RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1

RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Exercise 4.1.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 4
Chapter Name	Linear Equation and Inequalities in Two Variables
Exercise	4.1
Number of Questions Solved	4
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1

Ex 4.1 Class 10 RBSE Question 1
By comparing \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } ,\frac { { b }_{ 1 } }{ { b }_{ 2 } } \) and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } \) find,
whether the following (RBSESolutions.com) pair of linear equations is consistent or inconsistent.
(i) 2r – 3y = 8; 4c – 6y = 9
(ii) 3x – y = 2; 6x – 2y = 4
(iii) 2x – 2y = 2; 4x – 4y = 5
(iv) \(\frac { 4 }{ 3 } \) + 2y = 8; 2x + 3y = 12
Solution:
(i) Given linear pair of equations
23 – 3y = 8 or 2x – 3y – 8 = 0
and 4x – 6y = 9 or 4x – 6y – 9 = 0
Comparing above equations by a₁ x + b₁y + c1and a₂ x + b₂ y + c₂ = 0,
a₁ = 2, b₁ = – 3, c₁ = – 8
and a₂ = 4, b₂ = – 6, c₂ = – 9.
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } ,\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -3 }{ -6 } =\frac { 1 }{ 2 } ,\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -8 }{ -9 } =\frac { 8 }{ 9 } \)
∴ \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c2 }_{ } } \)
∴ Given linear pair has no solution.
So, given linear pair is inconsistent.

(ii) Given pair of (RBSESolutions.com) linear equations
3x – y= 2
or 3x – y – 2 = 0 …(i)
and 6x – 2y = 4
or 6x – 2y – 4 = 0
or 3x – y – 2 = 0…(ii)
Comparing equations (i) and (ii) by a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
a₁ = 3, b₁ = -1, and c₁ = -2
and a₂ = 3, b₂ = -1 and c₂= -2

Linear pair is coincident, so (RBSESolutions.com) linear pair has infinite solutions.
Thus given pair is consistant.

(iii) Given linear pair
2x – 2y = 2
or 2x – 2y – 2 = 0
or x – y – 1 = 0 …(i)
and 4x – 4y – 5 = 0 …(ii)
Comparing equations (i) and (ii) by a₁x + b₁y+c₁ = 0 and a₂x + b₂y + c₂ = 0
a₁ = 1, b₁ = -1, and c₁ = -1
and a₂ = 4, b₂ = -4 and c₂= -5

Given linear pair has no solution.
Thus given Linear pair is inconsistent.

(iv) Given (RBSESolutions.com) linear pair
\(\frac { 4 }{ 3 }\)x + 2y = 8
\(\frac { 4 }{ 3 }\)x + 2y – 8 = 0 …(i)
and 2x + 3y = 12
2x + 3y – 12 = 0 …(ii)
Comparing equation (i) and (ii) by pair a₁x + b₁y+c₁ = 0 and a₂x + b₂y + c₂ = 0

Given linear pair has infinite solutions, so linear pair is consistent.

RBSE Solutions For Class 10 Maths Chapter 4 Question 2
Solve the following pair of (RBSESolutions.com) linear equations graphically and write nature of solution.
(i) x + y = 3; 3x – 2y = 4
(ii) 2x – y = 4; x + y = -1
(iii) x + y = 5; 2x + 2y = 10
(iv) 3x + y = 2; 2x – 3y = 5
Solution:
(i) Given linear pair
x + y = 3
x + y -3 = 0 ….(i)
3x – 2y = 4
3x – 2y – 4 = 0 ………(ii)
Comparing equation (i) and (ii) by pair a₁x + b₁y+c₁ = 0 and a₂x + b₂y + c₂ = 0
a₁ = 1, b₁ = 1, and c₁ = -3
and a₂ = 3, b₂ = -2 and c₂= -4

Linear pair has (RBSESolutions.com) unique solution.
Thus, linear pair is consistent.

Graphical Method:
By equation (i),
x + y = 3
x = 3 – y
Putting y = 0, x = 3 – 0 =3
Putting y = 1, x = 3 – 1 = 2
Putting y = 2, x = 3 – 2 = 1
Table 1 for equation (i),

Table 2 for equation (ii),

Plot the points of Table (1) and (2) on graph (RBSESolutions.com) paper and by joining these points, two straight lines are obtained.

From above graph, it is clear that two straight lines cut at point P(2, 1). Thus x = 2 and y = 1 is required solution.

(ii) Given linear pair
2x – y = 4
2x – y – 4 = 0 …..(i)
x + y = -1
x + y + 1 = 0 ….(ii)
Comparing (RBSESolutions.com) equation (i) and (ii) by linear pair
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

Linear pair is consistent which will have unique solutions.

Graphical Method:
By equation (i),
2x – y = 4
Putting x = 0, 2 x 0 – y = 4
y = -4
Putting x = 1, 2 x – y = 4
-y = 4 – 2
Putting x = 2, 2 x 2 – y = 4
4 – y =4
y = 0

By equation (ii),
x + y = -1
Putting x = 0. 0 + y = -1
y = -1
Putting x = 1, +1 + y = -1
y = -1 – 1
y = -2
Putting x = 2,
2 + y = -1
y = -3

By plotting the (RBSESolutions.com) points of Table 1 and 2 we get two straight lines.

From above graph, It is clear that both the straight lines cut each other at point P( 1, – 2).
Thus, x = 1,y = – 2 are required solution

(iii) Given linear pair:
x + y = 5 or x + y – 5 = 0 …(i)
2x + 2y = 10 or 2x + 2y – 10 = 0 ……..(ii)
Comparing above (RBSESolutions.com) pair by general linear pair
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

∴ Lines represented linear pair will be coincident and linear pair will have infinite solutions.
Thus, given linear pair is consistent.

Graphical Method:
By equation (i),
x + y = 5
⇒ x = 5 – y
Putting y = 0, x = 5 – 0 = 5
Putting y = 3, x = 5 – 3 = 2
Putting y = 5, x =5 – 5 = 0

By joining (RBSESolutions.com) the points A(5, 0), B(2, 3) and C(0, 5) on graph paper.
We get a straight line which indicates the equation x + y = 5.
By equation (ii),
2x + 2y = 10
⇒ 2(x + y) = 10
⇒ x + y =5
⇒ x = 5 – y
Putting y = 0, x = 5 – 0 = 5
Putting y = 2, x = 5 – 2 = 3
Putting y = 5, x = 5 – 5 = 0

By joining the (RBSESolutions.com) points A(5, 0), 8(3,2) and C(0, 5) on graph paper we get a straight line which indicates the equation 2x + 2y = 10.

From graph, lis clear that given pair of linear equations are coincident. Thus they have infinitely many solutions.

(iv) Given pair of (RBSESolutions.com) linear equations
3x + y = 2 or 3x + y – 2 = 0
2x – 3y = 5 or 2x – 3y – 5 = 0 …(ii)
Comparing equation (i) and (ii) by pair
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

Thus given equation will have unique solutions.
∴ Given pair is consistent.

Graphical Method:
By equation (i)
3x + y = 2
y = 2 – 3x
Putting x = 0, y = -3 x 0
y = 2
Putting x = -1, y = 2 – 3 x (-1)
y = 2 + 3
y = 5

From Table (1) arid (2), plot the points on (RBSESolutions.com) graph paper and by joining them, we get two straight lines.

From above graph, It is clear that two straight lines intersect each other at point P( 1, – 1)
Thus, x = 1 and y = – 1 is required solution.

RBSE Solutions For Class 10 Maths Chapter 4 Miscellaneous Question 3
Solve the following pair of linear equations, (RBSESolutions.com) graphically and find the coordinates of that points where lines represented by these cuts y-axis.
(i) 2x – 5y + 4 = 0; 2x + y – 8 = 0
(ii) 3x + 2 = 12 ; 5x – 2y = 4
Solution:
(i) Given pair of linear equations
2x – 5y + 4 = 0 …(i)
and 2x + y – 8 = 0 …(ii)
From equation (i),

By equation (ii),
2x + y – 8 = 0
or y – 2x + 8 = 0
Putting x = 4, y = – 2 x 4 + 8
= -8 + 8
= 0
Putting x = 3, y = – 2 x 3 + 8
= -6 + 8
= 2
Putting x = 2, y = – 2 x 2 + 8
= -4 + 8
= 4

Plot the points (RBSESolutions.com) from Table (1) and (2) on graph.
By joining these points two straight lines are obtained.

From above graph ¡t is clear that two straight lines intersect each other at point P(3, 2).
∴ Its required solutions are x = 3 and y = 2
and two straight lines cuts the y-axis at (0, 0.8) and (0, 8).

(ii) Given pair of (RBSESolutions.com) linear equations
3x + 2y = 12 …(i)
and 5x – 2y = 4 …(ii)
From equation (i),
3x + 2y = 12


Plot the points from (RBSESolutions.com) Table (1) and (2) on graph paper. By joining these points two straight lines are obtained.

From above graph, it is clear that two straight lines intersect each other at point P(2, 3).
∴ x = 2 and y = 3 are required solutions and two straight lines cuts y-axis at (0, 6) and (0, – 2).

RBSE Solutions For Class 10 Maths Chapter 4.1 Question 4
Solve the following pair of (RBSESolutions.com) linear equations grapycally and find the coordinates of the triangle so formed with the y-axis and the lines.
4x – 5y = 20,
3x + 5y = 15
Solution:
Given, pair of linear equation
4x – 5y = 20 ………..(i)
and 3x + 5y = 15 ………(ii)
From equation (i),
4x – 5y = 20
5y = 4x – 20


Plot the points obtained from (RBSESolutions.com) Table (1) and (2) on graph paper. By joining these points two straight lines are obtained.

From the above graph it is (RBSESolutions.com) clear that two lines intersect each other at point P(5, 0)
∴ x = 5 and y = 0 are required solution.
(0, 3), (0, – 4) and (5, 0) are co-ordinates of vertices of ΔABP formed by two straight lines at y – axis.

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