RS Aggarwal Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables

Get RS Aggarwal Solutions for Class 10 Chapter 3 Linear Equations in Two Variables here. Solution for all linear equations in two variables exercises problems helps students in preparing for the exams in a better way. These RS Aggarwal solutions help students to ace their exams by preparing them well in advance. Get free RS Aggarwal Solutions for Class 10 Maths, Chapter 3 Linear Equations in Two Variables and clear all doubts on the concept. Download RS Aggarwal Maths Class 10 Solutions now for free.

Download PDF of RS Aggarwal Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables

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Access Solutions of Maths RS Aggarwal Chapter 3 – Linear Equations in Two Variables

Get detailed solutions for all the questions listed under the below exercises:

Exercise 3A Solutions: 29 Questions (Long Answers)

Exercise 3B Solutions: 20 Questions (Short Answers)

Exercise 3C Solutions: 13 Questions (Long Answers)

Exercise 3D Solutions: 14 Questions (Short Answers)

Exercise 3E Solutions: 20 Questions (Long Answers)

Exercise 3F Solutions: 10 Questions (6 Short Answers and 4 Long Answers)

Exercise 3A Page No: 87

Solve each of the following systems of equations graphically:

(Question 1 to Question 10)

Question 1:

2x + 3y = 2

x – 2y = 8

Solution:

2x + 3y = 2 …………(1)

x – 2y = 8 ..………(2)

Step 1: From Equation (1), isolate x

2x = 2 – 3y

Or x = (2 – 3y)/ 2

Choose any values for y and get corresponding values of x:

x 1 -2 4
y 0 2 -2

Plot above points on the graph and join them.

Step 2: From equation (2)

x – 2y = 8

or x = 8 + 2y

Choose any values for y and get corresponding values of x:

x 6 4 2
y -1 -2 -3

Plot above points on the graph and join them.

Step 3:

Graph:

rs aggarwal class 10 chapter 3 ex a img 1

Step 4: Both the lines intersect each other at point (4, -2).

So, x = 4, y = -2

Question 2:

3x + 2y = 4,

2x – 3y = 7.

Solution:

3x + 2y = 4 …..(1)

2x – 3y = 7 ……(2)

Isolate x from equation (1) and find the values of x and y.

3x + 2y = 4

or x = (4 – 2y)/ 3

x 0 2 4
y 2 -1 -4

Similarly, Isolate x from equation (2) and find the values of x and y.

2x – 3y = 7

x = (3y+ 7) / 2

x 5 2 -1
y 1 -1 -3

Graph:

rs aggarwal class 10 chapter 3 ex a img 2

Both the lines intersect each other at point (2, -1).

So, x = 2, y = -1

Question 3:

2x + 3y = 8,

x – 2y + 3 = 0.

Solution:

2x + 3y = 8 …(1)

x – 2y + 3 = 0 …(2)

Isolate x from equation (1) and find the values of x and y.

2x + 3y = 8

2x = 8 – 3y

x = ( 8 – 3y )/2

x 4 1 -2
y 0 2 4

Similarly, Isolate x from equation (2) and find the values of x and y.

x – 2y + 3 = 0

or x = 2y – 3

x -3 -1 1
y 0 1 2

Graph:

rs aggarwal class 10 chapter 3 ex a img 3

Both the lines intersect each other at the point (1, 2).

So, x = 1, y = 2

Question 4:

2x – 5y + 4 = 0,

2x + y – 8 = 0.

Solution:

2x – 5y + 4 = 0 …..(1)

2x + y – 8 = 0 …..(2)

Isolate x from equation (1) and find the values of x and y.

2x – 5y + 4 = 0

or 2x = 5y – 4

x = (5y – 4)/ 2

x -2 3 8
y 0 2 4

Similarly, Isolate x from equation (2) and find the values of x and y.

2x + y – 8 = 0

or y = 8 – 2x

x 1 2 3
y 6 1 2

Graph:

rs aggarwal class 10 chapter 3 ex a img 4

Both the lines intersect each other at the point (3, 2).

So, x = 3, y = 2

Question 5:

3x + 2y = 12,

5x – 2y = 4.

Solution:

3x + 2y = 12 …..(1)

5x – 2y = 4 ….(2)

Isolate x from equation (1) and find the values of x and y.

3x + 2y = 12

3x = 12 – 2y

Or x = (12 – 2y) / 2

x 4 2 0
y 0 3 6

Similarly, Isolate x from equation (2) and find the values of x and y.

5x – 2y = 4

5x = 4 + 2 y

or x = (4 + 2 y) / 5

x 0 2 4
y -2 3 8

Graph:

rs aggarwal class 10 chapter 3 ex a img 5

Both the lines intersect each other at the point (2, 3).

So, x = 2, y = 3

Question 6:

3x + y + 1 = 0

2x – 3y + 8 = 0

Solution:

3x + y + 1 = 0 ….(1)

2x – 3y + 8 = 0 ……(2)

Isolate x from equation (1) and find the values of x and y.

3x + y + 1 = 0

or y = -3x – 1

x 0 -1 -2
y -1 2 5

Similarly, Isolate x from equation (2) and find the values of x and y.

2x – 3y + 8 = 0

x = (3y – 8 ) / 2

x -4 -1 2
y 0 2 4

Graph:

rs aggarwal class 10 chapter 3 ex a img 6

Both the lines intersect each other at the points (-1, 2).

So, x = -1, y = 2

Question 7:

2x + 3y + 5 = 0

3x – 2y – 12 = 0

Solution:

2x + 3y + 5 = 0 …(1)

3x – 2y – 12 = 0 …(2)

Isolate x from equation (1) and find the values of x and y.

2x + 3y + 5 = 0

2x = -3y – 5

x = (-3y – 5) / 2

x -4 -1 2
y 1 -1 -3

Similarly, Isolate x from equation (2) and find the values of x and y.

3x – 2y – 12 = 0

or 3x = 2y + 12

or x = (2y + 12 )/ 3

x 4 0 2
y 0 -6 -3

Graph:

rs aggarwal class 10 chapter 3 ex a img 7

Both the lines intersect each other at the point (2, -3).

So, x = 2, y = -3

Question 8:

2x – 3y + 13 = 0

3x – 2y + 12 = 0

Solution:

2x – 3y + 13 = 0 ……………(1)

3x – 2y + 12 = 0 …………..(2)

Isolate x from equation (1) and find the values of x and y.

2x – 3y + 13 = 0

or 2x = 3y – 13

or x = (3y – 13) / 2

x -5 -2 1
y 1 3 5

Similarly, Isolate x from equation (2) and find the values of x and y.

3x – 2y + 12 = 0

or x = (2y – 12 ) / 3

x -4 -2 0
y 0 3 6

Graph:

rs aggarwal class 10 chapter 3 ex a img 8

Both the lines intersect each other at the point (-2, 3).

x = -2, y = 3

Question 9:

2x + 3y – 4 = 0

3x – y + 5 = 0

Solution:

2x + 3y – 4 = 0 ……(1)

3x – y + 5 = 0 …….(2)

Isolate x from equation (1) and find the values of x and y.

2x + 3y – 4 = 0

or x = (4 – 3y) / 2

x 2 -1 5
y 0 2 -2

Similarly, Isolate x from equation (2) and find the values of x and y.

3x – y + 5 = 0

or y = 5 + 3x

x 0 -1 -2
y 5 2 -1

Graph:

rs aggarwal class 10 chapter 3 ex a img 9

Both the lines intersect each other at the point (-1, 2).

So, x = -1, y = 2

Question 10:

x + 2y + 2 = 0

3x + 2y – 2 = 0

Solution:

x + 2y + 2 = 0 …………(1)

3x + 2y – 2 = 0 …………(2)

Isolate x from equation (1) and find the values of x and y.

x + 2y + 2 = 0

or x = -2y – 2

x -2 0 2
y 0 -1 -2

Similarly, Isolate x from equation (2) and find the values of x and y.

3x + 2y – 2 = 0

or x = (2 – 2y) / 3

x 0 2 -2
y 1 -2 4

Graph:

rs aggarwal class 10 chapter 3 ex a img 10

Both the lines intersect each other at the point (2, -2).

So, x = 2, y = -2

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:

(Question 11 to Question 15)

Question 11:

x – y + 3 = 0, 2x + 3y – 4 = 0.

Solution:

x – y + 3 = 0 ………..(1)

2x + 3y – 4 = 0 ………(2)

Isolate x from equation (1) and find the values of x and y.

x – y + 3 = 0

or x = y – 3

x y
-3 0
-1 2
0 3

Similarly, Isolate x from equation (2) and find the values of x and y.

2x + 3y – 4 = 0

or x = (4 – 3y) / 2

x y
2 0
-1 2
5 -2

Graph:

rs aggarwal class 10 chapter 3 ex a img 11

Both the lines intersect each other at (-1, 2) and x-axis at A (-3, 0) and D (2, 0).

Which form a triangle, BAD.

Now,

Find the area of triangle BAD:

Area of ∆BAD = 1/ 2 (base x altitude)

= 1/ 2 x 5 x 2

= 5 sq.units. Answer!

Question 12:

2x – 3y + 4 = 0, x + 2y – 5 = 0.

Solution:

2x – 3y + 4 = 0 …………(1)

x + 2y – 5 = 0 ……….(2)

Isolate x from equation (1) and find the values of x and y.

2x – 3y + 4 = 0

or x = (3y – 4) / 2

x y
-2 0
1 2
4 4

Similarly, Isolate x from equation (2) and find the values of x and y.

x + 2y – 5 = 0

or x = 5 – 2y

x y
5 0
3 1
1 2

Graph:

rs aggarwal class 10 chapter 3 ex a img 12

Now,

Find the area of triangle BAD:

Area of ∆BAD = 1/2 x base x altitude

= 1/2 x 7 x 2

= 7 sq. units

Question 13:

4x – 3y + 4 = 0, 4x + 3y – 20 = 0.

Solution:

4x – 3y + 4 = 0 ……..(1)

4x + 3y – 20 = 0 ………(2)

Isolate x from equation (1) and find the values of x and y.

4x – 3y + 4 = 0

or x = (3y – 4) / 4

x y
-1 0
2 4
-4 -4

Similarly, Isolate x from equation (2) and find the values of x and y.

4x + 3y – 20 = 0

or x = (20 – 3y) / 4

x y
5 0
2 4
-1 8

Graph:

rs aggarwal class 10 chapter 3 ex a img 13

Both the lines intersect each other at the point B (2, 4) and intersect x-axis at A (-1, 0) and D (5, 0).

To find the area of Triangle BAD:

Area ∆BAD = 1/2 x base x altitude

= ½ x 6 x 4

= 12 sq. units

Question 14:

x – y + 1 = 0, 3x + 2y – 12 = 0

Solution:

x – y + 1 = 0 ………….(1)

3x + 2y – 12 = 0 ………….(2)

Isolate x from equation (1) and find the values of x and y.

x – y + 1 = 0

or x = y – 1

x y
-1 0
0 1
1 2

Similarly, Isolate x from equation (2) and find the values of x and y.

3x + 2y – 12 = 0

or x = (12 – 2y) / 3

x y
4 0
2 3
0 6

Graph:

rs aggarwal class 10 chapter 3 ex a img 14

Both the lines intersect each other at the point E (2, 3) and intersect x- axis at A (-1, 0) and D (4, 0).

To find the area of Triangle EAD:

Area of ∆EAD = 1/2 x base x altitude 1

= ½ x 5 x 3

= 7.5 sq. units

Question 15:

x – 2y + 2 = 0, 2x + y – 6 = 0.

Solution:

x – 2y + 2 = 0 ………..(1)

2x + y – 6 = 0 ………..(2)

Isolate x from equation (1) and find the values of x and y.

x – 2y + 2 = 0

or x = 2y – 2

x y
-2 0
0 1
2 2

Similarly, Isolate y from equation (2) and find the values of x and y.

2x + y – 6 = 0

Or y = 6 – 2x

x y
0 6
2 2
3 0

Graph:

rs aggarwal class 10 chapter 3 ex a img 15

Both the lines intersect each other at the point C (2, 2) and intersect the x-axis at A (-2, 0) and F (3, 0).

Now, To find the area of Triangle CAF:

Area of ∆CAF = 1/2 x base x altitude

= 1/2 x 5 x 2

= 5 sq. units

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:

(Question 16 to Question 21)

Question 16:

2x – 3y + 6 = 0, 2x + 3y – 18 = 0.

Solution:

2x – 3y + 6 = 0 ……….(1)

2x + 3y – 18 = 0 ……….(2)

Isolate x from equation (1) and find the values of x and y.

2x – 3y + 6 = 0

or x = (3y – 6) / 2

x y
-3 0
0 2
3 4

Similarly, Isolate x from equation (2) and find the values of x and y.

2x + 3y – 18 = 0

or x = (18 – 3y) / 2

x y
0 6
3 4
6 2

Graph:

rs aggarwal class 10 chapter 3 ex a img 16

Both the lines intersect each other at C (3, 4) and intersect y-axis at B (0, 2) and D (0, 6).

Now, To find the area of Triangle CBD:

Area of ∆CBD = 1/2 x base x altitude

= 1/2 x 4 x 3 sq. units

= 6 sq. units

Question 17:

4x – y – 4 = 0, 3x + 2y – 14 = 0.

Solution:

4x – y – 4 = 0 ………..(1)

3x + 2y – 14 = 0 …………(2)

Isolate x from equation (1) and find the values of x and y.

4x – y – 4 = 0

or x = (y + 4) /2

x y
1 0
0 -4
2 4

Similarly, Isolate y from equation (2) and find the values of x and y.

3x + 2y – 14 = 0

or y = (14 – 3x) / 2

x y
0 7
2 4
4 1

Graph:

rs aggarwal class 10 chapter 3 ex a img 17

Both the lines intersect each other at C (2, 4) and y-axis at B (0, -4) and D (0, 7).

To find the area of Triangle CBD:

Area of ∆CBD = 1/2 x base x altitude

= 1/2 x 11 x 2

= 11 sq. units

Question 18:

x – y – 5 = 0, 3x + 5y – 15 = 0

Solution:

x – y – 5 = 0 …….(1)

3x + 5y – 15 = 0 ….(2)

Isolate x from equation (1) and find the values of x and y.

x – y – 5 = 0

or x = y + 5

x y
5 0
0 -5
1 -4

Similarly, Isolate x from equation (2) and find the values of x and y.

3x + 5y – 15 = 0

or x = (15 – 5y ) /3

x y
5 0
0 3
-5 6

Graph:

rs aggarwal class 10 chapter 3 ex a img 18

Both the lines intersect each other at A (5, 0) and y-axis at B (0, -5) and E (0, 3).

To find the area of Triangle ABE:

Area of ∆ABE = 1/2 x base x altitude

= 1/2 x 8 x 5

= 20 sq. units

Question 19.

2x – 5y + 4 = 0, 2x + y – 8 = 0.

Solution:

2x – 5y + 4 = 0 ……..(1)

2x + y – 8 = 0 …….(2)

Isolate x from equation (1) and find the values of x and y.

2x – 5y + 4 = 0

or x = (5y – 4) /2

x y
-2 0
3 2
-7 -2

Similarly, Isolate x from equation (2) and find the values of x and y.

2x + y – 8 = 0

or x = (8 – y) / 2

x y
4 0
3 2
2 4

Graph:

rs aggarwal class 10 chapter 3 ex a img 19

Both the lines intersect each other at B (3, 2) and y-axis at G (0, 1) and H (0, 8).

To find the area of Triangle EGH:

Now area of ∆EGH = 1/2 x base x altitude

= 1/2 x 7 x 3

= 10.5 sq. units

Question 20:

5x – y – 7 = 0, x – y + 1 = 0

Solution:

5x – y – 7 = 0 ………….(1)

x – y + 1 = 0 ………….(2)

Isolate y from equation (1) and find the values of x and y.

5x – y – 7 = 0

or y = 5x – 7

x y
0 -7
1 -2
2 3

Similarly, Isolate x from equation (2) and find the values of x and y.

x – y + 1 = 0

or x = y – 1

x y
-1 0
1 2
2 3

Graph:

rs aggarwal class 10 chapter 3 ex a img 20

Both the lines intersect each other at C (2, 3) and intersect y-axis at A (0, -7) and F (0, 1).

Now, To find the area of Triangle CAG.

Area of ∆CAG = 1/2 x base x altitude

= 1/2 x 8 x 2

= 8 sq.units

Question 21:

2x – 3y = 12, x + 3y = 6

Solution:

2x – 3y = 12 …………(1)

x + 3y = 6 …………(2)

Isolate x from equation (1) and find the values of x and y.

2x – 3y = 12

or x = (12 + 3y) /2

x y
6 0
3 -2
0 -4

Plot the points (6, 0), (3, -2) and (0, -4) on the graph and join them to get a line.

Similarly, Isolate x from equation (2) and find the values of x and y.

x + 3y = 6

or x = 6 – 3y

x y
6 0
0 2
-6 4

Graph:

rs aggarwal class 10 chapter 3 ex a img 21

Both the lines intersect each other at the points A (6, 0) and intersect the y-axis at C (0, -4) and E (0, 2).

To find the area of Triangle ACE:

Now area of ∆ACE = 1/2 x base x altitude

= 1/2 x CE x AO

= 1/2 x 6 x 6

= 18 sq.units

Show graphically that each of the following given systems of equations has infinitely many solutions:

Question 22:

2x + 3y = 6, 4x + 6y = 12

Solution:

2x + 3y = 6 …………(1)

4x + 6y = 12 …………(2)

Isolate x from equation (1) and find the values of x and y.

2x + 3y = 6

or x = (6 – 3y) / 2

x y
3 0
0 2
-3 4

Isolate x from equation (2) and find the values of x and y.

4x + 6y = 12

or x = (12 – 6y) / 2

x y
3 0
3/2 1
-3/2 3

Graph:

rs aggarwal class 10 chapter 3 ex a img 22

From the graph, we can see that, all the points lie on the same straight line.

Which shows that, system has infinite many solutions.

Question 23:

3x – y = 5, 6x – 2y = 10.

Solution:

3x – y = 5 …………..(1)

6x – 2y = 10 ………….(2)

Isolate y from equation (1) and find the values of x and y.

3x – y = 5

or y = 3x – 5

x y
0 -5
1 -2
2 1

Similarly, Isolate x from equation (2) and find the values of x and y.

6x – 2y = 10

Or x = (10 + 2y)/ 6

x y
2 1
4 7
3 4

Graph:

rs aggarwal class 10 chapter 3 ex a img 23

Both the lines coincide each other. Which shows that system has infinite many solutions.

Question 24:

2x + y = 6, 6x + 3y = 18

Solution:

2x + y = 6 ……….(1)

6x + 3y = 18 ….…..(2)

Isolate y from equation (1) and find the values of x and y.

2x + y = 6

or y = 6 – 2x

x y
0 6
2 2
4 -2

Similarly, Isolate x from equation (2) and find the values of x and y.

6x + 3y = 18

or x = (18 – 3y )/ 2

x y
3 0
1 4
5 -4

Graph:

rs aggarwal class 10 chapter 3 ex a img 24

Both the lines coincide each other.

This system has infinitely many solutions

Question 25:

x – 2y = 5, 3x – 6y = 15.

Solution:

x – 2y = 5 ……(1)

3x – 6y = 15 …….(2)

Isolate x from equation (1) and find the values of x and y.

x – 2y = 5

or x = 5 + 2y

x y
5 0
3 -1
1 -2

Similarly, Isolate x from equation (2) and find the values of x and y.

3x – 6y = 15

or x = (15 + 6y) / 3

x y
7 1
-1 -3
-3 -4

Graph:

rs aggarwal class 10 chapter 3 ex a img 25

All the points lie on the same line.

Which shows that, lines coincide each other. Hence, the system has infinite many solutions.

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:

(Question 26 to Question 29)

Question 26:

x – 2y = 6, 3x – 6y = 0.

Solution:

x – 2y = 6 ………….(1)

3x – 6y = 0 …….(2)

Isolate x from equation (1) and find the values of x and y.

x – 2y = 6

or x = 6 + 2y

x y
6 0
4 -1
0 -3

Similarly, Isolate x from equation (2) and find the values of x and y.

3x – 6y = 0

or x = 2y

x y
0 0
2 1
4 2

Graph:

rs aggarwal class 10 chapter 3 ex a img 26

From the graph, we can see that both the lines are parallel to each other.

This system has no solution.

Question 27:

2x + 3y = 4, 4x + 6y = 12.

Solution:

2x + 3y = 4 …………(1)

4x + 6y = 12 ……….(2)

Isolate x from equation (1) and find the values of x and y.

2x + 3y = 4

or x = (4 – 3y )/ 2

x y
2 0
-1 2
5 -2

Similarly, Isolate x from equation (2) and find the values of x and y.

4x + 6y = 12

or x = (12 – 6y )/ 4

x y
3 0
0 2
-3 4

Graph:

rs aggarwal class 10 chapter 3 ex a img 27

From the graph, both the lines are parallel.

Therefore given system has no solution.

Question 28.

2x + y = 6, 6x + 3y = 20.

Solution:

2x + y = 6 …….(1)

6x + 3y = 20 ………(2)

Isolate y from equation (1) and find the values of x and y.

2x + y = 6

or y = 6 – 2x

x y
0 6
1 4
3 0

Similarly, Isolate x from equation (2) and find the values of x and y.

6x + 3y = 20

or x = 1/6(20 – 3y)

x y
10/3 0
7/3 2
4/3 4

Graph:

rs aggarwal class 10 chapter 3 ex a img 28

From the graph, both the lines are parallel.

Therefore this system has no solution.

Question 29: Draw the graphs of the following equations on the same graph paper:

2x + y = 2, 2x + y = 6.

Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.

Solution:

2x + y = 2 ………….(1)

2x + y = 6 ………….(2)

Isolate y from equation (1) and find the values of x and y.

2x + y = 2

or y = 2 – 2x

x y
0 2
1 0
-2 6

Similarly, Isolate y from equation (2) and find the values of x and y.

2x + y = 6

or y = 6 – 2x

x y
0 6
2 2
3 0

Graph:

rs aggarwal class 10 chapter 3 ex a img 29

From above graph: ABFD is the trapezium formed by lines whose vertices are A (0, 2), B (1, 0), F (3, 0), D (0, 6).

Now, find the area of trapezium :

Area of trapezium ABFD = Area ∆DOF – Area ∆AOB

= 1/2 (DO x OF) – 1/2 (AO x OB)

= 1/2 (6 x 3) – 1/2 (2 x 1)

= 8 sq.units. Answer!


Exercise 3B Page No: 103

Solve for x and y:

Question 1:

x + y = 3,

4x – 3y = 26.

Solution:

x + y = 3 ……..(1)

4x – 3y = 26 ……(2)

Isolate x from equation (1), we get

x = 3 – y

Substituting the value of x in equation (2),

4(3 – y) – 3y = 26

12 – 4y – 3y = 26

-7y = 14

y = -2

Solve for x:

x = 3 – y = 3 – (-2) = 5

Answer: x = 5 and y = -2

Question 2:

x – y = 3,

x/3 + y /2 = 6

Solution:

x – y = 3 …………(1)

x/3 + y /2 = 6 ……..(2)

Find the value of x from equation and substitute in equation (2).

x = 3 + y

Now,

rs aggarwal class 10 chapter 3 ex b img 1

The value of x is 9 and the value of y is 6.

Question 3:

2x + 3y = 0,

3x + 4y = 5.

Solution:

2x + 3y= 0 ……..(1)

3x + 4y = 5 …….(2)

Let us use elimination method to solve the given system of equations.

Multiply (1) by 4 and (2) by 3. And subtract both the equations.

rs aggarwal class 10 chapter 3 ex b img 2

From (1), y = -10

Hence: x = 15 and y = -10

Question 4:

2x – 3y = 13,

7x – 2y = 20.

Solution:

2x – 3y = 13 ……(1)

7x – 2y = 20 ……….(2)

Let us use elimination method to solve the given system of equations.

Multiply (1) by 2 and (2) by 3. And subtract both the equations.

rs aggarwal class 10 chapter 3 ex b img 3

Substitute the value of x in equation (1), we have

y = -3

Hence: x = 2 and y = -3

Question 5.

3x – 5y – 19 = 0,

-7x + 3y + 1 = 0.

Solution:

3x – 5y – 19 = 0 …………..(1)

-7x + 3y + 1 = 0 ……….(2)

Let us use elimination method to solve the given system of equations.

Multiply (1) by 3 and (2) by 5. And add both the equations.

rs aggarwal class 10 chapter 3 ex b img 4

Substitute the value of x in equation (1), we have

3 × (-2) -5y = 19 ⇨ -6 -5y = 19

y = -5

Hence: x = -2 and y = -5

⇨ x = -2, y = -5

Question 6:

2x – y + 3 = 0,

3x – 7y + 10 = 0.

Solution:

2x – y + 3 = 0 …………..(1)

3x – 7y + 10 = 0 ………..(2)

Let us use substitution method to solve the given system of equations.

Find the value of y form equation (1) and substitute the value in (2).

From (1)

-y = -3 – 2x ⇨ y = 2x + 3

And,

rs aggarwal class 10 chapter 3 ex b img 5

y = 1

Again, y = 2x + 3

1 = 2x + 3

x = -1

Answer: x = -1 and y = 1

Question 7:

x/2 – y/9 = 6,

x/7 + y/3 = 5.

Solution:

x/2 – y/9 = 6

x/7 + y/3 = 5

Simplify both the equations:

rs aggarwal class 10 chapter 3 ex b img 6

9x – 2y = 108 ………..(1)

3x + 7y = 105 …………(2)

Let us use elimination method to solve the given system of equations.

Multiply (2) by 3. And subtract both the equations.

rs aggarwal class 10 chapter 3 ex b img 7

From (1); 9x – 2(9) = 108

x = 14

Answer: x = 14 and y = 9

Question 8:

x/3 + y/4 = 11,

5x/6 – y/3 = -7.

Solution:

x/3 + y/4 = 11

5x/6 – y/3 = -7

Simplify both the equations:

rs aggarwal class 10 chapter 3 ex b img 8

We have,

4x + 3y = 132 …………(1)

5x – 2y = -42 …….……(2)

Let us use elimination method to solve the given system of equations.

Multiply (1) by 2 and (2) by 3. And add both the equations.

rs aggarwal class 10 chapter 3 ex b img 9

From (1): 4(6) + 3y = 132

y = 108/3 = 36

Answer: x = 6 and y = 36

Question 9:

4x – 3y = 8,

6x – y = 29 / 3

Solution:

4x – 3y = 8 ………….(1)

6x – y = 29 / 3 ……..(2)

Let us use elimination method to solve the given system of equations.

Multiply (2) by 3. And subtract both the equations.

rs aggarwal class 10 chapter 3 ex b img 10

Or x = 3/2

From (1);

4(3/2) – 3y = 8

Or y = -2/3

Answer: x = 3/ 2 and y = -2 / 3

Question 10:

2x – 3y / 4 = 3,

5x = 2y + 7.

Solution:

2x – 3y/4 = 3 or 8x – 3y = 12

5x = 2y + 7

Given set of equations can be written as:

8x – 3y = 12 ……..(1)

5x – 2y = 7 ………..(2)

Let us use elimination method to solve the given system of equations.

Multiply (1) by 2 and (2) by 3. And subtract both the equations.

rs aggarwal class 10 chapter 3 ex b img 11

From (1); 8(3) – 3y = 12

y = 4

Answer: x = 3 and y = 4

Question 11:

2x + 5y = 8 / 3,

3x – 2y = 5 / 6 .

Solution:

2x + 5y = 8 /3 …….(1)

3x – 2y = 5 /6 ……..(2)

Let us use elimination method to solve the given system of equations.

Multiply (1) by 4 and (2) by 5. And add both the equations.

rs aggarwal class 10 chapter 3 ex b img 12

x = ½

Substitute the value of x in equation (1), we have

2(1/2) + 5y = 8 /3

y = 1/3

Answer: x = ½ and y = 1/3

Question 12:

2x + 3y + 1 = 0,

(7 – 4x) /3 = y

Solution:

2x + 3y + 1 = 0 ……..(1)

(7 – 4x) /3 = y ……….(2)

Put value of y in (1), we get

2x + 3((7 – 4x) /3 ) + 1 = 0

2x + 7 – 4x + 1 = 0

x = 4

from (2):

(7 – 4(4)) /3 = y

y = -3

Answer: x = 4 and y = -3

Question 13:

0.4x + 0.3y = 1.7,

0.7x – 0.2y = 0.8.

Solution:

0.4x + 0.3y = 1.7

0.7x – 0.2y = 0.8

Multiply both the equations by 10, we get

4x + 3y = 17 …………(1)

7x – 2y = 8 ..………..(2)

Multiply (1) by 2 and (2) by 3,

8x + 6y = 34

21x – 6y = 24

Adding both the equations

29x = 58

x = 2

From (1); 4 x 2 + 3y = 17

⇨ 8 + 3y = 17

⇨ 3y = 17 – 8 = 9

y = 3

Answer: x = 2, y = 3

Question 14:

0.3x + 0.5y = 0.5,

0.5x + 0.7y = 0.74.

Solution:

0.3x + 0.5y = 0.5,

0.5x + 0.7y = 0.74.

Multiply both the equations by 10, we get

3x + 5y = 5 ….(1)

5x + 7y = 7.4 ….(2)

Multiply (1) by 7 and (2) by 5. And subtract both the equations.

rs aggarwal class 10 chapter 3 ex b img 13

Substitute the value of x in equation (1), we have

3(0.5) + 5y = 5

y = 0.7

Answer: x = 0.5 and y = 0.7

Question 15:

7(y + 3) – 2(x + 2) = 14,

4(y – 2) + 3(x – 3) = 2.

Solution:

Simplify given equations

7(y + 3) – 2(x + 2) = 14

or 7y – 2x = -3

and 4(y – 2) + 3(x – 3) = 2

or 4y + 3x = 19

new set of equations is:

7y – 2x = -3 ……(1)

4y + 3x = 19 ……(2)

Let us use elimination method to solve the given system of equations.

Multiply (1) by 3 and (2) by 2. And add both the equations.

rs aggarwal class 10 chapter 3 ex b img 14

Substitute the value of x in equation (1), we have

7(1) – 2x = -3

x = 5

Answer: x = 5 and y = 1

Question 16:

6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)

Solution:

6x + 5y = 7x + 3y + 1 = 2(x + 6y – 1)

6x + 5y = 7x + 3y + 1

⇨ 6x + 5y – 7x – 3y = 1

⇨ -x + 2y = 1

⇨ 2y – x = 1

Again, 7x + 3y + 1 = 2(x + 6y – 1)

7x + 3y + 1 = 2x + 12y – 2

⇨ 7x + 3y – 2x – 12y = -2 – 1

⇨ 5x – 9y = -3

New set of equations is:

2y – x = 1 ……(1)

5x – 9y = -3 ……….(2)

Using substitution method;

From (1), x = 2y – 1

Substituting the value of x in (2),

5(2y – 1) – 9y = -3

⇨ 10y – 5 – 9y = -3

⇨ y = -3 + 5

⇨ y = 2

And, x = 2y – 1 = 2(2) – 1 = 3

Answer: x = 3, y = 2

Question 17:

rs aggarwal class 10 chapter 3 ex b img 15

Solution:

Simply equations:

rs aggarwal class 10 chapter 3 ex b img 16

And

rs aggarwal class 10 chapter 3 ex b img 17

rs aggarwal class 10 chapter 3 ex b img 18

New set of equations is:

x – y = -4 ……….(1)

2x +19y = 118 …………(2)

Using substitution method:

From (1): x = y – 4

Put x in (2)

rs aggarwal class 10 chapter 3 ex b img 19

Again, x = y – 4 = 6 – 4 = 2

Answer: x = 2, y = 6

Question 18:

5/x + 6y = 13,

3/x + 4y = 7 (x ≠ 0)

Solution:

5/x + 6y = 13 ……..(1)

3/x + 4y = 7 …….(2)

Let us use elimination method to solve the given system of equations.

Multiply (1) by 2 and (2) by 3. And subtract both the equations.

rs aggarwal class 10 chapter 3 ex b img 20

Substitute the value of x in equation (1), we have

Answer: x = and y =

Question 19.

x + 6 / y = 6,

3x – 8 / y = 5 (y ≠ 0)

Solution:

Question 20.

2x – 3 / y = 9,

3x + 7 / y = 2 (y ≠ 0)

Solution:

x = 3, y = -1


Exercise 3C Page No: 111

Solve each of the following systems of equations by using the method of cross multiplication:

Question 1:

x + 2y + 1 = 0,

2x – 3y – 12 = 0.

Solution:

x + 2y + 1 = 0 …(1)

2x – 3y – 12 = 0 …(2)

From equation (1): a_1 = 1, b_1 = 2 and c_1 = 1

And from equation (2): a_2 = 2, b_2 = – 3 and c_2 = – 12

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 1

Answer: x = 3 and y = – 2

Question 2:

3x – 2y + 3 = 0,

4x + 3y – 47 = 0.

Solution:

3x – 2y + 3 = 0 …(1)

4x + 3y – 47 = 0 …(2)

From equation (1): a_1 = 3, b_1 = – 2 and c_1 = 3

From equation (2): a_2 = 4, b_2 = 3 and c_2 = – 47

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 2

Question 3:

6x – 5y – 16 = 0,

7x – 13y + 10 = 0.

Solution:

6x – 5y – 16 = 0 …(1)

7x – 13y + 10 = 0 …(2)

From equation (1): a_1 = 6, b_1 = – 5 and c_1 = – 16

From equation (2): a_2 = 7, b_2 = – 13 and c_2 = 10

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 3

Question 4:

3x + 2y + 25 = 0,

2x + y + 10 = 0.

Solution:

3x + 2y + 25 = 0 …(1)

2x + y + 10 = 0 …(2)

From equation (1): a_1 = 3, b_1 = 2 and c_1 = 25

From equation (2): a_2 = 2, b_2 = 1 and c_2 = 10

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 4

Question 5:

2x + 5y = 1,

2x + 3y = 3.

Solution:

2x + 5y – 1 = 0 …(1)

2x + 3y – 3 = 0 …(2)

From equation (1): a_1 = 2, b_1 = 5 and c_1 = – 1

From equation (2): a_2 = 2, b_2 = 3 and c_2 = – 3

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 5

Question 6:

2x + y = 35,

3x + 4y = 65.

Solution:

2x + y – 35 = 0 …(1)

3x + 4y – 65 = 0 …(2)

From equation (1): a_1 = 2, b_1 = 1 and c_1 = – 35

From equation (2): a_2 = 3, b_2 = 4 and c_2 = – 65

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 6

Question 7:

7x – 2y = 3,

22x – 3y = 16.

Solution:

7x – 2y – 3 = 0 …(1)

22x – 3y – 16 = 0 …(2)

From equation (1): a_1 = 7, b_1 = – 2 and c_1 = – 3

From equation (2): a_2 = 22, b_2 = – 3 and c_2 = – 16

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 7

Question 8.

x/6 + y/15 = 4

x/3 – y/12 = 19/4

Solution:

Simplify given equations:

x/6 + y/15 = 4

(5x + 2y)/30 = 4

5x + 2y – 120 = 0

And x/3 – y/12 = 19/4

(4x – y)/12 = 19/4

4x – y – 57 = 0

New set of equations is:

5x + 2y – 120 = 0 …..(1)

4x – y – 57 = 0 ……….(2)

From equation (1): a_1 = 5, b_1 = 2 and c_1 = – 120

From equation (2): a_2 = 4, b_2 = – 1 and c_2 = – 57

Using cross multiplication,

rs aggarwal class 10 chapter 3 ex c img 8

Question 9:

1/x + 1/y = 7

2/x + 3/y = 17 (x≠0 and y≠ 0)

Solution:

Simplify given equations:

Let 1/x = u and 1/y = v

1/x + 1/y = 7 implies

u + v = 7

2/x + 3/y = 17 implies:

2u + 3v = 17

New set of equations is:

u + v = 7 ……(1)

2u + 3v = 17 ………(2)

From equation (1): a_1 = 1, b_1 = 1 and c_1 = – 7

From equation (2): a_2 = 2, b_2 = 3 and c_2 = – 17

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 9

Question 10:

rs aggarwal class 10 chapter 3 ex c img 10a

Solution:

Consider 1/(x + y) = a and 1/(x – y) = b

Given equations turn as:

5a – 2b + 1 = 0 ….(1)

15a + 7b – 10 = 0 …(2)

From equation (1): a_1 = 5, b_1 = -2 and c_1 = 1

From equation (2): a_2 = 15, b_2 = 7 and c_2 = – 10

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 10b

So a = 13/65 = 1/5 and b = 65/65 = 1

Putting back values and write equations in the form of x and y, we get

x + y = 5 and x – y = 1

Again we get a set of two linear equations with two variables. Solve again to find the value of x and y.

Rearranging them again,

x + y – 5 = 0 …(3)

x – y – 1 = 0 …(4)

From equation (3): a_1 = 1, b_1 = 1 and c_1 = – 5

From equation (4): a_2 = 1, b_2 = – 1 and c_2 = – 1

Using cross multiplication method,

rs aggarwal class 10 chapter 3 ex c img 10c

Answer: x = 3 and y = 2

Question 11:

rs aggarwal class 10 chapter 3 ex c img 11a

From equation (2): a_2 = a, b_2 = – b and c_2 = – 2ab

Using cross multiplication method, we get

rs aggarwal class 10 chapter 3 ex c img 11b

Answer: x = b and y = -a

Question 12.

2ax + 3by = (a + 2b),

3ax + 2by = (2a + b).

Solution:

rs aggarwal class 10 chapter 3 ex c img 12

Question 13:

rs aggarwal class 10 chapter 3 ex c img 13


Exercise 3C Page No: 111

Solve each of the following systems of equations by using the method of cross multiplication:

Question 1:

x + 2y + 1 = 0,

2x – 3y – 12 = 0.

Solution:

x + 2y + 1 = 0 …(1)

2x – 3y – 12 = 0 …(2)

From equation (1): a_1 = 1, b_1 = 2 and c_1 = 1

And from equation (2): a_2 = 2, b_2 = – 3 and c_2 = – 12

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 1

Answer: x = 3 and y = – 2

Question 2:

3x – 2y + 3 = 0,

4x + 3y – 47 = 0.

Solution:

3x – 2y + 3 = 0 …(1)

4x + 3y – 47 = 0 …(2)

From equation (1): a_1 = 3, b_1 = – 2 and c_1 = 3

From equation (2): a_2 = 4, b_2 = 3 and c_2 = – 47

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 2

Question 3:

6x – 5y – 16 = 0,

7x – 13y + 10 = 0.

Solution:

6x – 5y – 16 = 0 …(1)

7x – 13y + 10 = 0 …(2)

From equation (1): a_1 = 6, b_1 = – 5 and c_1 = – 16

From equation (2): a_2 = 7, b_2 = – 13 and c_2 = 10

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 3

Question 4:

3x + 2y + 25 = 0,

2x + y + 10 = 0.

Solution:

3x + 2y + 25 = 0 …(1)

2x + y + 10 = 0 …(2)

From equation (1): a_1 = 3, b_1 = 2 and c_1 = 25

From equation (2): a_2 = 2, b_2 = 1 and c_2 = 10

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 4

Question 5:

2x + 5y = 1,

2x + 3y = 3.

Solution:

2x + 5y – 1 = 0 …(1)

2x + 3y – 3 = 0 …(2)

From equation (1): a_1 = 2, b_1 = 5 and c_1 = – 1

From equation (2): a_2 = 2, b_2 = 3 and c_2 = – 3

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 5

Question 6:

2x + y = 35,

3x + 4y = 65.

Solution:

2x + y – 35 = 0 …(1)

3x + 4y – 65 = 0 …(2)

From equation (1): a_1 = 2, b_1 = 1 and c_1 = – 35

From equation (2): a_2 = 3, b_2 = 4 and c_2 = – 65

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 6

Question 7:

7x – 2y = 3,

22x – 3y = 16.

Solution:

7x – 2y – 3 = 0 …(1)

22x – 3y – 16 = 0 …(2)

From equation (1): a_1 = 7, b_1 = – 2 and c_1 = – 3

From equation (2): a_2 = 22, b_2 = – 3 and c_2 = – 16

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 7

Question 8.

x/6 + y/15 = 4

x/3 – y/12 = 19/4

Solution:

Simplify given equations:

x/6 + y/15 = 4

(5x + 2y)/30 = 4

5x + 2y – 120 = 0

And x/3 – y/12 = 19/4

(4x – y)/12 = 19/4

4x – y – 57 = 0

New set of equations is:

5x + 2y – 120 = 0 …..(1)

4x – y – 57 = 0 ……….(2)

From equation (1): a_1 = 5, b_1 = 2 and c_1 = – 120

From equation (2): a_2 = 4, b_2 = – 1 and c_2 = – 57

Using cross multiplication,

rs aggarwal class 10 chapter 3 ex c img 8

Question 9:

1/x + 1/y = 7

2/x + 3/y = 17 (x≠0 and y≠ 0)

Solution:

Simplify given equations:

Let 1/x = u and 1/y = v

1/x + 1/y = 7 implies

u + v = 7

2/x + 3/y = 17 implies:

2u + 3v = 17

New set of equations is:

u + v = 7 ……(1)

2u + 3v = 17 ………(2)

From equation (1): a_1 = 1, b_1 = 1 and c_1 = – 7

From equation (2): a_2 = 2, b_2 = 3 and c_2 = – 17

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 9

Question 10:

rs aggarwal class 10 chapter 3 ex c img 10a

Solution:

Consider 1/(x + y) = a and 1/(x – y) = b

Given equations turn as:

5a – 2b + 1 = 0 ….(1)

15a + 7b – 10 = 0 …(2)

From equation (1): a_1 = 5, b_1 = -2 and c_1 = 1

From equation (2): a_2 = 15, b_2 = 7 and c_2 = – 10

Using cross multiplication method:

rs aggarwal class 10 chapter 3 ex c img 10b

So a = 13/65 = 1/5 and b = 65/65 = 1

Putting back values and write equations in the form of x and y, we get

x + y = 5 and x – y = 1

Again we get a set of two linear equations with two variables. Solve again to find the value of x and y.

Rearranging them again,

x + y – 5 = 0 …(3)

x – y – 1 = 0 …(4)

From equation (3): a_1 = 1, b_1 = 1 and c_1 = – 5

From equation (4): a_2 = 1, b_2 = – 1 and c_2 = – 1

Using cross multiplication method,

rs aggarwal class 10 chapter 3 ex c img 10c

Answer: x = 3 and y = 2

Question 11:

rs aggarwal class 10 chapter 3 ex c img 11a

From equation (2): a_2 = a, b_2 = – b and c_2 = – 2ab

Using cross multiplication method, we get

rs aggarwal class 10 chapter 3 ex c img 11b

Answer: x = b and y = -a

Question 12.

2ax + 3by = (a + 2b),

3ax + 2by = (2a + b).

Solution:

rs aggarwal class 10 chapter 3 ex c img 12

Question 13:

rs aggarwal class 10 chapter 3 ex c img 13


Exercise 3D Page No: 122

Show that each of the following systems of equations has a unique solution and solve it:

Question 1:

3x + 5y = 12, 5x + 3y = 4

Solution:

3x + 5y = 12 ……(1)

5x + 3y = 4 ……(2)

Here,

a_1 = 3, b_1 = 5, c_1 = 12

a_2 = 5, b_2 = 3, c_2 = 4

\(\frac{a_{1}}{a_{2}} = \frac{3}{5} , \frac{b_{1}}{b_{2}} = \frac{5}{3}\)

Which shows,

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

The system of equations have unique solution.

Solve the equations using substitution method:

From (1): x = (12-5y)/3

From (2): 5x + 3y – 4 = 0

5((12-5y)/3) + 3y – 4 = 0

60 – 25y + 9y = 12

60 – 16y = 12

16y = 48

y = 3

Now, substitute y in (1)

3x + 5(3) = 12

3x + 15 = 12

3x = 12 – 15

x = – 1

Answer: x = – 1 and y = 3

Question 2.

2x – 3y = 17, 4x + y = 13.

Solution:

2x – 3y = 17 ……..(1)

4x + y = 13 ………(2)

Here

a_1 = 2, b_1 = – 3, c_1 = 17

a_2 = 4, b_2 = 1, c_2 = 13

\(\frac{a_{1}}{a_{2}} = \frac{2}{4} , \frac{b_{1}}{b_{2}} = \frac{-3}{1} = -3 \)

Which shows:

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

System has unique solutions.

Solve the equations using substitution method:

From (1): x = (17 + 3y)/2

Putting value of x in (2)

4((17 + 3y)/2) + y = 13

68 + 12y + 2y = 26

68 + 14y = 26

y = -3

Again, x = (17 + 3(-3))/2

2x = 8

x = 4

Answer: x = 4 and y = – 3

Question 3:

x/2 + y/2 = 3, x – 2y = 2

Solution:

2x + 3y = 18 …(1)

x – 2y = 2 …(2)

a_1 = 2, b_1 = 3, c_1 = 18

a_2 = 1, b_2 = – 2, c_2 = 2

\(\frac{a_{1}}{a_{2}} = \frac{2}{1} = 2 , \frac{b_{1}}{b_{2}} = \frac{3}{-2} \) \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

This system has a unique solution.

Solve the equations:

From (2), x = 2 + 2y

Substituting the value of x in (1),

2(2 + 2y) + 3y = 18

=> 4 + 4y + 3y = 18

=> 7y = 18 – 4 = 14

=> y = 2

and x = 2 + 2 x 2 = 2 + 4 = 6

Answer: x = 6 and y = 2

Find the value of k for which each of the following systems of equations has a unique solution:

Question 4:

2x + 3y – 5 = 0, kx – 6y – 8 = 0.

Solution:

2x + 3y – 5 = 0 ………….(1)

kx – 6y – 8 = 0 ………..(2)

a_1 = 2, b_1 = 3, c_1 = – 5

a_2 = k, b_2 = – 6, c_2 = – 8

\(\frac{a_{1}}{a_{2}} = \frac{2}{k} , \frac{b_{1}}{b_{2}} = \frac{3}{-6} \)

Which shows:

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

System has a unique solution.

Now, Find the value of k:

2/k ≠ 3/-6

k ≠ -4

Question 5.

x – ky = 2, 3x + 2y + 5 = 0.

Solution:

x – ky = 2 ….(1)

3x + 2y + 5 = 0 ….(2)

Here,

a_1 = 1, b_1 = – k, c_1 = – 2

a_2 = 3, b_2 = 2, c_2 = 5

\(\frac{a_{1}}{a_{2}} = \frac{1}{3} = 2 , \frac{b_{1}}{b_{2}} = \frac{-k}{2} , \frac{c_{1}}{c_{2}} = \frac{2}{-5} \)

Systems has a unique solution.

Now,

\( \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{1}{3} \neq \frac{k}{2} \)

k ≠ 2/3

Question 6.

5x – 7y – 5 = 0, 2x + ky – 1 = 0.

Solution:

5x – 7y – 5 = 0 ….(1)

2x + ky – 1 = 0 …..(2)

Here,

a_1 = 5, b_1 = – 7, c_1 = – 5

a_2 = 2, b_2 = k, c_2 = – 1

\(\frac{a_{1}}{a_{2}} = \frac{5}{2} , \frac{b_{1}}{b_{2}} = \frac{-7}{+k} = \frac{-7}{k} \)

Systems has a unique solution.

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{5}{2} \neq \frac{-7}{k}\)

k ≠ -14/5

Question 7.

4x + ky + 8 = 0, x + y + 1 = 0.

Solution:

4x + ky + 8 = 0 ………(1)

x + y + 1 = 0 ………….(2)

Here,

a_1 = 4, b_1 = k, c_1 = 8

a_2 = 1, b_2 = 1, c_2 = 1

\(\frac{a_{1}}{a_{2}} = \frac{4}{1} , \frac{b_{1}}{b_{2}} = \frac{k}{1}\)

Systems has a unique solution.

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{4}{1} \neq \frac{k}{1}\)

k ≠ 4

Question 8.

4x – 5y = k, 2x – 3y = 12.

Solution:

4x – 5y = k …….(1)

2x – 3y = 12 …………(2)

Here,

a_1 = 4, b_1 = – 5, c_1 = – k

a_2 = 2, b_2 = – 3, c_2 = – 12

rs aggarwal class 10 chapter 3 ex d img 1a

Which shows that,

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

2 ≠ \(\frac{5}{7}\)

Therefore, k is any real number.

Question 9.

kx + 3y = (k – 3), 12x + ky = k.

Solution:

kx + 3y = (k – 3), 12x + ky = k

Here,

a_1 = k, b_1 = 3, c_1 = k – 3

a_2 = 12, b_2 = k, c_2 = k

\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) \Rightarrow \(\frac{k}{12} \neq \frac{3}{4}\)

Systems has a unique solution.

rs aggarwal class 10 chapter 3 ex d img 1

Question 10:

Show that the system of equations 2x – 3y = 5, 6x – 9y = 15 has an infinite number of solutions.

Solution:

2x – 3y = 5, 6x – 9y = 15

rs aggarwal class 10 chapter 3 ex d img 2

Question 11.

Show that the system of equations 6x + 5y = 11, 9x + 15 /2 y = 21 has no solution.

Solution:

6x + 5y = 11

9x + 15/2 y = 21

Here,

a_1 = 6, b_1 = 5, c_1 = – 11

a_2 = 9, b_2 = 15/2, c_2 = -21

rs aggarwal class 10 chapter 3 ex d img 3

Question 12: For what value of k does the system of equations kx + 2y = 5, 3x – 4y = 10 have

(i) a unique solution,

(ii) no solution?

Solution:

System of equations kx + 2y = 5 and 3x – 4y = 10

Here,

a_1 = k, b_1 = 2, c_1 = 5

a_2 = 3, b_2 = – 4, c_2 = 10

rs aggarwal class 10 chapter 3 ex d img 4

Question 13:

For what value of k does the system of equations x + 2y = 5, 3x + ky + 15 = 0 have

(i) a unique solution,

(ii) no solution?

Solution:

x + 2y = 5, 3x + ky + 15 = 0

Here,

a_1 = 1, b_1 = 2, c_1 = – 5

a_2 = 3, b_2 = k, c_2 = 15

rs aggarwal class 10 chapter 3 ex d img 5

Question 14.

For what value of k does the system of equations x + 2y = 3, 5x + ky + 7 = 0 have

(i) a unique solution,

(ii) no solution? Also, show that there is no value of k for which the given system of equations has infinitely many solutions.

Solution:

x + 2y = 3, 5x + ky + 7 = 0

Here

a_1 = 1, b_1 = 2, c_1 = – 5

a_2 = 3, b_2 = k, c_2 = 15

rs aggarwal class 10 chapter 3 ex d img 6


Exercise 3E Page No: 145

Question 1:

5 chairs and 4 tables together cost ₹ 5600, while 4 chairs and 3 tables together cost ₹ 4340. Find the cost of a chair and that of a table.

Solution:

Let us consider,

Cost of one chair = ₹ x and

Cost of one table = ₹ y

As per the statement,

5x + 4y = ₹ 5600 …(1)

4x + 3y = ₹ 4340 …(2)

Using elimination method:

Multiply (1) by 3 and (2) by 4. And subtract both the equations.

rs aggarwal class 10 chapter 3 ex e 1

So, x = 560

From (1):

5 x 560 + 4y = 5600

2800 + 4y = 5600

4y = 5600 – 2800

4y = 2800

y = 700

Answer:

Cost of one chair = ₹ 560

and cost of one table = ₹ 700

Question 2:

23 spoons and 17 forks together cost ₹ 1770, while 17 spoons and 23 forks together cost ₹ 1830. Find the cost of a spoon and that of a fork.

Solution:

Let us consider,

Cost of one spoon = ₹ x and

Cost of one fork = ₹ y

As per the statement,

23x + 17y = 1770 …(1)

17x + 23y = 1830 …(2)

Adding both the equations, we get

40x + 40y = 3600

Dividing by 40,

x + y = 90 …(3)

and subtracting (2) from (1)

6x – 6y = -60

Dividing by 6,

x – y = -10 …(4)

Now, Adding (3) and (4)

2x = 80

or x = 40

From (3): x + y = 90

40 + y = 90

y = 50

Therefore, Cost of one spoon = ₹ 40

and cost of one fork = ₹ 50

Question 3:

A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all totalling ₹ 19.50, how many coins of each kind does she have?

Solution:

Let us consider,

Number of 25-paisa coins = x and

Number 50-paisa coins = y

Total number of coins = 50

and total amount = ₹ 19.50 or 1950 paisa

x + y = 50 …(1)

25x + 50y = 1950

x + 2y = 78 …(2)

Subtracting (1) from (2),

y = 28

And, x = 50 – y = 50 – 28 = 22

Number of 25-paisa coins = 22

and 50-paisa coins = 28

Question 4:

The sum of two numbers is 137 and their difference is 43. Find the numbers.

Solution:

Given: Sum of two numbers is 137 and difference is 43

Let us consider, first number = x and

Second number = y

x + y = 137 …..(1)

x – y = 43 ……(2)

Adding both the equations, we get

2x = 180

or x = 90

On subtracting (2) from (1),

2y = 94

y = 47

So,

First number = 90 and

Second number = 47

Question 5:

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.

Solution:

Let us consider, first number = x and

Second number = y

As per the statement,

2x + 3y = 92 …(1)

4x – 7y = 2 …(2)

Using elimination method:

Multiply (1) by 2 and (2) by 1, we get

4x + 6y = 184 …..(3)

4x – 7y = 2 …….(4)

Subtracting (3) from (4),

13y = 182

y = 14

From (1), 2x + 3y = 92

2x + 3 x 14 = 92

2x + 42 = 92

2x = 92 – 42 = 50

x = 25

First number = 25

Second number = 14

Question 6:

Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.

Solution:

Let us consider,

First number = x and

Second number = y

As per the statement,

3x + y=142 …(1)

4x – y = 138 …(2)

Using elimination method:

Add (1) and (2), we get

7x = 280

x = 40

From (1): 3 x 40 + y = 142

120 + y = 142

y = 142 – 120 = 22

Answer:

First number = 40,

Second number = 22

Question 7:

If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the numbers.

Solution:

Let us consider,

First greater number = x and

Second smaller number = y

As per the statement,

2x – 45 = y …(1)

2y – 21 = x …(2)

Using Substitution method:

Substituting the value of y in (2),

2 (2x – 45) – 21 = x

4x – 90 – 21 = x

4x – x = 111

3x = 111

x = 37

From (1): y = 2 x 37 – 45 = 29

Answer:

The numbers are 37 and 29.

Question 8:

If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.

Solution:

Let us consider, larger number = x and

Smaller number = y

As per the statement,

3x = 4y + 8 …….(1)

5y = 3x + 5 …(2)

Using Substitution method:

Substitute the value of 3x in (2), we get

5y = 4y + 8 + 5

5y – 4y = 13

or y = 13

From (1): 3x = 4 x 13 + 8 = 60

x = 20

Answer:

Larger number = 20

Smaller number = 13

Question 9:

If 2 is added to each of two given numbers, their ratio becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.

Solution:

Let us consider,

First number = x and

Second number = y

As per the statement,

(x+2)/(y+2) = 1/2

or y + 2 = 2x + 4

or y = 2x + 2 ………….(1)

And

(x-4)/(y-4) = 5/11

or 11(x-4) = 5(y – 4)

11x – 44 = 5(2x + 2) – 20 (using result of equation (1) )

11x – 44 = 10x + 10 – 20

11x – 10x = 10 – 20 + 44

or x = 34

From (1): y = 2 x 34 + 2 = 68 + 2 = 70

Answer:

Numbers are 34 and 70

Question 10:

The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.

Solution:

Let us consider,

First number = x and

Second number = y

(Assume y is smaller)

As per the statement,

x – y = 14 …..(1)

and x² – y² = 448

(x + y) (x – y) = 448

(x + y) x 14 = 448

x + y = 32 ……(2)

Adding (1) and (2),

2x = 46 or x = 23

Subtracting (1) from (2)

2y = 18 or y = 9

Answer:

Numbers are 23 and 9

Question 11:

The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.

Solution:

Let us consider,

One’s digit of a two digit number = x and

Ten’s digit = y

So, the number is x + 10y

By interchanging the digits,

One’s digit = y and

Ten’s digit = x

Number is y + 10x

As per the statement,

x + y = 12 ………. (1)

y + 10x = x + 10y + 18

y + 10x – x – 10y = 18

x – y = 2 …(2)

Adding (1) and (2), we have

2x = 14 or x = 7

On subtracting (1) from (2),

2y = 10 or y = 5

Answer:

Number = 7 + 10 x 5 = 57

Question 12:

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.

Solution:

Let us consider, one’s digit of a two digit number = x and

ten’s digit = y

The number is x + 10y

After reversing the digits,

One’s digit = y

and ten’s digit = x

The number is y + 10x

As per the statement,

x + 10y – 27 = y + 10x

y + 10x – x – 10y = -27

9x – 9y = -27

x – y = -3 …(1)

Again,

7 (x + y) = x + 10y

7x + 7y = x+ 10y

7x – x = 10y – 7y

6x = 3y

2x = y …(2)

Using Substitution method:

Substituting the value of y in (1)

x – 2x = -3

-x = -3

or x = 3

From (2): y = 2(3) = 6

Answer:

Number = x + 10y = 3 + 10( 6) = 63

Question 13:

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.

Solution:

Let us consider, one’s digit of a two digit number = x and

ten’s digit = y

The number is x + 10y

After interchanging the digits,

One’s digit = y and

ten’s digit = x

The number is y + 10x

As per the statement,

x + y= 15 ………….(1)

And,

y + 10x = x + 10y + 9

y + 10x – x – 10y = 9

9x – 9y = 9

x – y = 1 …(2)

On adding, we get

2x = 16

Or x = 8

On subtracting (2) from (1)

2y = 14

y = 7

Answer:

Number = x + 10y = 8 + 10( 7) = 78

Question 14:

A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Solution:

Let us consider, one’s digit of the two digit number = x and

Ten’s digit = y

The number is x + 10y

By reversing the digits,

One’s digit = y and

ten’s digit = x

The number is y + 10x

Now, As per the statement,

4(x + y) + 3 = x + 10y

4x + 4y + 3 = x + 10y

3x – 6y = -3

x – 2y = -1 ……..(1)

And,

x + 10y + 18 = y + 10x

18 = y + 10x – x – 10y

9x – 9y = 18

x – y = 2 …(2)

On subtracting (1) form (2)

y = 3

Form (1): x = 2y – 1 = 2(3 ) – 1 = 5

Answer:

Number = x + 10y = 5 + 10 (3) = 35

Question 15:

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.

Solution:

Let us consider, ones digit of a two digit number = x and

Tens digit = y

The number is x + 10y

By reversing the digits,

One’s digit = y and ten’s digit = x

The number is y + 10x

As per the statement,

(x + 10y)/(x+y) = 6

x + 10y = 6(x + y)

5x = 4y

or x = 4/5 y …………(1)

And,

x + 10y – 9 = y + 10x

x + 10y – y – 10x = 9

-9x + 9y = 9

x – y = -1

using equation (1)

4/5 y – y = -1

y = 5

From (1): x = 4/5(5) = 4

Answer:

The number is: x + 10y = 4 + 10(5) = 54

Question 16:

A two-digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Solution:

Let us consider,

One’s digit of a two digit number = x and

Ten’s digit = y

The number is x + 10y

By interchanging the digits,

One’s digit = y and ten’s digit = x

The number is y + 10x

As per the statement,

xy = 35 ………(1)

And, x + 10y + 18 = y + 10x

18 = y + 10x – x – 10y

x – y = 2 …(2)

using algebraic identity: (x + y)² = (x – y)² + 4xy

(x + y)² = (2)2 + 4(35)

= 144

= (12)2

Or x + y = 12 …(3)

Subtracting (2) from (3), we get

2x =14

or x = 7

Form (1) : xy = 35

7(y) = 35

y = 5

Answer: The number is

x + 10y = 7 + 10(5) = 57

Question 17:

A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Solution:

Let us consider, one’s digit of a two digit number = x and

Ten’s digit = y

The number is x + 10y

After interchanging the digits One’s digit = y

Ten’s digit = x

The number is y + 10x

As per the statement,

xy = 18 ………(1)

And, x + 10y -63 = y + 10x

9y – 9x – y – 10x = 63

y – x = 7 …..(2)

using algebraic identity: (x + y)^2 = (x – y)^2 + 4xy

(x + y)^2 = (-7)^2 + 4(18) = 121 (Using (1))

(x + y)^2 = 11^2

or x + y = 11 …..(3)

Add (1) and (2)

2y = 18

or y = 9

From (1): xy = 18

9x = 18

x = 2

Answer:

The number is: x + 10y = 2 + 10(9) = 92

Question 18:

The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.

Solution:

Let us consider, one’s digit of a two digit number = x and

Ten’s digit = y

Number is x + 10y

By reversing the digits,

One’s digit = y and

Ten’s digit = x

Number is y + 10x

As per the statement,

x + 10y + y + 10x = 121

11x + 11y = 121

x + y = 11 …(1)

And,

x – y = 3 …(2)

On adding, we get

2x = 14 or x = 7

On subtracting (2) form (1),

2y = 8 or y = 4

Answer:

Number = 7 + 10( 4) = 47

or 4 + 10( 7) = 4 + 70 = 74

Question 19.

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes 3/4. Find the fraction.

Solution:

Let us consider, a fraction x/y, where x is numerator and y is denominator.

As per the statement,

x + y = 8 …(1)

And,

(x+3)/(y+3) = 3/4

4(x+3) = 3(y+3)

4x – 3y = -3 …..(2)

Using Substitution method:

Substitute the value of x from (1) in (2), we get

4(8 – y) – 3y = -3

32 – 4y – 3y = -3

-7y = -35

or y = 5

From (1): x + 5 = 8

or x = 3

Answer: Fraction is 3/5

Question 20:

If 2 is added to the numerator of a fraction, it reduces to 1/2 and if 1 is subtracted from the denominator, it reduces to 1/3 . Find the fraction.

Solution:

Let us consider, a fraction x/y, where x is numerator and y is denominator.

As per the statement,

(x + 2)/y = ½

x/y – 1 = 1/3

2x + 4 = y …(1)

And, 3x = y – 1 …(2)

Using Substitution method:

Using (1) in (2)

3x = 2x + 4 – 1

3x = 2x + 3

x = 3

Form (1): y = 2(3) + 4 = 10

Answer: Required fraction is 3/10


Exercise 3F Page No: 155

Question 1:

Write the number of solutions of the following pair of linear equations:

x + 2y – 8 = 0, 2x + 4y = 16.

Solution:

rs aggarwal class 10 chapter 3 ex f 1

Question 2:

Find the value of k for which the following pair of linear equations have infinitely many solutions:

2x + 3y = 7, (k – 1) x + (k + 2) y = 3k.

Solution:

2x + 3y = 7

(k – 1)x + (k + 2)y = 3k

These given equations are in the form:

a_1x + b_1y + c_1 = 0 and

a_2x + b_2y + c_2 = 0

where,

rs aggarwal class 10 chapter 3 ex f 2

Question 3:

For what value of k does the following pair of linear equations have infinitely many solutions?

10x + 5y – (k – 5) = 0 and 20x + 10y – k = 0.

Solution:

10x + 5y – (k – 5) = 0 and 20x + 10y – k = 0.

These given equations are in the form:

a_1x + b_1y + c_1 = 0 and

a_2x + b_2y + c_2 = 0

where,

rs aggarwal class 10 chapter 3 ex f 3

Question 4:

For what value of k will the following pair of linear equations have no solution?

2x + 3y = 9, 6x + (k – 2) y = (3k – 2).

Solution:

2x + 3y = 9, 6x + (k – 2) y = (3k – 2).

These given equations are in the form:

a_1x + b_1y + c_1 = 0 and

a_2x + b_2y + c_2 = 0

where,

rs aggarwal class 10 chapter 3 ex f 4

Question 5:

Write the number of solutions of the following pair of linear equations:

x + 3y – 4 = 0 and 2x + 6y – 7 = 0.

Solution:

x + 3y – 4 = 0 and 2x + 6y – 7 = 0.

These given equations are in the form:

a_1x + b_1y + c_1 = 0 and

a_2x + b_2y + c_2 = 0

where,

rs aggarwal class 10 chapter 3 ex f 5

Question 6:

Write the value of k for which the system of equations 3x + ky = 0, 2x – y = 0 has a unique solution.

Solution:

rs aggarwal class 10 chapter 3 ex f 6

Question 7:

The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.

Solution:

Let x be the first number and y be second number.

x – y = 5

x^2 – y^2 = 65 …(2)

Now, by dividing (2) by (1) we get:

x + y = 13 …(3)

On adding (1) and (2) we get

2x = 18

or x = 9

From (3): 9 + y = 13

or y = 4

Two numbers are 4 and 9.

Question 8:

The cost of 5 pens and 8 pencils is f 120, while the cost of 8 pens and 5 pencils is ₹ 153. Find the cost of 1 pen and that of 1 pencil.

Solution:

Let the cost of one pen is ₹ x and cost of one pencil is ₹ y, then

As per statement,

5x + 8y = 120 …(1)

8x + 5y = 153 …(2)

Adding both the equations, we get

13x + 13y = 273

x + y = 21 …(3)

On subtracting (1) from (2),

3x – 3y = 33

x – y = 11 …….(iv)

Again,

Adding (3) and (iv),

2x = 32 or x = 16

On subtracting,

2y = 10 or y = 5

Answer:

Cost of 1 pen = ₹ 16

and cost of 1 pencil = ₹ 5

Question 9:

The sum of two numbers is 80. The larger number exceeds four times the smaller one by 5. Find the numbers.

Solution:

Let x be the first number and y be the second number.

As per statement,

x + y = 80 and

x – 4y = 5

on subtracting both the equations, we get

y = 15

From (1): x + 15 = 80

x = 80 – 15 = 65

Answer:

Required numbers are 15 and 65.

Question 10:

A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its digits are reversed. Find the number.

Solution:

Let one’s digit of a two digits number is x and ten’s digit is y, then the number is x + 10y

By reversing its digits One’s digit = y and ten’s digit = x

Then the number is y + 10x

As per statement,

x + y = 10 …(1)

And,

x + 10y – 18 = y + 10x

x+ 10y – y – 10x = 18

-9x + 9y = 18

x – y = -2 (Dividing by -9) …..(2)

Adding (1) and (2), we have

2x = 8 or x = 4

From (1): 4 + y = 10

y = 6

Answer:

Number = x + 10y = 4 + 10( 6) = 64


RS Aggarwal Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables

In this chapter students will study important concepts on linear equations in two variables as listed below:

  • Solution of a linear equation
  • Simultaneous linear equations in two variables
  • Solution of a given system of two Simultaneous equations
  • Consistent and Inconsistent systems of linear equations
  • Solving simultaneous linear equations using the graphical method
  • Solving simultaneous linear equations using algebraic methods (substitution method, elimination method,
  • Method of cross multiplication
  • Conditions for the solvability of linear equations (Consistent and Inconsistent systems of linear equations/conditions for the solvability of linear equations_
  • Word problems on linear equations

Key Features of RS Aggarwal Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables

1. RS Aggarwal Solutions consist of a set of solutions to all the important questions.

2. Linear Equations in Two Variables study material prepared based on the latest CBSE syllabus by subject experts.

3. Easy for quick revision.

4. Step by step solving approach helps students to clear their basics.

5. Helps students to solve complex problems at their own pace.