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.

Inequality solver is an application help you to solve linear inequality and tracing linear equations and root point.

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

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 a_{1} x + b_{1}y + c1and a_{2} x + b_{2} y + c_{2} = 0,

a_{1} = 2, b_{1} = – 3, c_{1} = – 8

and a_{2} = 4, b_{2} = – 6, c_{2} = – 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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0

a_{1} = 3, b_{1} = -1, and c_{1} = -2

and a_{2} = 3, b_{2} = -1 and c_{2}= -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 a_{1}x + b_{1}y+c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0

a_{1} = 1, b_{1} = -1, and c_{1} = -1

and a_{2} = 4, b_{2} = -4 and c_{2}= -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 a_{1}x + b_{1}y+c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0

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

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 a_{1}x + b_{1}y+c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0

a_{1} = 1, b_{1} = 1, and c_{1} = -3

and a_{2} = 3, b_{2} = -2 and c_{2}= -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

a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 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

a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 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

a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 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.

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).

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.

We hope the given RBSE Solutions for Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Ex 4.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 4 Linear Equation and Inequalities in Two Variables Exercise 4.1, drop a comment below and we will get back to you at the earliest.

## Leave a Reply