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NCERT Solutions for Exercise 3.3 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

NCERT Solutions for Exercise 3.3 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Edited By Ramraj Saini | Updated on Nov 13, 2023 02:54 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3

NCERT Solutions for class 10 maths ex 3.3 Pair of Linear Equations in two variables is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 3.3 class 10 deals with the graphical and algebraic representations of linear equations in two variables and their different type of solutions such as a unique solution, no solutions or infinitely many solutions, using the five methods namely the Graphical method, substitution method, cross Multiplication method, elimination method and determinant method.

This Story also Contains
  1. NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.3
  2. Access Exercise 1.3 Class 10 Maths Answers
  3. Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.3
  4. More About NCERT Solutions for Class 10 Maths Exercise 3.3:
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 3.3 :
  6. NCERT Solutions of Class 10 Subject Wise
  7. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 3.3 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables
NCERT Solutions for Exercise 3.3 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

These class 10 maths ex 3.3 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Access Exercise 1.3 Class 10 Maths Answers

Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.3

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

\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, Solution of the given equations is s = 9 and t = 6.

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

\\ 3 x - y = 3\\ 9x - 3y = 9

Answer:

Given, two equations,

\\ 3 x - 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 the equation 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\left(\frac{1.3-0.2x}{0.3}\right)=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=\left ( \frac{1.3-0.2x}{0.3} \right )=\left ( \frac{1.3-0.4}{0.3} \right )=\frac{0.9}{0.3}=3

Hence, Solution of the given equations is,

x=2\:and\:y=3 .

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

\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0

Answer:

Given, two equations,

\\\sqrt2x + \sqrt3y = 0\\ \sqrt3x - \sqrt8y= 0

Now, from (1), we have

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

Substituting this in (2), we get

\sqrt{3}x-\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )=0

\Rightarrow \sqrt{3}x=\sqrt{8}\left ( -\frac{\sqrt{2}x}{\sqrt{3}} \right )

\Rightarrow \3x=-4x

\Rightarrow7x=0

\Rightarrow x=0

Substituting this value of x in (3)

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

Hence, 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{2}{3}\left ( \frac{5y}{3}-2 \right )........(3)

Putting this in (2) we get,

\frac{1}{3}\times\frac{2}{3}\left ( \frac{5y}{3}-2 \right )+\frac{y}{2}=\frac{13}{6}

\frac{10y}{27}-\frac{4}{9}+\frac{y}{2}=\frac{13}{6}

\frac{20y}{54}-\frac{4}{9}+\frac{27y}{54}=\frac{13}{6}

\frac{47y}{54}=\frac{13}{6}+\frac{4}{9}

\frac{47y}{54}=\frac{117}{54}+\frac{24}{54}

47y=117+24

47y=141

y=\frac{141}{47}

y=3

putting this value in (3) we get,

x=\frac{2}{3}\left ( \frac{5y}{3}-2 \right )

x=\frac{2}{3}\left ( \frac{5\times3}{3}-2 \right )

x=\frac{2}{3}\left (5-2 \right )

x=\frac{2}{3}\times 3

x=2

Hence x=2\:\:and\:\:y=3.

Q2 Solve 2x + 3y = 11and 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\left ( \frac{11-2x}{3} \right )=-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=\left ( \frac{11-2x}{3} \right )=\frac{11-2\times(-2)}{3}=\frac{15}{3}=5

Hence, 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 Value of m is -1.

Q3(i) Form the pair of linear equations for the following problem and find their solution by 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 let the bigger number is y.

Now, According to the question,

y-x=26......(1)

And

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

Now, the substituting 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 substitution method

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

Answer:

Let the larger angle be x and smaller angle be y

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

x+y=180.......(1)

Also given in the question,

x-y=18.......(2)

Now, From (2) we have,

y=x-18.......(3)

Substituting this value in (1)

x+x-18=180

\Rightarrow 2x=180+18

\Rightarrow 2x=198

\Rightarrow x=99

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

\Rightarrow y=x-18=99-18=81

Hence the two supplementary angles are

99^0\:and\:81^0.

Q4 Form the pair of linear equations for the following problems and find their solution by 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\left ( \frac{3800-7x}{6} \right )=1750

\Rightarrow 18x+19000-35x=1750\times6

\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}=\frac{3800-7\times500}{6}=\frac{3800-3500}{6}=\frac{300}{6}=50

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

Q5 Form the pair of linear equations for the following problems and find their solution by 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 traveling 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.

Q6 Form the pair of linear equations for the following problems and find their solution by substitution method. (v) A fraction becomes \frac{9}{11 }, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \frac{5 }{6} . Find the fraction.

Answer:

Let the numerator of the fraction be x and denominator of the fraction is 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\left ( \frac{11x+4}{9} \right )=-3

\Rightarrow 54x-55x-20=-27

\Rightarrow -x=20-27

\Rightarrow x=7

Substituting this value of x in (3)

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

Hence the required fraction is

\frac{x}{y}=\frac{7}{9}.

Q7 Form the pair of linear equations for the following problems and find their solution by 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, Present age of Jacob is 40 years and the present age of Jacob's son is 10 years.

More About NCERT Solutions for Class 10 Maths Exercise 3.3:

In exercise 3.3 Class 10 Maths, The method of substitution involves three main steps. First, we need to solve one equation for one of the variables. Then we need to substitute the expression (value) into the other equation and solve. Finally, we need to substitute the value into the original equation to find the corresponding variable. The solution to the system of linear equations is written in the form of (x, y). The NCERT solutions for Class 10 Maths exercise 3.3 mainly focused on solving the pair of linear equations in two variables by using a substitution method. All the three questions related to the substitution method are given in exercise 3.3 Class 10 Maths. Also students can get access of Pair of linear equations in two variables notes to revise all the concepts discussed in this chapter.

Benefits of NCERT Solutions for Class 10 Maths Exercise 3.3 :

  • NCERT solutions for Class 10 Maths exercise 3.3 helps the students who wish to score high in CBSE term exams
  • n exercise 3.3 Class 10 Maths, the substitution method is the easy method to solve equations and it takes away an equation to make it easier for you to do and also the method gives you only one equation to work on instead of having two.
  • By solving the NCERT solution for Class 10th Maths chapter 3 exercise 3.3 exercises, students can attain a good score in term exams as well as in competitive examinations.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. Find the value of x in 6x-y=10 if y=2 .

Putting y=2 in 6x-y=10 

6x-2=10 

6x=10+2

Thus x =12/6=2

2. 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.

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. 

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

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) 

4. Mention the different methods to solve the system of equations linear equations in two variables.

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

5. What is the general form of linear equation?

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

6. Whether the substitution method can be used to solve the system of equations in three variables?

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

7. Define substitution method according to NCERT solutions for Class 10 Maths chapter 3 exercise 3.3?

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

8. In the NCERT solutions for Class 10 Maths chapter 3 exercise 3.3, how many questions are there?

In the NCERT solutions for Class 10 Maths chapter 3 exercise 3.3, there are three questions.

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