs of linear equations in two variables using various algebraic methods such as substitution, elimination, and cross-multiplication. 

  • This exercise deals with equations represented in the general form of ax + by = c, where a, b, and c are constants, and x and y are the variables.

  • Solve systems of linear equations using the substitution method.

  • Interpreting the solutions obtained, which can be unique solutions, infinitely many solutions, or no solutions.

  • This exercise is crucial as it not only sharpens students' problem-solving and analytical skills but also illustrates the practical application of mathematical principles in everyday situations.

  • In Class 10th Maths, Chapter 3, Exercise 3.2 Pair of Linear Equations in Two Variables there are 2 Solved Questions.

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NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Ex 3.2
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Pair of Linear Equations in Two Variables L-1 | Consistency of a System by Graphical Method |CBSE 10
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Class 10 Ex 3.2

1. Solve the following pair of linear equations by the substitution method.

(i) \[x+y=14;x-y=4\]

Ans: The given equations are:

\[x+y=14\]           …… (i)

\[x-y=4\]             …… (ii)

From equation (i):

\[x=14-y\]          …… (iii)

Substituting (iii) in equation (ii), we get

\[\left( 14-y \right)-y=4\]

\[14-2y=4\]

\[10=2y\]

\[y=5\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=9\]

Therefore, \[x=9\] and \[y=5\].

(ii) \[s-t=3;\frac{s}{3}+\frac{t}{2}=6\]

Ans: The given equations are:

\[s-t=3\]           …… (i)

\[\frac{s}{3}+\frac{t}{2}=6\]       …… (ii)

From equation (i):

\[s=t+3\]          …… (iii)

Substituting (iii) in equation (ii), we get

\[\frac{t+3}{3}+\frac{t}{2}=6\]

\[2t+6+3t=36\]

\[5t=30\]

\[t=6\]                  …… (iv)

Substituting (iv) in (iii), we get

\[s=9\]

Therefore, \[s=9\] and \[t=6\].


(iii) \[3x-y=3;9x-3y=9\]

Ans: The given equations are:

\[3x-y=3\]           …… (i)

\[9x-3y=9\]             …… (ii)

From equation (i):

\[y=3x-3\]          …… (iii)

Substituting (iii) in equation (ii), we get

\[9x-3\left( 3x-3 \right)=9\]

\[9x-9x+9=9\]

\[9=9\]

For all \[x\] and \[y\].

Therefore, the given equations have infinite solutions. One of the solution is \[x=1,y=0\].


(iv) \[0.2x+0.3y=1.3;0.4x+0.5y=2.3\]

Ans: The given equations are:

\[0.2x+0.3y=1.3\]           …… (i)

\[0.4x+0.5y=2.3\]             …… (ii)

From equation (i):

\[x=\frac{1.3-0.3y}{0.2}\]          …… (iii)

Substituting (iii) in equation (ii), we get

\[0.4\left( \frac{1.3-0.3y}{0.2} \right)+0.5y=2.3\]

\[2.6-0.6y+0.5y=2.3\]

\[2.6-2.3=0.1y\]

\[y=3\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=\frac{1.3-0.3\left( 3 \right)}{0.2}\]

\[x=2\]

Therefore, \[x=2\] and \[y=3\].


(v) \[\sqrt{2}x+\sqrt{3}y=0;\sqrt{3}x-\sqrt{8}y=0\]

Ans: The given equations are:

\[\sqrt{2}x+\sqrt{3}y=0\]           …… (i)

\[\sqrt{3}x-\sqrt{8}y=0\]             …… (ii)

From equation (i):

\[x=\frac{-\sqrt{3}y}{\sqrt{2}}\]          …… (iii)

Substituting (iii) in equation (ii), we get

\[\sqrt{3}\left( \frac{-\sqrt{3}y}{\sqrt{2}} \right)-\sqrt{8}y=0\]

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

\[y=0\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=0\]

Therefore, \[x=0\] and \[y=0\].


(vi) \[\frac{3x}{2}-\frac{5y}{3}=-2;\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\]

Ans: The given equations are:

\[\frac{3x}{2}-\frac{5y}{3}=-2\]           …… (i)

\[\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\]             …… (ii)

From equation (i):

\[x=\frac{-12+10y}{9}\]          …… (iii)

Substituting (iii) in equation (ii), we get

\[\frac{\left( \frac{-12+10y}{9} \right)}{3}+\frac{y}{2}=\frac{13}{6}\]

\[\frac{-12+10y}{27}+\frac{y}{2}=\frac{13}{6}\]

\[\frac{-24+20y+27y}{54}=\frac{13}{6}\]

\[47y=141\]

\[y=3\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=2\]

Therefore, \[x=2\] and \[y=3\].


2. Solve \[\mathbf{2x}+\mathbf{3y}=\mathbf{11}\] and \[\mathbf{2x}-\mathbf{4y}=-\mathbf{24}\] and hence find the value of ‘\[m\]’ for which \[\mathbf{y}=\mathbf{mx}+\mathbf{3}\].

Ans: The given equations are:

\[2x+3y=11\]           …… (i)

\[2x-4y=-24\]             …… (ii)

From equation (i):

\[x=\frac{11-3y}{2}\]          …… (iii)

Substituting (iii) in equation (ii), we get

\[2\left( \frac{11-3y}{2} \right)-4y=-24\]

\[11-3y-4y=-24\]

\[-7y=-35\]

\[y=5\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=-2\]

Therefore, \[x=-2\] and \[y=5\].

Calculating the value of \[m\]:

\[y=mx+3\]

\[5=-2m+3\]

\[m=-1\]


3. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is \[\mathbf{26}\] and one number is three times the other. Find them.

Ans: Assuming one number be \[x\] and another number be \[y\] such that \[y>x\],

Writing the algebraic representation using the information given in the question:

\[y=3x\]             …… (i)

\[y-x=26\]    …… (ii)

Substituting the value of \[y\] from equation (i) in equation (ii), we get

\[3x-x=26\]

\[2x=26\]

\[x=13\]                  …… (iii)

Substituting (iii) in (i), we get

\[y=39\]

Therefore, numbers are $13$ & $39$.


(ii) The larger of two supplementary angles exceeds the smaller by \[\mathbf{18}\] degrees. Find them.

Ans: Assuming the larger angle be \[x\] and smaller angle be \[y\]. 

The sum of a pair of supplementary angles is always \[{{180}^{\circ }}\].

Writing the algebraic representation using the information given in the question:

\[x+y=180\]             …… (i)

\[x-y=18\]    …… (ii)

Substituting the value of \[x\] from equation (i) in equation (ii), we get

\[180-y-y=18\]

\[162=2y\]

\[y=81\]                  …… (iii)

Substituting (iii) in (i), we get

\[x=99\]

Therefore, the two angles are  \[x={{99}^{\circ }}\] and \[y={{81}^{\circ }}\].


(iii) The coach of a cricket team buys \[\mathbf{7}\] bats and 6 balls for \[\mathbf{Rs}\text{ }\mathbf{3800}\]. Later, she buys \[\mathbf{3}\] bats and \[\mathbf{5}\] balls for \[\mathbf{Rs}\text{ }\mathbf{1750}\]. Find the cost of each bat and each ball.

Ans: Assuming the cost of a bat is \[x\] and the cost of a ball is \[y\].

Writing the algebraic representation using the information given in the question:

\[7x+6y=3800\]             …… (i)

\[3x+5y=1750\]    …… (ii)

From equation (i):

\[y=\frac{3800-7x}{6}\]       …… (iii)

Substituting (iii) in equation (ii): 

\[3x+5\left( \frac{3800-7x}{6} \right)=1750\]

\[3x+\frac{9500}{3}-\frac{35x}{6}=1750\]

\[3x-\frac{35x}{6}=1750-\frac{9500}{3}\]

\[\frac{18x-35x}{6}=\frac{5250-9500}{3}\]

\[\frac{-17x}{6}=\frac{-4250}{3}\]

\[x=500\]                  …… (iv)

Substituting (iv) in (iii), we get

\[y=\frac{3800-7\left( 500 \right)}{6}\]

\[y=50\]

Therefore, the bat costs \[Rs\text{ }500\] and the ball costs \[Rs\text{ }50\].


(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of \[\mathbf{10}\text{ }\mathbf{km}\], the charge paid is \[\mathbf{Rs}\text{ }\mathbf{105}\] and for a journey of \[\mathbf{15}\text{ }\mathbf{km}\], the charge paid is \[\mathbf{Rs}\text{ }\mathbf{155}\]. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of \[\mathbf{25}\text{ }\mathbf{km}\] ?.

Ans: Assuming the fixed charge be \[Rs\text{ }x\] and the per km charge be \[Rs\text{ y}\].

Writing the algebraic representation using the information given in the question:

\[x+10y=105\]             …… (i)

\[x+15y=155\]    …… (ii)

From equation (i):

\[x=105-10y\]       …… (iii)

Substituting (iii) in equation (ii): 

\[105-10y+15y=155\]

\[5y=50\]

\[y=10\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=105-10\left( 10 \right)\]

\[x=5\]

Therefore, the fixed charge is \[Rs\text{ }5\] and the per km charge is \[Rs\text{ 10}\].

So, charge for \[25\text{ }km\] will be:

\[=Rs\text{ }\left( x+25y \right)\]

\[=Rs\text{ 255}\] 


(v) A fraction becomes \[\frac{9}{11}\], if \[\mathbf{2}\] is added to both the numerator and the denominator. If, \[\mathbf{3}\] is added to both the numerator and the denominator it becomes \[\frac{5}{6}\]. Find the fraction.

Ans: Assuming the fraction be \[\frac{x}{y}\].

Writing the algebraic representation using the information given in the question:

\[\frac{x+2}{y+2}=\frac{9}{11}\]

\[11x+22=9y+18\]

\[11x-9y=-4\]             …… (i)

\[\frac{x+3}{y+3}=\frac{5}{6}\]

\[6x+18=5y+15\]

\[6x-5y=-3\]    …… (ii)

From equation (i):

\[x=\frac{-4+9y}{11}\]       …… (iii)

Substituting (iii) in equation (ii): 

\[6\left( \frac{-4+9y}{11} \right)-5y=-3\]

\[-24+54y-55y=-33\]

\[y=9\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=\frac{-4+9\left( 9 \right)}{11}\]

\[x=7\]

Therefore, the fraction is \[\frac{7}{9}\].


(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?

Ans: Assuming the age of Jacob be \[x\] and the age of his son be \[y\].

Writing the algebraic representation using the information given in the question:

\[\left( x+5 \right)=3\left( y+5 \right)\]

\[x-3y=10\]             …… (i)

\[\left( x-5 \right)=7\left( y-5 \right)\]

\[x-7y=-30\]    …… (ii)

From equation (i):

\[x=3y+10\]       …… (iii)

Substituting (iii) in equation (ii): 

\[3y+10-7y=-30\]

\[-4y=-40\]

\[y=10\]                  …… (iv)

Substituting (iv) in (iii), we get

\[x=3\left( 10 \right)+10\]

\[x=40\]

Therefore, Jacob’s present age is \[40\] years and his son’s present age is \[10\] years.


Conclusion

NCERT Solutions for Class 10 Maths Chapter 3, Exercise 3.2 act as your perfect guide to conquering linear equations with two variables. This exercise is essential for mastering this fundamental concept. The key understanding from class 10th exercise 3.2 is to make use of methods like elimination and substitution to solve these equations and interpret solutions based on how the equations' graph lines intersect, run parallel, or coincide. By diligently practicing these methods with NCERT Solutions for Class 10 Maths Chapter 3.2 Students can score good marks in their exams.


Class 10 Maths Chapter 3: Exercises Breakdown

Chapter 3 - Pair of Linear Equations in Two Variables Exercises in PDF Format

Exercise 3.1

7 Questions & Solutions

Exercise 3.3

2 Questions & Solutions



CBSE Class 10 Maths Chapter 3 Other Study Materials


Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

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FAQs on NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables Ex 3.2

1. What is the main approach used in NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2?

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 primarily use two algebraic methods to solve pairs of linear equations in two variables: substitution method and elimination method. Both methods follow stepwise CBSE guidelines to ensure clarity of concepts and accuracy in answers as per the latest syllabus (2025–26).

2. How do you solve linear equations by the substitution method in Class 10 Maths Chapter 3?

In the substitution method as explained in the NCERT Solutions for Class 10 Maths Chapter 3, you first solve one equation for one variable in terms of the other. This expression is then substituted into the second equation, reducing it to one variable and making it easier to solve both variables.

3. When should you use the elimination method instead of substitution in NCERT Solutions for Class 10 Maths Chapter 3?

Use the elimination method when the coefficients of a variable in both equations are the same or easily made identical (possibly by multiplying an equation by a constant). This allows you to add or subtract the equations to eliminate one variable quickly, aligning with the CBSE stepwise pattern.

4. What types of solutions can you encounter in pairs of linear equations in two variables?

According to NCERT Solutions Class 10 Maths Chapter 3, pairs of linear equations in two variables can have:

  • A unique solution (lines intersect at one point)
  • Infinite solutions (lines coincide)
  • No solution (lines are parallel and never meet)

5. Which real-life situations can be modeled using pairs of linear equations in Class 10 Chapter 3?

Common real-life scenarios such as calculating ages, determining costs given fixed and variable charges (like taxi fares), mixing solutions, and comparing profits or losses are modeled and solved using linear equations, as reflected in NCERT Solutions for Class 10 Maths Chapter 3 as per CBSE 2025–26 syllabus.

6. Why do some pairs of linear equations have infinite solutions, according to NCERT pattern?

Infinite solutions occur when both equations describe the same straight line—mathematically, this happens when the ratios of coefficients of x, y, and constants are equal (a1/a2 = b1/b2 = c1/c2), as per the concept clarified in NCERT Solutions for Class 10 Maths Chapter 3.

7. How can you identify if a pair of linear equations in two variables has no solution?

If the ratios of the coefficients of x and y are the same but do not equal the ratio of the constants (a1/a2 = b1/b2 ≠ c1/c2), the equations represent parallel lines that never meet, resulting in no solution as per CBSE standards for linear equations in two variables.

8. What should you do if substitution leads to an identity like 0 = 0 or a contradiction as per Class 10 Maths NCERT Solutions?

  • If substitution leads to an identity such as 0 = 0, it means there are infinite solutions (the equations are dependent).
  • If substitution leads to a contradiction, such as 5 = 0, it means the pair has no solution (the equations are inconsistent or parallel).

9. How are word problems involving pairs of linear equations solved in NCERT Solutions for Class 10 Maths Chapter 3?

Word problems are approached by:

  • Translating the given situation into two algebraic equations
  • Assigning variables to unknown quantities
  • Solving the equations using substitution or elimination as demonstrated in the NCERT Solutions
  • Interpreting the mathematical result in the context of the original problem

10. What are common student errors while applying the elimination or substitution method as per NCERT Class 10?

  • Arithmetic mistakes during transposition or simplification
  • Wrong substitution of values
  • Not aligning coefficients properly in elimination
  • Misinterpreting infinite or no solutions
Careful stepwise practice using the NCERT Solutions for Class 10 Maths Chapter 3 helps avoid such errors.

11. Are graphical solutions included in NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2?

No, Exercise 3.2 in NCERT Solutions for Class 10 Maths Chapter 3 focuses on algebraic methods only; graphical representation is typically addressed in earlier exercises or theory as per CBSE 2025–26 syllabus.

12. How does practicing NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 help for board exams?

Regular practice with NCERT Solutions ensures you understand the stepwise CBSE marking scheme, build conceptual clarity in methods (elimination, substitution), and become efficient in solving both direct problems and application-based word problems, helping to maximize your marks in the Class 10 board exam.

13. What is a linear equation in two variables as defined in Class 10 Maths Chapter 3?

A linear equation in two variables takes the form ax + by = c, where a, b, and c are real numbers and at least one of a or b is non-zero. Every solution (x, y) makes the equation true and graphs as a straight line according to NCERT Solutions for Class 10 Maths Chapter 3.

14. How do you check your answer after solving a pair of linear equations in two variables?

After obtaining the values of x and y, substitute them back into both original equations. If both equations are satisfied, your solution is correct, as advised in NCERT Solutions for Class 10 Maths Chapter 3.

15. What is the importance of learning to solve pairs of linear equations in two variables in Class 10 Maths?

Mastering this concept is vital for further maths studies and everyday applications, including financial calculations, mixture problems, and scientific analysis. Proficiency in the methods given in the NCERT Solutions for Class 10 Maths Chapter 3 is crucial for excelling in school exams and competitive tests as per CBSE requirements.