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

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

Edited By Ramraj Saini | Updated on Nov 08, 2023 08:07 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.2

NCERT Solutions for class 10 maths ex 3.2 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 3.2 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.

These class 10 maths ex 3.2 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 3.2 Class 10 Maths Answers

Pair of Linear Equations in Two Variables Class 10 Chapter 3 Exercise: 3.2

Q1 Form the pair of linear equations in the following problems and find their solutions graphically.

(i) 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.

Let the number of boys is x and the number of girls is y.

Now, According to the question,

Total number of students in the class = 10, i.e.

$\Rightarrow x+y=10.....(1)$

And

the number of girls is 4 more than the number of boys,i.e.

$x=y+4$

$\Rightarrow x-y=4..........(2)$

Different points (x, y) for equation (1)

 X 5 6 4 Y 5 4 6

Different points (x,y) satisfying (2)

 X 5 6 7 y 1 2 3

Graph,

As we can see from the graph, both lines intersect at the point (7,3). that is x= 7 and y = 3 which means the number of boys in the class is 7 and the number of girls in the class is 3.

Q1 Form the pair of linear equations in the following problems and find their solutions graphically.

(ii) 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.

Let x be the price of 1 pencil and y be the price of 1 pen,

Now, According to the question

$5x+7y=50......(1)$

And

$7x+5y=46......(2)$

Now, the points (x,y), that satisfies the equation (1) are

 X 3 -4 10 Y 5 10 0

And, the points(x,y) that satisfies the equation (2) are

 X 3 8 -2 Y 5 -2 12

The Graph,

As we can see from the Graph, both line intersects at point (3,5) that is, x = 3 and y = 5 which means cost of 1 pencil is 3 and the cost of 1 pen is 5.

Give, Equations,

$\\5x - 4y + 8 = 0\\ 7x + 6y - 9 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{5}{7},\:\frac{b_1}{b_2}=\frac{-4}{6}\:and\:\frac{c_1}{c_2}=\frac{8}{-9}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means that both lines intersect at exactly one point.

Given, Equations,

$\\9x + 3y + 12 = 0\\ 18x + 6y + 24 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\:and \\\:\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means that both lines are coincident.

Give, Equations,

$\\6x - 3y + 10 = 0\\ 2x - y+ 9 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{6}{2}=3,\:\frac{b_1}{b_2}=\frac{-3}{-1}=3\:and\:\frac{c_1}{c_2}=\frac{10}{9}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means that both lines are parallel to each other.

Give, Equations,

$\\3x + 2y = 5;\qquad\\ 2x - 3y = 7$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{3}{2},\:\frac{b_1}{b_2}=\frac{2}{-3}\:and\:\frac{c_1}{c_2}=\frac{5}{7}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Given, Equations,

$\\2x - 3y = 8;\qquad \\4x - 6y = 9$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{8}{9}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Given, Equations,

$\\\frac{3}{2}x + \frac{5}{3}y = 7;\qquad\\ \\ 9x -10y = 14$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{3/2}{9}=\frac{3}{18}=\frac{1}{6},\\\:\frac{b_1}{b_2}=\frac{5/3}{-10}=\frac{5}{-30}=-\frac{1}{6}\:and\\\:\frac{c_1}{c_2}=\frac{7}{14}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Given, Equations,

$5x - 3y = 11;\qquad \\-10x + 6y =-22$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{11}{-22}=-\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

Given, Equations,

$\\\frac{4}{3}x + 2y = 8; \qquad\\\\ 2x + 3y = 12$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{4/3}{2}=\frac{4}{6}=\frac{2}{3},\\\:\frac{b_1}{b_2}=\frac{2}{3}\:\:and\\\:\frac{c_1}{c_2}=\frac{8}{12}=\frac{2}{3}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

Given, Equations,

$\\x + y = 5 \qquad\\ 2x + 2 y = 10$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{5}{10}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

It means the given equations have an infinite number of solutions and thus pair of linear equations is consistent.

The points (x,y) which satisfies in both equations are

 X 1 3 5 Y 4 2 0

Given, Equations,

$\\x - y = 8,\qquad\\ 3x - 3y = 16$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{1}{3},\\\:\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\:and\\\:\frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Given, Equations,

$\\2x + y - 6 =0, \qquad \\4x - 2 y - 4 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}$

As we can see

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

It means the given equations have exactly one solution and thus pair of linear equations is consistent.

Now The points(x, y) satisfying the equation are,

 X 0 2 3 Y 6 2 0

And The points(x,y) satisfying the equation $\\4x - 2 y - 4 = 0$ are,

 X 0 1 2 Y -2 0 2

GRAPH:

As we can see both lines intersects at point (2,2) and hence the solution of both equations is x = 2 and y = 2.

Given, Equations,

$\\2x - 2y - 2 =0, \qquad\\ 4x - 4y -5 = 0$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\\\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\\\:\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\:and\\\:\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}$

As we can see

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

It means the given equations have no solution and thus pair of linear equations is inconsistent.

Let $l$be the length of the rectangular garden and $b$ be the width.

Now, According to the question, the length is 4 m more than its width.i.e.

$l=b+4$

$l-b=4....(1)$

Also Given Half Parameter of the rectangle = 36 i.e.

$l+b=36....(2)$

Now, as we have two equations, on adding both equations, we get,

$l+b+l-b=4+36$

$\Rightarrow 2l=40$

$\Rightarrow l=20$

Putting this in equation (1),

$\Rightarrow 20-b=4$

$\Rightarrow b=20-4$

$\Rightarrow b=16$

Hence Length and width of the rectangle are 20m and 16 respectively.

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the intersection of lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , is ,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}$

So Any line with this condition can be $4x+3y-16=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{3}=1$

As

$\frac{1}{2}\neq1$ the line satisfies the given condition.

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , for being parallel is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

So Any line with this condition can be $4x+6y-8=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-8}{-8}=1$

As

$\frac{1}{2}=\frac{1}{2}\neq1$ the line satisfies the given condition.

Given the equation,

$2x + 3y -8 =0$

As we know that the condition for the coincidence of the lines $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So any line with this condition can be $4x+6y-16=0$

Here,

$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}$

As

$\frac{1}{2}=\frac{1}{2}=\frac{1}{2}$ the line satisfies the given condition.

Given, two equations,

$x - y + 1=0.........(1)$

And

$3x +2 y - 12=0.........(2)$

The points (x,y) satisfying (1) are

 X 0 3 6 Y 1 4 7

And The points(x,y) satisfying (2) are,

 X 0 2 4 Y 6 3 0

GRAPH:

As we can see from the graph that both lines intersect at the point (2,3), And the vertices of the Triangle are ( -1,0), (2,3) and (4,0). The area of the triangle is shaded with a green colour.

## More About NCERT Solutions for Class 10 Maths Exercise 3.2

The emphasis of NCERT solutions for Class 10 Maths exercise 3.2 was on graphical representations of linear equations in two variables, types of solutions, and associated graphs. Two systems of equations can be parallel at times. They have no solution in such situations. In rare situations, the two systems of equations may coincide, resulting in an infinite number of solutions for the given system. If the two systems of equations overlap, the given system is said to be consistent because it has a unique solution. NCERT solutions for Class 10 Maths exercise 3.2 was on the graphical depiction of the system of equations. NCERT syllabus Exercise 3.2 Class 10 Maths contains all three questions relating to the graphical depiction.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 3.2

• NCERT solutions for Class 10 Maths exercise 3.2 help us solve and revise all of the problems in these exercises, as well as develop a deeper grasp of the pair of linear equations in two variables.
• If you thoroughly practise the NCERT solution for Class 10th Maths chapter 3 exercise 3.2, which contains all significant questions from the exam point and all questions are revisited, you will be able to acquire higher marks in the test.
• Exercise 3.2 Class 10 Maths, is based on graphical and algebraic representations of two-variable linear equations, types of solutions, and their graphs, all of which are significant concepts in the chapter.

Also see-

## Subject Wise NCERT Exemplar Solutions

1. Check whether the equation xy - 9 = 3 a linear equation in two variables ?

The concepts related to linear equation is discussed in ex 3.2 class 10. Practice the problems discussed in this exercise to command the concepts. For this question, because of the term xy is of degree 2,  xy - 9 = 3 is not a linear equation in two variables.

2. In a graph, how many quadrants are there?

To understand the concepts of quadrants go through the problems discussed in class 10 maths ex 3.2. In a graph, there are four quadrants. a point can be represent in in a plain using the (x, y) coordinates.

3. When two lines intersect on a plane, how many choices are there, and what are they?

To get in depth understanding of related concepts practice problems enumerated in the class 10 ex 3.2.  as per these concepts, when two lines are in a plane, there are three alternative solutions. They really are.

• Two lines may intersect at times.
• Two lines may not intersect at times, and they may be parallel to each other.
• Two lines may be coincident at times.
4. In the NCERT solutions for Class 10 Maths chapter 3 exercise 3.2 How many questions and what types of questions are covered ?

In Class 10th Maths chapter 3 exercise 3.2, there are seven questions based on the notion of graphical representation of a system of equations.

5. How many solutions are there for linear equations in two variables if the equations are inconsistent and independent?

If the equations are consistent and dependent, there are no solutions to linear equations in two variables.

6. What is the criterion for two-variable linear equations that are both consistent and dependent?

You can go through the 10th class maths exercise 3.2 answers to get deeper understanding of the concepts related to equations are consistent and dependent. The requirement for linear equations in two variables is:

a1/a2 = b1/b2 = c1/c2

7. According to NCERT solutions for Class 10 Maths chapter 3 exercise 3.2 , What is the graphical method of solution of a pair of linear equations?

The basic strategy to represent the linear equations on the graph and determine the point of intersection is the graphical method of solution of a pair of linear equations.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.

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 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

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