Maths Class 10 RS Aggarwal Solutions

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

MCQ
Question 1.
Solution:
(a)

Question 2.
Solution:
(d)

Question 3.
Solution:
(a)

Question 4.
Solution:
(d)

Short-Answer Questions
Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:
Let first number = x
and second number = y
According to the conditions, x – y = 26 …(i)
and x = 3y …..(ii)
From (i),
3y – y = 26
⇒ 2y = 26
⇒ y = 13
and x = 3 x 13 = 39
Numbers are 39 and 13

Short-Answer Questions (3 marks)
Question 11.
Solution:
23x + 29y = 98 …..(i)
29x + 23y = 110 …..(ii)
Adding, we get 52x + 52y = 208
x + y = 4 …..(iii) (Dividing by 52)
and subtracting,
-6x + 6y = -12
x – y = 2. …..(iv) (Dividing by -6)
Adding (iii) and (iv),
2x = 6 ⇒ x = 3
Subtracting,
2x = 2 ⇒ y = 1
Hence, x = 3, y = 1

Question 12.
Solution:


x = 1, y = \(\frac { 3 }{ 2 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to the condition,
5x + 7y = 195 …(i)
7x + 5y= 153 …(ii)
Adding, (i) and (ii)
12x + 12y = 348
x + y = 29 ….(iii) (Dividing by 12)
and subtracting,
-2x + 2y = 42
-x + y = 21 …..(iv) (Dividing by -2)
Now, Adding (iii) and (iv),
2y = 50 ⇒ y = 25
and from (iv),
-x + 25 = 21 ⇒ -x = 21 – 25 = -4
x = 4
Cost of one pencil = ₹ 4
and cost of one pen = ₹ 25

Question 16.
Solution:
2x – 3y = 1, 4x – 3y + 1 = 0
2x – 3y = 1
2x = 1 + 3y
x = \(\frac { 1 + 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below

Now plot the points (2, 1), (5, 3) and (-1, -1) on the graph and join them to get a line.
Similarly,
4x – 3y + 1 = 0
⇒ 4x = 3y – 1
⇒ x = \(\frac { 3y – 1 }{ 4 }\)

Now plot the points (-1, -1), (-4, -5) and (2, 3) on the graph and join them to get another line which intersects the first line at the point (-1, -1).
Hence, x = -1, y = -1

Long-Answer Questions
Question 17.
Solution:
We know that opposite angles of a cyclic quadrilateral are supplementary.
∠A + ∠C = 180° and ∠B + ∠D = 180°
Now, ∠A = 4x° + 20°, ∠B = 3x° – 5°, ∠C = 4y° and ∠D = 7y° + 5°
But ∠A + ∠C = 180°
4x + 20° + 4y° = 180°
⇒ 4x + 4y = 180° – 20 = 160°
x + y = 40° …(i) (Dividing by 4)
and ∠B + ∠D = 180°
⇒ 3x – 5 + 7y + 5 = 180°
⇒ 3x + 7y = 180° …(ii)
From (i), x = 40° – y
Substituting the value of x in (ii),
3(40° – y) + 7y = 180°
⇒ 120° – 3y + 7y = 180°
⇒ 4y = 180°- 120° = 60°
y = 15°
and x = 40° – y = 40° – 15° = 25°
∠A = 4x + 20 = 4 x 25 + 20 = 100 + 20= 120°
∠B = 3x – 5 = 3 x 25 – 5 = 75 – 5 = 70°
∠C = 4y = 4 x 15 = 60°
∠D = 7y + 5 = 7 x 15 + 5 = 105 + 5 = 110°

Question 18.
Solution:

Question 19.
Solution:
Let numerator of a fraction = x
and denominator = y
Fraction = \(\frac { x }{ y }\)
According to the conditions,

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

Choose the correct answer in each of the following questions.
Question 1.
Solution:
(c)

Question 2.
Solution:
(c)

Question 3.
Solution:
(a)

Question 4.
Solution:
(d)

Question 5.
Solution:
(a)

Question 6.
Solution:
(b)

Question 7.
Solution:
(c) 4x + 6y = 3xy, 8x + 9y = 5xy
Dividing each term by xy,

Question 8.
Solution:
(a)

Question 9.
Solution:
(c)

Question 10.
Solution:
(b)

Question 11.
Solution:
(d)

Question 12.
Solution:
(b)

Question 13.
Solution:
(a)

Question 14.
Solution:
(d)

Question 15.
Solution:
(d)

Question 16.
Solution:
(d)

Question 17.
Solution:
(d)

Question 18.
Solution:
(d) The system of equations is consistent then their graph lines will be either intersecting or coincident.

Question 19.
Solution:
(a) The pair of lines of equation is inconsistent then the system will not have no solution i.e., their lines will be parallel.

Question 20.
Solution:
(b)

Question 21.
Solution:
(b) ABCD is a cyclic quadrilateral
∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°
∠A + ∠C = 180°
Now, x + y + 10°+ x + y – 30° = 180°
⇒ 2x + 2y – 20 = 180°
⇒ 2x + 2y = 180° + 20° = 200°
⇒ x + y = 100° …(i)
and ∠B + ∠D = 180°
⇒ y + 20° + x + y = 180°
⇒ x + 2y = 180° – 20° = 160° …(ii)
Subtracting,
-y = -60° ⇒ y = 60°
and x + 60° = 100°
⇒ x = 100° – 60° = 40°
Now, ∠B = y + 20° = 60° + 20° = 80°

Question 22.
Solution:
(d) Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + y = 15 …(i)
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(ii)
Adding (i) and (ii),
2x = 16 ⇒ x = 8
and x + y = 15
⇒ 8 + y = 15
⇒ y = 15 – 8 = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

Question 23.
Solution:
(b) Let the numerator of a fractions = x
and denominator = y

Question 24.
Solution:
(d) Let present age of man = x years
and age of his son = y years
5 years hence,
Age of man = (x + 5) years
and age of son = y + 5 years
(x + 5) = 3 (y + 5)
⇒ x + 5 = 3y + 15
x = 3y + 15 – 5
x = 3y + 10 ……(i)
and 5 years earlier
Age of man = x – 5 years
and age of son = y – 5 years
x – 5 = 7 (y – 5)
x – 5 = 7y – 35
⇒ x = 7y – 35 + 5
x = 7y – 30 ……….(ii)
From (i) and (ii),
7y – 30 = 3y + 10
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
y = 10
x = 3y + 10 = 3 x 10 + 10 = 30 + 10 = 40
Present age of father = 40 years

Question 25.
Solution:
(b)

Question 26.
Solution:
(c)

Question 27.
Solution:
(a)

The system has infinitely many solutions.
The lines are coincident.

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

Very-Short and Short-Answer Questions
Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Let first, number = x
and second number = y
x – y = 5

Question 8.
Solution:
Let cost of one pen = ₹ x
and cost of one pencil = ₹ y
According to the conditions,
5x + 8y = 120 …(i)
8x + 5y = 153 …(ii)
Adding, we get
13x + 13y = 273
x + y = 21 …(iii) (Dividing by 13)
and subtracting (i) from (ii),
3x – 3y = 33
⇒ x – y = 11 …….(iv) (Dividing by 3)
Again adding (iii) and (iv),
2x = 32 ⇒ x = 16
Subtracting,
2y = 10 ⇒ y = 5
Cost of 1 pen = ₹ 16
and cost of 1 pencil = ₹ 5

Question 9.
Solution:
Let first number = x
and second number = y
According to the conditions,

and x + y = 80
⇒ x + 15 = 80
x = 80 – 15 = 65
Numbers are : 65, 15

Question 10.
Solution:
Let one’s digit of a two digits number = x
and ten’s digit = y
Number = x + 10y
By reversing its digits One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + y = 10 …(i)
x + 10y – 18 = y + 10x
x+ 10y – y – 10x = 18
⇒ -9x + 9y = 18
⇒ x – y = -2 (Dividing by -9) …..(ii)
Adding (i) and (ii),
2x = 8 ⇒ x = 4
and by subtracting,
2y = 12 ⇒ y = 6
Number = x + 10y = 4 + 10 x 6 = 4 + 60 = 64

Question 11.
Solution:
Let number of stamps of 20p = x
and stamps of 25 p = y
According to the conditions,
x + y = 47 …..(i)
20x + 25y = 1000
4x + 5y = 200 …(ii)
From (i), x = 47 – y
Substituting the value of x in (ii),
4 (47 – y) + 5y = 200
188 – 4y + 5y = 200
⇒ y = 200 – 188 = 12
and x + y = 47
⇒ x + 12 = 47
⇒ x = 47 – 12 = 35
Hence, number of stamps of 20 p = 35
and number of stamps of 25 p = 12

Question 12.
Solution:
Let number of hens = x
and number of cows = y
According to the conditions,
x + y = 48 …..(i)
x x 2 + y x 4 = 140
⇒ 2x + 4y = 140
⇒ x + 2y = 70 ……(ii)
Subtracting (i) from (ii),
y = 22
and x + y = 48
⇒ x + 22 = 48
⇒ x = 48 – 22 = 26
Number of hens = 26
and number of cows = 22

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
12x + 17y = 53 …(i)
17x + 12y = 63 …(ii)
Adding, 29x + 29y = 116
Dividing by 29,
x + y = 4 …(iii)
Subtracting,
-5x + 5y = -10
⇒ x – y = 2 …(iv) (Dividing by -5)
Adding (iii) and (iv)
2x = 6 ⇒ x = 3
Subtracting,
2y = 2 ⇒ y = 1
x = 3, y = 1
x + y = 3 + 1 = 4

Question 16.
Solution:

Question 17.
Solution:
kx – y = 2
6x – 2y = 3

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

Solve each of the following systems of equations by using the method of cross multiplication:
Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

x = 15, y= 5

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

Solve each of the following systems of equations graphically.
Question 1.
Solution:
2x + 3y = 2 …..(i)
x – 2y = 8 …(ii)
From Eq. (i),
⇒ 2x = 2 – 3y
⇒x = \(\frac { 2 – 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (1, 0), (-2, 2) and (4, -2) on the graph and join them to get a line.
Similarly x – 2y = 8 ⇒ x = 8 + 2y

Now plot the points (6, -1), (4, -2) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at point(4, -2).
x = 4, y = -2

Question 2.
Solution:
3x + 2y = 4
⇒ 3x = 4 – 2y
⇒ x = \(\frac { 4 – 2y }{ 3 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now plot the points (0, 2), (2, -1) and (4, -4) on the graph and join them to get a line. Similarly,
2x – 3y = 7
2x = 3y+ 7
x = \(\frac { 3y+ 7 }{ 2 }\)

Now plot the points on the graph and join them to get another line.
We see that these two lines intersect each other at point (2, -1).
x = 2, y = -1

Question 3.
Solution:
2x + 3y = 8
⇒ 2x = 8 – 3y
x = \(\frac { 8 – 3y }{ 2 }\)
Now, giving some different values to y, we get corresponding values of x as given below

Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
x – 2y + 3 = 0
x = 2y – 3

Now, plot the points (-3, 0), (-1, 1) and (1, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (1, 2).
x = 1, y = 2

Question 4.
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x = \(\frac { 5y – 4 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-2, 0) (3, 2) and (8,4) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
⇒ y = 8 – 2x

Plot the points (1, 6), (2, 1) and (3, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (3, 2).
x = 3, y = 2

Question 5.
Solution:
3x + 2y = 12
⇒ 3x = 12 – 2y
x = \(\frac { 12 – 2y }{ 2 }\)
Giving some different values to y, we get corresponding the values of x as given below:

Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
5x – 2y = 4
⇒ 5x = 4 + 2 y
⇒ x = \(\frac { 4 + 2 y }{ 5 }\)

Plot the points (0, -2), (2, 3) and (4, 8) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, 3).
x = 2, y = 3

Question 6.
Solution:
3x + y + 1 = 0 ⇒y = -3x – 1
Giving some different values to x, we get corresponding values of y as given below:

Now, plot the points (0, -1), (-1, 2) and (-2, 5) on the graph and join them to get a line.
Similarly,
2x – 3y + 8 = 0
2x = 3y – 8
x = \(\frac { 3y – 8 }{ 2 }\)

Now, plot the points (-4, 0), (-1, 2) and (2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the points (-1, 2).
x = -1, y = 2

Question 7.
Solution:
2x + 3y + 5 = 0
2x = -3y – 5
x = \(\frac { -3y – 5 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-4, 1), (-1, -1) and (2, -3) on the graph and join them to get a line. Similarly,
3x – 2y – 12 = 0
⇒ 3x = 2y + 12
x = \(\frac { 2y + 12 }{ 3 }\)

Plot the points (4, 0), (0, -6) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -3).
x = 2, y = -3

Question 8.
Solution:
2x – 3y + 13 = 0
⇒ 2x = 3y – 13
⇒ x = \(\frac { 3y – 13 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-5, 1), (-2, 3) and (1, 5) on the graph and join them to get a line.
Similarly,
3x – 2y + 12 = 0
3x = 2y – 12
x = \(\frac { 2y – 12 }{ 3 }\)

Now, plot the points (-4, 0), (-2, 3) and (0, 6) on the graph and join them to get another line.
We see that these lines intersect each other at the point (-2, 3).
x = -2, y = 3

Question 9.
Solution:
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
⇒ x = \(\frac { 4 – 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (2, 0) (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
3x – y + 5 = 0
⇒ -y = -5 – 3x
⇒ y = 5 + 3x

Plot the points (0, 5), (-1, 2) and (-2, -1) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (-1, 2).
x = -1, y = 2

Question 10.
Solution:
x + 2y + 2 = 0
⇒ x = – (2y + 2)
Giving some different values to y, we get corresponding the values of x as given below:

Plot the points (-2,0), (0, -1) and (2, -2) on the graph and join them to get a line.
Similarly,
3x + 2y – 2 = 0
⇒ 3x = 2 – 2y
x = \(\frac { 2 – 2y }{ 3 }\)

Plot the points (0, 1), (2, -2) and (-2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -2).
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.
Solution:
x – y + 3 = 0
⇒ x = y – 3
Giving some different value to y, we get corresponding values of x as given below:

Plot the points (-3, 0), (-1, 2) and (0, 3) on the graph and join them to get a line.
Similarly,
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
x = \(\frac { 4 – 3y }{ 2 }\)

Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get another line.
We see that these line intersect each other at (-1, 2) and x-axis at A (-3, 0) and D (2, 0).

Area of ∆BAD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AD x BL
= \(\frac { 1 }{ 2 }\) x 5 x 2 = 5 sq.units

Question 12.
Solution:
2x – 3y + 4 = 0
⇒ 2x = 3y – 4
x = \(\frac { 3y – 4 }{ 2 }\)
Giving some different values to y, we get corresponding value of x as given below:

Plot the points (-2, 0), (1, 2) and (4, 4) on the graph and join them to get a line.
Similarly,
x + 2y – 5 = 0
x = 5 – 2y

Now, plot the points (5, 0), (3, 1) and (1, 2) on the graph and join them to get another line.
We see that there two lines intersect each other at the point B (1, 2) and intersect x- axis at A (-2, 0) and D (5, 0) respectively.

Now, area of ∆BAD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AD x BL
= \(\frac { 1 }{ 2 }\) x 7 x 2
= 7 sq. units

Question 13.
Solution:
4x – 3y + 4 = 0
⇒ 4x = 3y – 4
x = \(\frac { 3y – 4 }{ 4 }\)
Giving some different values to y, we get corresponding value of x as given below:

Plot the points (-1, 0), (2, 4) and (-4, -4) on the graph and join them, to get a line.
Similarly,
4x + 3y – 20 = 0
4x = 20 – 3y
x = \(\frac { 20 – 3y }{ 4 }\)

Plot the points (5, 0), (2, 4) and (-1, 8) on the graph and join them to get a line.

We see that these two lines intersect each other at the point B (2, 4) and intersect x-axis at A (-1, 0) and D (5, 0).
Area ∆BAD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AD x BL
= \(\frac { 1 }{ 2 }\) x 6 x 4 = 12 sq. units

Question 14.
Solution:
x – y + 1 = 0 ⇒ x = y – 1
Giving some different values toy, we get the values of x as given below:

Plot the points (-1, 0), (0, 1) and (1, 2) on the graph and join them to get a line.
Similarly,
3x + 2y – 12 = 0
⇒ 3x = 12 – 2y
x = \(\frac { 12 – 2y }{ 3 }\)

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

We see that these two lines intersect each ohter at the point E (2, 3) and intersect x- axis at A (-1, 0) and D (4, 0).
Area of ∆EAD = \(\frac { 1 }{ 2 }\) x base x altitude 1
= \(\frac { 1 }{ 2 }\) x AD x EL
= \(\frac { 1 }{ 2 }\) x 5 x 3
= \(\frac { 15 }{ 2 }\)
= 7.5 sq. units

Question 15.
Solution:
x – 2y + 2 = 0 ⇒ x = 2y – 2
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (-2, 0), (0, 1), (2, 2) on the graph and join them to get a line.
Similarly,
2x + y – 6 = 0
⇒ y = 6 – 2x

Plot the points on the graph and join them to get a line.

We see that these two lines intersect each other at the point C (2, 2) and intersect the x-axis at A (-2, 0) and F (3, 0).
Now, area of ∆CAF = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AF x CL
= \(\frac { 1 }{ 2 }\) 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.
Solution:
2x – 3y + 6 = 0
⇒ 2x = 3y – 6
x = \(\frac { 3y – 6 }{ 2 }\)
Giving some different values to y, we get corresponding values of x, as given below:

Now, plot the points (-3, 0), (0, 2), (3, 4) on the graph and join them to get a line.
Similarly,
2x + 3y – 18 = 0
2x = 18 – 3y
x = \(\frac { 18 – 3y }{ 2 }\)

Plot the points (0, 6), (3, 4) and (6, 2) on the graph and join them to get a line.
We see that these two lines intersect each other at C (3, 4) and intersect y-axis at B (0, 2) and D (0, 6).

Now, area of ∆CBD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x BD x CL
= \(\frac { 1 }{ 2 }\) x 4 x 3 sq. units = 6 sq. units

Question 17.
Solution:
4x – y – 4 = 0
⇒ 4x = y + 4
x = \(\frac { y + 4 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (1, 0), (0, -4) and (2, 4) on the graph and join them to get a line.
Similarly,
3x + 2y – 14 = 0
2y = 14 – 3x
y = \(\frac { 14 – 3x }{ 2 }\)

Now, plot the points (0, 7), (2, 4) and (4, 1) on the graph and join them to get another line.
We see that these two lines intersect each other at C (2, 4) and y-axis at B (0, -4) and D (0, 7) respectively.

Area of ∆CBD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x BD x CL
= \(\frac { 1 }{ 2 }\) x 11 x 2 sq. units = 11 sq. units

Question 18.
Solution:
x – y – 5 = 0 ⇒ x = y + 5
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (5, 0), (0, -5) and (1, -4) on the graph and join these to get a line.
Similarly,
3x + 5y – 15 = 0
3x = 15 – 5y
x = \(\frac { 15 – 5y }{ 3 }\)

Plot the points (5, 0), (0, 3) and (-5, 6) on the graph and join them to get a line.
We see that these two lines intersect each other at A (5, 0) and y-axis at B (0, -5) and E (0, 3) respectively.

Now area of ∆ABE = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x BE x AL
= \(\frac { 1 }{ 2 }\) x 8 x 5 sq. units = 20 sq. units

Question 19.
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x = \(\frac { 5y – 4 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (-2, 0), (3, 2) and (-7, -2) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
2x = 8 – y
x = \(\frac { 8 – y }{ 2 }\)

Plot the points (4, 0), (3, 2) and (2, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at B (3, 2) and y-axis at G (0, 1) and H (0, 8) respectively.

Now area of ∆EGH = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x GH x EL
= \(\frac { 1 }{ 2 }\) x 7 x 3 sq. units
= \(\frac { 21 }{ 2 }\) = 10.5 sq. units

Question 20.
Solution:
5x – y – 7 = 0 ⇒ y = 5x – 7
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, -7), (1, -2), (2, 3) on the graph and join them to get a line.
Similarly,
x – y + 1 = 0 ⇒ x = y – 1

Plot the points (-1, 0), (1, 2) and (2, 3) on the graph and join them to get a line.

We see that these two lines intersect each other at C (2, 3) and intersect y-axis at A (0, -7) and F (0, 1) respectively.
Now, area of ∆CAF = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AF x CL
= \(\frac { 1 }{ 2 }\) x 8 x 2 = 8 sq.units

Question 21.
Solution:
2x – 3y – 12 ⇒ 2x = 12 + 3y
x = \(\frac { 12 + 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (6, 0), (3, -2) and (0, -4) on the graph and join them to get a line.
Similarly,
x + 3y = 6 ⇒ x = 6 – 3y

Plot the points (6, 0), (0, 2) and (-6, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at the points A (6, 0) and intersect the y-axis at C (0, -4) and E (0, 2) respectively.

Now area of ∆ACE = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x CE x AO
= \(\frac { 1 }{ 2 }\) x 6 x 6 sq. units = 18 sq.units

Show graphically that each of the following given systems of equations has infinitely many solutions:
Question 22.
Solution:
2x + 3y = 6
2x = 6 – 3y
x = \(\frac { 6 – 3y }{ 2 }\)
Giving some different values to y, we get the corresponding val ues of x as given below:

Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get a line.
Similarly,
4x + 6y = 12
4x = 12 – 6y
x = \(\frac { 12 – 6y }{ 2 }\)

on the graph and join them to get a line.
We see that all the points lie on the same straight line.
This system has infinite many solutions.

Question 23.
Solution:
3x – y = 5
⇒ y = 3x – 5
Giving some different values to x, we get corresponding values of y as shown below:

Now, plot the points (0, -5), (1, -2), (2, 1) on the graph and join them to get a line:
Similarly,
6x – 2y = 10
⇒ 6x = 10 + 2y
x = \(\frac { 10 + 2y }{ 6 }\)

Plot the points (2, 1), (4, 7), (3, 4) on the graph and join them to get another line.
We see that these lines coincide each other.
This system has infinite many solutions.

Question 24.
Solution:
2x + y = 6
y = 6 – 2x
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, 6), (2, 2), (4, -2) on the graph and join them to get a line.
Similarly,
6x + 3y = 18
⇒ 6x = 18 – 3y
⇒ x = \(\frac { 18 – 3y }{ 2 }\)

Now, plot the points (3, 0), (1, 4) and (5, -4) on the graph and join them to get another line.
We see that these two lines coincide each other.
This system has infinitely many solutions

Question 25.
Solution:
x – 2y = 5 ⇒ x = 5 + 2y
Giving some different values to y, we get corresponding values of x as shown below:

Plot the points (5, 0), (3, -1), (1, -2) on the graph and join them to get a line.
similarly,
3x – 6y = 15
⇒ 3x = 15 + 6y
x = \(\frac { 15 + 6y }{ 3 }\)

Plot the points (7, 1), (-1, -3) and (-3, -4) on the graph and join them to get another line.
We see that all the points lie on the same line.
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.
Solution:
x – 2y = 6 ⇒ x = 6 + 2y
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (6, 0), (4, -1) and (0, -3) on the graph and join them to get a line.
Similarly,
3x – 6y = 0
⇒ 3x = 6y
⇒ x = 2y

Plot the points (0, 0), (2, 1), (4, 2) on the graph and join them to get a line.

We see that these two lines are parallel i.e., do not intersect each other.
This system has no solution.

Question 27.
Solution:
2x + 3y = 4
⇒ 2x = 4 – 3y
⇒ x = \(\frac { 4 – 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
4x + 6y = 12 ⇒ 4x = 12 – 6y
x = \(\frac { 12 – 6y }{ 4 }\)

Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get another line.
We see that two lines are parallel i.e., these do not intersect each other at any point.
Therefore the system has no solution.

Question 28.
Solution:
2x + y = 6, y = 6 – 2x
Giving some different values to x, we get corresponding values ofy as given below:

Plot the points (0, 6), (1, 4) and (3, 0) on the graph and join them to get a line.
Similarly,
6x + 3y = 20 ⇒ 6x = 20 – 3y

We see that these two lines are parallel and do not intersect each other.
Therefore this system has no solution.

Question 29.
Solution:
2x + y = 2 ⇒ y = 2 – 2x
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, 2), (1, 0) and (-2, 6) on the graph and join them to get a line.
Similarly,
2x + y = 6 ⇒ y = 6 – 2x

Plot the points (0, 6), (2, 2) and (3, 0) on the graph and join them to get another line.
ABFD is the trapezium whose vertices are A (0, 2), B (1, 0), F (3, 0), D (0, 6).

Area of trapezium ABFD = Area ∆DOF – Area ∆AOB
= \(\frac { 1 }{ 2 }\) (DO x OF) – \(\frac { 1 }{ 2 }\) (AO x OB)
= \(\frac { 1 }{ 2 }\) (6 x 3) – \(\frac { 1 }{ 2 }\) (2 x 1) sq.units
= 9 – 1 = 8 sq.units

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.