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

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

Very-Short and Short-Answer Questions
Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 1
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 2

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 3
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 4

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 5

Question 4.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 6
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 7

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 8

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 9

Question 7.
Solution:
Let first, number = x
and second number = y
x – y = 5
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 10

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,
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 11
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 12
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 13

Question 14.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 14
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 15

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 16

Question 17.
Solution:
kx – y = 2
6x – 2y = 3
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 17

Question 18.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 18

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 19
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 20

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 21

Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 22
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F 23

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

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

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

Solve each of the following systems of equations by using the method of cross multiplication:
Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 1
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 2

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 3

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 4

Question 4.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 5

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 6

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 7
x = 15, y= 5

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 8

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 9

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 10

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 11
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 12

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 13
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 14

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 15
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 16

Question 13.
Solution:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C 17

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

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

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 1
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 2
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 3

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 4
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 = \(\frac { 3y+ 7 }{ 2 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 5
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 6

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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 7
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 8
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 9

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 10
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 11
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 12

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 13
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 = \(\frac { 4 + 2 y }{ 5 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 14
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 15

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 16
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 = \(\frac { 3y – 8 }{ 2 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 17
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 18

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 19
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 = \(\frac { 2y + 12 }{ 3 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 20
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 21

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 22
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 = \(\frac { 2y – 12 }{ 3 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 23
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 24

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 25
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 26
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 27

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 28
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 = \(\frac { 2 – 2y }{ 3 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 29
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 30

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 31
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 = \(\frac { 4 – 3y }{ 2 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 32
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).
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 33
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 34
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 35
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 36
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 37
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 = \(\frac { 20 – 3y }{ 4 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 38
Plot the points (5, 0), (2, 4) and (-1, 8) on the graph and join them to get a line.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 39
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 40
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 = \(\frac { 12 – 2y }{ 3 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 41
Plot the points (4, 0), (2, 3) and (0, 6) on the graph and join them to get another line.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 42
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 43
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 44
Plot the points on the graph and join them to get a line.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 45
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 46
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 = \(\frac { 18 – 3y }{ 2 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 47
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).
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 48
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 49
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 = \(\frac { 14 – 3x }{ 2 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 50
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 51
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 52
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 = \(\frac { 15 – 5y }{ 3 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 53
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 54
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 55
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 = \(\frac { 8 – y }{ 2 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 56
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 57
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 58
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 59
Plot the points (-1, 0), (1, 2) and (2, 3) on the graph and join them to get a line.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 60
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 61
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 62
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 63
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 64
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 = \(\frac { 12 – 6y }{ 2 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 65
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 66

Question 23.
Solution:
3x – y = 5
⇒ y = 3x – 5
Giving some different values to x, we get corresponding values of y as shown below:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 67
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 = \(\frac { 10 + 2y }{ 6 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 68
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 69

Question 24.
Solution:
2x + y = 6
y = 6 – 2x
Giving some different values to x, we get corresponding values of y as given below:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 70
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 = \(\frac { 18 – 3y }{ 2 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 71
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 72

Question 25.
Solution:
x – 2y = 5 ⇒ x = 5 + 2y
Giving some different values to y, we get corresponding values of x as shown below:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 73
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 = \(\frac { 15 + 6y }{ 3 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 74
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 75

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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 76
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 77
Plot the points (0, 0), (2, 1), (4, 2) on the graph and join them to get a line.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 78
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:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 79
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 = \(\frac { 12 – 6y }{ 4 }\)
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 80
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.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 81

Question 28.
Solution:
2x + y = 6, y = 6 – 2x
Giving some different values to x, we get corresponding values ofy as given below:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 82
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 83
We see that these two lines are parallel and do not intersect each other.
Therefore this system has no solution.
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 84

Question 29.
Solution:
2x + y = 2 ⇒ y = 2 – 2x
Giving some different values to x, we get corresponding values of y as given below:
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 85
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
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 86
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).
RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A 87
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

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