NCERT Solutions for class 10 maths chapter 3

Pair of Linear Equations in Two Variables

NCERT Solutions for class 10 maths chapter 3 Exercise 3.2

Question 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.

Answer

(i)Let the number of girls is x and number of boys is y
As per the given question
x +y = 10 .......(i)
x= 4 + y
x - y = 4 --------(ii)
Now, for x+y = 10 the solutions are; For x = 4 + y, the solutions are;


From the graph, lines intersecting each other at point (7, 3). So there are 7 girls and 3 boys in the class.


(ii) Let costs of 1 pencil is Rs.x and costs 1 pen is Rs.y.
According to the question,
5x + 7y = 50
7x + 5y = 46
For, 5x + 7y = 50 the solutions are;
For 7x + 5y = 46 the solutions are;

Hence, the graphical representation is as follows;


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

Question 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

Answer

(i) Given expressions;
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 have a unique solution

(ii) Given expressions;
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 have infinite possible solutions and the lines are coincident.

(iii) Given Expressions;
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 are parallel to each other so there is no possible solution for the given pair of equations.

Question 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

Answer

(i) Given : 3x + 2y = 5 or 3x + 2y -5 = 0
and 2x – 3y = 7 or 2x – 3y -7 = 0
Comparing these equations with a₁x+b₁y+c₁ = 0
And a₁x+b₁y+c₁ = 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 have only one possible solution. The equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9
Therefore,
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)Given (3/2)x + (5/3)y = 7 and 9x – 10y = 14
Therefore,
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) Given, 5x – 3y = 11 and – 10x + 6y = –22
Therefore,
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)
Infinite number of possible solutions. Hence, the equations are consistent.

(v)Given, (4/3)x +2y = 8 and 2x + 3y = 12
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)
Infinite number of possible solutions. Hence, the equations are consistent.

Question 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

Answer

(i)Given, x + y = 5 and 2x + 2y = 10
(a₁/a₂) = 1/2
(b₁/b₂) = 1/2
(c₁/c₂) = 1/2
Since (a₁/a₂) = (b₁b₂) = (c₁/c₂)
∴The equations are coincident and they have infinite number of possible solutions.
So, the equations are consistent.
For, x + y = 5 For 2x + 2y = 10


The lines are overlapping each other.Therefore, It have infinite possible solutions.

(ii) Given, x – y = 8 and 3x – 3y = 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) Given, 2x + y – 6 = 0 and 4x – 2y – 4 = 0
(a₁/a₂) = 2/4 = 1/2
(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.
For 2x + y – 6 = 0 And for 4x – 2y – 4 = 0

The lines are intersecting each at point,(2,2).

(iv) Given, 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
(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.

Question 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.

Answer

Let the width of the garden is x and length is y.
Now, according to the question
y = 4 + x
y + x = 36
Now, taking y = x + 4 For y + x = 36


The lines intersects each other at a point(16, 20)
width of the garden is 16
length is 20.

Question 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

Answer

(i) Given the linear equation 2x + 3y – 8 = 0.
For Intersecting lines

(a₁/a₂) ≠ (b₁/b₂)
Thus, another equation could be 3x – 7y + 9 = 0, such that;
(a₁/a₂) = 2/3 and (b₁/b₂) = 3/-7
Clearly, equation satisfies the condition.

(ii) Given the linear equation 2x + 3y – 8 = 0.
For Parallel lines
(a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)
Thus, another equation could be 8x + 12y + 9 = 0, such that;
(a₁/a₂) = 2/8 = 1/4
(b₁/b₂) = 3/12= 1/4
(c₁/c₂) = -8/9
Clearly, equation satisfies the condition.

(iii) Given the linear equation 2x + 3y – 8 = 0.
For Coincident lines
(a₁/a₂) = (b₁/b₂) = (c₁/c₂)
Thus, another equation could be 8x + 12y – 24 = 0, such that;
(a₁/a₂) = 2/8 = 1/4 ,(b₁/b₂) = 3/12 = 1/4, (c₁/c₂) = -8/-24 = 1/4
Clearly, equation satisfies the condition.

Question 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.

Answer

Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.
For, x – y + 1 = 0 For, 3x + 2y – 12 = 0


The lines are intersecting each other at point (2, 3).