NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

NCERT Solutions Class 10 Maths 
Chapter – 3 (Pair of Linear Equations in Two Variables) 

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 3 Pair of Linear Equations in Two Variables Exercise 3.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 3 Pair of Linear Equations in Two Variables

• NCERT Class 10 Maths Solution Ex – 3.1
• NCERT Class 10 Maths Solution Ex – 3.3
• NCERT Class 10 Maths Solution Ex – 3.4
• NCERT Class 10 Maths Solution Ex – 3.5
• NCERT Class 10 Maths Solution Ex – 3.6
• NCERT Class 10 Maths Solution Ex – 3.7

Exercise – 3.2 

1. 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.
(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

Solution –
(i) Let the number of girls be x and the number of boys be y.
According to the question, the algebraic representation is
x + y = 10
x − y = 4
For x + y = 10,
x = 10 − y

For x − y = 4,
x = 4 + y

The graphical representation is as follows;

From the figure, it can be observed that these lines intersect each other at point (7, 3).
Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.
According to the question, the algebraic representation is
5x + 7y = 50
7x + 5y = 46
For 5x + 7y = 50,

x =

7x + 5y = 46

x =

Hence, the graphic representation is as follows.

From the graph, it is can be seen that the given lines cross each other at point (3, 5).
So, the cost of a pencil is 3/- and cost of a pen is 5/-.

2. On comparing the ratios a₁/a₂ , b₁/b₂ , c₁/c₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0

Solutions –
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 5, b₁ = -4, c₁ = 8
a₂ = 7, b₂ = 6, c₂ = -9
(a₁/a₂) = 5/7
(b₁/b₂) = -4/6 = -2/3
(c₁/c₂) = 8/-9
Since, (a₁/a₂) ≠ (b₁/b₂)
So, the pairs of equations given in the question have a unique solution and the lines cross each other at exactly one point.

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 9, b₁ = 3, c₁ = 12
a₂ = 18, b₂ = 6, c₂ = 24
(a₁/a₂) = 9/18 = 1/2
(b₁/b₂) = 3/6 = 1/2
(c₁/c₂) = 12/24 = 1/2
Since (a₁/a₂) = (b₁/b₂) = (c₁/c₂)
So, the pairs of equations given in the question have infinite possible solutions and the lines are coincident.

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 6, b₁ = -3, c₁ = 10
a₂ = 2, b₂ = -1, c₂ = 9
(a₁/a₂) = 6/2 = 3/1
(b₁/b₂) = -3/-1 = 3/1
(c₁/c₂) = 10/9
Since (a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)
So, the pairs of equations given in the question are parallel to each other and the lines never intersect each other at any point, and there is no possible solution for the given pair of equations.

3. On comparing the ratio, (a₁/a₂) , (b₁/b₂) , (c₁/c₂), find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) (3/2)x+(5/3)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) (4/3)x+2y = 8 ; 2x + 3y = 12

Solutions – 

(i) 3x + 2y = 5 ; 2x – 3y = 7
3x + 2y – 5 = 0;
2x – 3y -7 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 3, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = -3, c₂ = -7
(a₁/a₂) = 3/2
(b₁/b₂) = 2/-3
(c₁/c₂) = -5/-7 = 5/7
Since, (a₁/a₂) ≠ (b₁/b₂)
So, the given equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(ii) 2x – 3y = 8 ; 4x – 6y = 9
2x – 3y – 8 = 0;
4x – 6y – 9 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 2, b₁ = -3, c₁ = -8
a₂ = 4, b₂ = -6, c₂ = -9
(a₁/a₂) = 2/4 = 1/2
(b₁/b₂) = -3/-6 = 1/2
(c₁/c₂) = -8/-9 = 8/9
Since , (a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)
So, the equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

(iii) (3/2)x+(5/3)y = 7; 9x – 10y = 14
(3/2)x + (5/3)y – 7 = 0;
9x – 10y – 14 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 3/2, b₁ = 5/3, c₁ = -7
a₂ = 9, b₂ = -10, c₂ = -14
(a₁/a₂) = 3/(2×9) = 1/6
(b₁/b₂) = 5/(3× -10)= -1/6
(c₁/c₂) = -7/-14 = 1/2
Since, (a₁/a₂) ≠ (b₁/b₂)
So, the equations are intersecting  each other at one point and they have only one possible solution. Hence, the equations are consistent.

(iv) 5x – 3y = 11 ; – 10x + 6y = –22
5x – 3y – 11 = 0;
– 10x + 6y + 22 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 5, b₁ = -3, c₁ = -11
a₂ = -10, b₂ = 6, c₂ = 22
(a₁/a₂) = 5/(-10) = -5/10 = -1/2
(b₁/b₂) = -3/6 = -1/2
(c₁/c₂) = -11/22 = -1/2
Since (a₁/a₂) = (b₁/b₂) = (c₁/c₂)
These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

(v) (4/3)x+2y = 8 ; 2x + 3y = 12
(4/3)x + 2y – 8 = 0;
2x + 3y – 12 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 4/3 , b₁ = 2 , c₁ = -8
a₂ = 2, b₂ = 3 , c₂ = -12
(a₁/a₂) = 4/(3×2)= 4/6 = 2/3
(b₁/b₂) = 2/3
(c₁/c₂) = -8/-12 = 2/3
Since (a₁/a₂) = (b₁/b₂) = (c₁/c₂)
These linear equations are coincident lines and have infinite number of possible solutions. Hence, the equations are consistent.

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solutions – 

(i) x + y = 5, 2x + 2y = 10
x + y – 5 = 0,
2x + 2y – 10 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 1 , b₁ = 1 , c₁ = – 5
a₂ = 2, b₂ = 2 , c₂ = -10
(a₁/a₂) = 1/2
(b₁/b₂) = 1/2
(c₁/c₂) = -5/-10 = 1/2
Since (a₁/a₂) = (b₁/b₂) = (c₁/c₂)
∴ The equations are coincident and they have infinite number of possible solutions.
So, the equations are consistent.
For, x + y = 5
x = 5 – y

For 2x + 2y = 10
x = (10 – 2y)/2 = 5 – y

So, the equations are represented in graphs as follows:

From the figure, we can see, that the lines are overlapping each other. Therefore, the equations have infinite possible solutions.

(ii) x – y = 8, 3x – 3y = 16
x – y – 8 = 0,
3x – 3y – 16 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 1 , b₁ = – 1 , c₁ = – 8
a₂ = 3, b₂ = -3 , c₂ = – 16
(a₁/a₂) = 1/3
(b₁/b₂) = -1/-3 = 1/3
(c₁/c₂) = 8/16 = 1/2
Since, (a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)
The equations are parallel to each other and have no solutions. Hence, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 2 , b₁ = 1 , c₁ = – 6
a₂ = 4, b₂ = -2 , c₂ = – 4
(a₁/a₂) = 2/4 = ½
(b₁/b₂) = 1/-2
(c₁/c₂) = -6/-4 = 3/2
Since, (a₁/a₂) ≠ (b₁/b₂)
The given linear equations are intersecting each other at one point and have only one solution. Hence, the pair of linear equations is consistent.
Now, for 2x + y – 6 = 0
y = 6 – 2x

And for 4x – 2y – 4 = 0
y = (4x – 4)/2 = 2x – 2

So, the equations are represented in graphs as follows:

From the graph, it can be seen that these lines are intersecting each other at only one point (2,2).

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Comparing these equations with a₁x + b₁y + c₁ = 0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 2 , b₁ = -2 , c₁ = – 2
a₂ = 4, b₂ = -4 , c₂ = – 5
(a₁/a₂) = 2/4 = ½
(b₁/b₂) = -2/-4 = 1/2
(c₁/c₂) = 2/5
Since, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Thus, these linear equations have parallel and have no possible solutions. Hence, the pair of linear equations are inconsistent.

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.

Solutions –
Let the width of the garden be x and length be y.
Now, according to the question, we can express the given condition as;
y – x = 4
and
y + x = 36
Now, taking y – x = 4
y = x + 4

For y + x = 36,
y = 36 – x

The graphical representation of both the equation is as follows;

From the graph you can see, the lines intersect each other at a point (16, 20). Hence, the width of the garden is 16 and length is 20.

6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) Intersecting lines
(ii) Parallel lines
(iii) Coincident lines

Solutions –

(i) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy the below condition;
(a₁/a₂) ≠ (b₁/b₂)
Thus, another equation could be 2x – 7y + 9 = 0, such that;
(a₁/a₂) = 2/2 = 1 and (b₁/b₂) = 3/-7
Clearly, you can see another equation satisfies the condition.

(ii) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is parallel lines, it should satisfy the below condition;
(a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)
Thus, another equation could be 6x + 9y + 9 = 0, such that;
(a₁/a₂) = 2/6 = 1/3
(b₁/b₂) = 3/9= 1/3
(c₁/c₂) = -8/9
Clearly, you can see another equation satisfies the condition.

(iii) Given the linear equation 2x + 3y – 8 = 0.
To find another linear equation in two variables such that the geometrical representation of the pair so formed is coincident lines, it should satisfy the below condition;
(a₁/a₂) = (b₁/b₂) = (c₁/c₂)
Thus, another equation could be 4x + 6y – 16 = 0, such that;
(a₁/a₂) = 2/4 = 1/2 ,
(b₁/b₂) = 3/6 = 1/2,
(c₁/c₂) = -8/-16 = 1/2
Clearly, you can see another equation satisfies the condition.

7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution –
Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.
For, x – y + 1 = 0
x = -1 + y

For, 3x + 2y – 12 = 0
x = (12 – 2y)/3

Hence, the graphical representation of these equations is as follows;

From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

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