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

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
From the pair of linear equations in the following problems and find their solutions graphically.
(i) 10 students of Class X took part in 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 ₹ 45. Find the cost of one pencil and one pen.
Solution:
(i) Let the number of boys participating in mathematics quiz is x and the number of girls participating in mathematics quiz is y.
Therefore the equations form in this situations are

(ii) Let the cost of 1 pencil is ₹ x
and the cost of 1 pen is ₹ y
Therefore the linear equations on the given situations are
5x + 7y = 50 … (i)
and 7x + 5y = 46 … (ii)

Again from equation (ii)

7x + 5y = 46
∴ x = \(\frac{46-5 y}{7}\)
Multiply equation. (1) by 7 and (ii) by 5 the get
35x + 49y = 350 … (i)
35x + 25y = 230 …(ii)
Substracting equation (ii) from equation (i) we get
24y = 120
y = 5
Putting the value of y in equation (j)
We get
5x + 35 = 50
5x = 15
x = 5
Cost of one pencil = ₹ 3
Cost of one pen = ₹ 5

Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\) find out whether the lines representing the following pairs of linear equations intersect at a point, 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
Solution:
(i) We have,
5x – 4y + 8 = 0
and 7x + 6y – 9 = 0
Here, a₁ = 5, b₁ = – 4, c₁ = 8
and a₂ = 7, b₂ = 6 and c₂ = – 9
Therefore, \(\frac{5}{7} \neq \frac{-4}{6}\)
So, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
Therefore the following pair of linear equations intersect each other at point.

(ii) We have,
9x + 3y + 12 = 0
18x + 6y + 24 = 0
Here, a₁ = 9, b₁ = 3, c₁ = 12
and a₂ = 18, b₂ = 6 and c₂ = 24
Therefore, \(\frac{9}{18}=\frac{3}{6}=\frac{12}{24}=\frac{1}{2}\)
So, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Therefore the lines representing the given pair of linear equations are coincident.

(iii) We have given,
6x – 3y + 10 = 0
2x – y + 9 = 0
Here, a₁ = 6, b₁ = – 3, c₁ = 10
and a₂ = 2, b₂ = – 1 and c₂ = 9
Therefore,
\(\frac{6}{2}=\frac{-3}{-1} \neq \frac{10}{9}\)
So, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Therefore the lines representing the given pair of linear equations are coincident.

Question 3.
On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\), find out whether the following pair of linear equations are consistent or inconsistent.
(i) 3% + 2y = 5
2x – 3y = 7

(ii) 2x – 3y = 8
4x – 6y = 9

(iii) \(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 3 }\)y = 7
9x – 10y = 14

(iv) 5x – 3y = 11
– 10x + 6y = – 22

(v) \(\frac { 4 }{ 3 }\)x + 2y = 8
2x + 3y = 12
Solution:
(i) We have given,
3x + 2y = 5
2x – 3y = 7
We can also write
3x + 2y – 5 = 0
2x – 3y – 7 = 0
Here, a₁ = 3, b₁ = 2, c₁ = – 5
and a₂ = 2, b₂ = – 3 and c₂ = – 7
So, \(\frac { 3 }{ 2 }\) ≠ \(\frac { 2 }{ – 3 }\)
Therefore, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
So the given pair of linear equations are consistent.

(ii) We have given,
2x – 3y = 8
4x – 4y = 9
We can also write
2x – 3y – 8 = 0
4x – 6y – 3 = 0
Here, a₁ = 2, b₁ = – 3, c₁ = – 8
and a₂ = 4, b₂ = – 6 and c₂ = – 3
So, \(\frac{2}{4}=\frac{-3}{-6} \neq \frac{-8}{-9}\)
Therefore,
So, the given pair of linear equations are inconsistent.

(iii) We have given,
\(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 3 }\)y = 7
9x – 10y = 14
We can also write,
\(\frac { 3 }{ 2 }\)x + \(\frac { 5 }{ 3 }\)y – 7 = 0
9x – 10y – 14 = 0
Here, a₁ = \(\frac { 3 }{ 2 }\), b₁ = \(\frac { 5 }{ 3 }\), c₁ = – 7
and a₂ = 9, b₂ = – 10 and c₂ = – 14

So, the given pair of linear equations consistent.

(iv) We have given,
5x – 3y = 11
– 10x + 6y = – 22
We can also write,
5x – 3y – 11 = 0
– 10x + 6y + 22 = 0
So, \(\frac { 5 }{ -10 }\) = \(\frac { – 3 }{ 6 }\) = \(\frac { -11 }{ 22 }\)
Here, a₁ = 5, b₁ = – 3, c₁ = – 11
and a₂ = – 10, b₂ = 6 and c₂ = 22
Therefore, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
So, the given pair of linear equations consistent.

(v) We have given
\(\frac { 4 }{ 3 }\)x + 2y = 8
2x + 3y = 12
We can also write,
\(\frac { 4 }{ 3 }\)x + 2y – 8 = 0
2x + 3y – 12 = 0
Here, a₁ = \(\frac { 4 }{ 3 }\), b₁ = 2, c₁ = 8
and a₂ = 2, b₂ = 3 and c₂ = – 12

So, the given pair of linear equations 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
Solution:
We have given
x + y = 5
2x + 2y = 10
We can also write,
x + y – 5 = 0
2x + 2y – 10 = 0
Here, a₁ = 1, b₁ = 1, c₁ = – 5
and a₂ = 2, b₂ = 2 and c₂ = – 10
So, \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\) = \(\frac { – 5 }{ – 10 }\)
Therefore, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
So, the given pair of linear equations are consistent from equation (i) we have,

So the given pair of linear equations are inconsistent.

(iii) We have given
2x + y -6 = 0 … (i)
4x – 2y – 4 = 0 … (ii)
Here, a₁ = 2, b₁ = 1, c₁ = – 6
and a₂ = 4, b₂ = – 2 and c₂ = – 4
Therefore, \(\frac{2}{4} \neq \frac{1}{-2}\)
So, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)
Therefore, the given linear equations are consistent From equation (i)

Therefore, the given linear equations are inconsistent.

Question 5.
Half the perimeter of rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let the length of the rectangular garden by y
and the breadth of the rectangular garden by y
According to question
x = y + 4
x – y = 4 … (i)
Perimeter of rectangular garden = 72
2(x + y)= 72
2x + 2y = 72 … (ii)
Multiplying equation (i) and (ii) and adding both equations, we get
2x = 40
x = 20 m
Putting this value in equation (i) we get
20 – y = 4
y = 16 m
Hence the length of garden is 20 m and the width of garden is 16 m.

Question 6.
Given the linear equation 2x + 3y – 8 = 0 write another linear equaton in two variables such that the geometrical representation of the pair so formed is (i) intersecting lines (ii) parallel lines, (iii) coincident lines.
Solution:
Given equation is 2x + 3y – 8 = 0 i.e., a₁ = 2, b₁ = 3, c₁ = 8
(i)

one such equation can be 5x + 4y + 1 = 0 (Try forming other equations. How many such equation can be ?)

One such equation can be 2x + 3y + 5 = 0 (Try forming other equations. How many such equation can be ?]

(iii) Try yourself.

Question 7.
Draw the graphs of the equation x- y +1 = of the vertices of the triangle formed by these line
Solution:

Verticles of the triangle are (-1, 0) (4, 0) and (2, 3).