CBSE Class 7

RS Aggarwal Class 7 Solutions Chapter 7 Linear Equations in One Variable Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7B
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7C
• RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable CCE Test Paper

Solve the following equations. Check your result in each case.
Question 1.
Solution:
3x – 5 = 0
Adding 5 to both sides
3x – 5 + 5 = 0 + 5
⇒ 3x = 5
⇒ x = \(\frac { 5 }{ 3 }\)
Check:
L.H.S. = 3x – 5
= 3 x \(\frac { 5 }{ 3 }\) – 5
= 5 – 5
= 0
= R.H.S.
Hence x = \(\frac { 5 }{ 3 }\)

Question 2.
Solution:
8x – 3 = 9 – 2x
⇒ 8x + 2x = 9 + 3 (By transposing)
⇒ 10x = 12

Question 3.
Solution:
7 – 5x = 5 – 7x
⇒ – 5x + 7x = 5 – 7 (By transposing)
⇒ 2x = -2
x = -1
Check:
L.H.S. = 7 – 5x = 7 – 5(-1) = 7 + 5 = 12
R.H.S. = 5 – 7x = 5 – 7(-1) = 5 + 7 = 12
L.H.S. = R.H.S.
Hence x = -1

Question 4.
Solution:
3 + 2x = 1 – x
⇒ 2x + x = 1 – 3 (By transposing)
⇒ 3x = -2

Question 5.
Solution:
2(x – 2) + 3(4x – 1) = 0
⇒ 2x – 4 + 12x – 3 = 0
⇒ 2x + 12x = 4 + 3 (By transposing)
⇒ 14x = 7
⇒ x = \(\frac { 7 }{ 14 }\) = \(\frac { 1 }{ 2 }\)
Check : L.H.S. = 2(x – 2) + 3 (4x -1)

Question 6.
Solution:
5 (2x – 3) – 3(3x – 7) = 5
⇒ 10x – 15 – 9x + 21 = 5
⇒ 10x – 9x – 15 + 21 = 5
⇒ 10x – 9x = 5 + 15 – 21 (By transposing)
⇒ x = 20 – 21 = -1
⇒ x = -1
Check:
L.H.S. = 5 (2x – 3) – 3(3x – 7)
= 5[2 x (-1) -3] -3[3 (-1) -7] = 5[-2 – 3] – 3[-3 – 7]
= 5 x (-5) -3 x (-10)
= -25 + 30
= 5 = R.H.S.
Hence x = -1

Question 7.
Solution:

Question 8.
Solution:

L.H.S. = R.H.S.
Hence x = 48

Question 9.
Solution:

Question 10.
Solution:
3x + 2(x + 2) = 20 – (2x – 5)
⇒ 3x + 2x + 4 = 20 – 2x + 5
⇒ 5x + 4 = 25 – 2x
⇒ 5x + 2x = 25 – 4 (By transposing)
⇒ 7x = 21
⇒ x = 3
Check:
L.H.S.= 3x + [2(x + 2)] = 3 x 3 + 2(3 + 2) = 9 + 2 x 5 = 9 + 10 = 19
R.H.S. = 20 – (2x – 5) = 20 – (2 x 3 – 5) = 20 – (6 – 5) = 20 – 1 = 19
L.H.S. = R.H.S.
Hence x = 3

Question 11.
Solution:
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
⇒ 13y – 52 – 3y + 27 – 5y – 20 = 0
⇒ 13y – 3y – 5y – 52 + 27 – 20 = 0
⇒ 13y – 8y – 72 + 27 = 0
⇒ 5y – 45 = 0
⇒ 5y = 45 (By transposing)
⇒ y = 9
Check:
L.H.S. = 13(y – 4) – 3(y – 9) – 5(y + 4)
= 13(9 – 4) – 3(9 – 9) – 5(9 + 4)
= 13 x 5 – 3 x 0 – 5 x 13
= 65 – 0 – 65 = 0 = R.H.S.
Hence y = 9

Question 12.
Solution:
\(\frac { 2m + 5 }{ 3 }\) = 3m – 10
⇒ 2m + 5 = 3 (3m – 10) (By cross multiplication)
⇒ 2m + 5 = 9m – 30
⇒ 2m – 9m = -30 – 5
⇒ -7m = -35
⇒ m = 5
m = 5
Check:

R.H.S. = 3m – 10 = 3 x 5 – 10 = 15 – 10 = 5
L.H.S. = R.H.S.
Hence m = 5

Question 13.
Solution:
6(3x + 2) – 5(6x – 1) = 3(x – 8) – 5(7x – 6) + 9x
⇒ 18x + 12 – 30x + 5 = 3x – 24 – 35x + 30 + 9x
⇒ 18x – 30x + 12 + 5 = 3x – 35x + 9x – 24 + 30
⇒ -12x + 17 = -23x + 6
⇒ – 12x + 23x = 6 – 17
⇒ 11x = -11
x = – 1
Check:
L.H.S. = 6(3x + 2) – 5(6x – 1)
= 6[3x (-1) + 2] – 5[6 x (-1) x -1]
= 6[-3 + 2] – 5[-6 – 1]
= 6 x (-1) – 5 x (-7)
= -6 + 35 = 29
R.H.S. = 3(x – 8) – 5 (7x – 6) + 9x
= 3[-1 – 8] -5 [7 x (-1) – 6] + 9 (-1)
= 3 x (-9) – 5 [-7 – 6] – 9
= -27 – 5(-13) – 9
= -27 + 65 – 9
= 65 – 36 = 29 .
L.H.S. = R.H.S.
Hence x = -1

Question 14.
Solution:
t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t – 4)
⇒ t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12t
⇒ t – 2t + 10t – 10 = 8t – 3t + 18
⇒ 9t – 10 = 5t + 18
⇒ 9t – 5t = 18 + 10 (By transposing)
⇒ 4t = 28
⇒ t = 7
Check:
L.H.S. = t – [2t + 5] -5[1 – 2t]
= 7 – [2 x 7 + 5] – 5[1 – 2 x 7]
= 7 – [14 + 5] – 5 [1 – 14]
= 7 – 19 – 5(-13)
= 7 – 19 + 65
= 72 – 19 = 53
R.H.S. = 2[3 + 4t) – 3(t – 4)
= 2 (3 + 4 x 7) – 3(7 – 4)
= 2(3 + 28) – 3(3)
= 2(31) – 9 = 62 – 9 = 53
L.H.S. = R.H.S.
Hence t = 7 Ans.

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
0.18 (5x – 4) = 0.5x + 0.8

Question 29.
Solution:
2.4 (3 – x) – 0.6 (2x – 3) = 0
⇒ 7.2 – 2.4x – 1.2x + 1.8 = 0
⇒ -2.4x – 1.2x = – (7.2 + 1.8).
L.H.S. = 2.4 (3 – x) – 0.6 (2x – 3)
⇒ 2.4 (3 – 2.5) – 0.6 (2 x 2.5 – 3)
⇒ 2.4 (0.5) – 0.6 (5 – 3)
⇒ 1.2 – 0.6 x 2 = 1.2 – 1.2 = 0 = R.H.S.
Hence x = 2.5

Question 30.
Solution:
0.5x – (0.8 – 0.2x) = 0.2 – 0.3x
⇒ 0.5x – 0.8 + 0.2x = 0.2 – 0.3x
⇒ 0.5x + 0.2x + 0.3x = 0.2 + 0.8
⇒ 1.0x = 1.0
⇒ x = 1
Check :
L.H.S. = 0.5x – (0.8 – 0.2x)
= 0.5 x 1 – (0.8 – 0.2 x 1)
= 0.5 – (0.8 – 0.2) = 0.5 – 0.6 = -0.1
R.H.S. = 0.2 – 0.3x = 0.2 – 0.3 x 1 = 0.2 – 0.3 = -0.1
L.H.S. = R.H.S.
Hence x = 1

Question 31.
Solution:

Question 32.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 7 Linear Equations in One Variable Ex 7A 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 7 Solutions Chapter 6 Algebraic Expressions Ex 6D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D

Find each of the following products.
Question 1.
Solution:
(5x + 7) (3x + 4)
= 5x (3x + 4) + 7 (3x + 4)
= 5x x 3x + 5x x 4 + 7 x 3x + 7 x 4
= 15x² + 20x + 21x + 28
= 15x² + 41x + 28

Question 2.
Solution:
(4x – 3) (2x + 5)
= 4x (2x + 5) – 3 (2x + 5)
= 4x x 2x + 4x x 5 – 3 x 2x -3 x 5
= 8x² + 20x – 6x – 15
= 8x² + 14x – 15

Question 3.
Solution:
(x – 6) (4x + 9)
= x (4x + 9) – 6 (4x + 9)
= x x 4x + x x 9 – 6 x 4x – 6 x 9
= 4x² + 9x – 24x – 54
= 4x² – 15x – 54

Question 4.
Solution:
(5y – 1) (3y – 8)
= 5y x 3y – 5y x 8 – 1 x 3y – 8 x (-1)
= 15y² – 40y – 3y + 8
= 15y² – 43y + 8

Question 5.
Solution:
(7x + 2y) (x + 4y)
= 7x (x + 4y) + 2y (x + 4y)
= 7x x x + 7x x 4y + 2y x x + 2y x 4y
= 7x² + 28xy + 2xy + 8y²
= 7x² + 30xy + 8y²

Question 6.
Solution:
(9x + 5y) (4x + 3y)
= 9x (4x + 3y) + 5y (4x + 3y)
= 9x x 4x + 9x x 3y + 5y x 4x + 5y x 3y
= 36x² + 27xy + 20xy + 15y²
= 36x² + 47xy +15y²

Question 7.
Solution:
(3m – 4n) (2m – 3n)
= 3m (2m – 3n) – 4n (2m – 3n)
= 3m x 2m – 3m x 3n – 4n x 2m – 4n x (-3n)
= 6m² – 9mn – 8mn + 12n²
= 6m² – 17mn + 12n²

Question 8.
Solution:
(0.8x – 0.5y) (1.5x – 3y)
= 0.8x (1.5x – 3y) – 0.5y (1.5x – 3y)
= 0.8x x 1.5x – 0.8x x 3y – 0.5y x 1.5x – 0.5y x (-3y)
= 1.20x² – 2.4xy – 0.75xy + 1.5y²
= 1.2x² – 3.15xy + 1.5y²

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
(x² – a²) (x – a)
= x² (x – a) – a² (x – a)
= x² x x – x² x a – a² x x – a² (-a)
= x³ – x² a – xa² + a³

Question 13.
Solution:
(3p² + q²) (2p² – 3q²)
= 3p² (2p² – 3q²) + q² (2p² – 3q²)
= 3p² x 2p² – 3p² x 3q² + q² x 2p² – q² x 3q²
= 6q⁴ – 9p²q² + 2p²q² – 3q⁴
= 6p⁴ – 7p²q² – 3q⁴

Question 14.
Solution:
(2x² – 5y²) (x² + 3y²)
= 2x² (x² + 3y²) – 5y² (x² + 3y²)
= 2x² x x² + 2x² x 3y² – 5y² x x² – 5y² x 3y²
= 2x⁴ + 6x² y² – 5x² y² – 15y⁴
= 2y⁴ + x² y² – 15y⁴

Question 15.
Solution:
(x³ – y³) (x² + y²)
= x³ (x² + y²) – y³ (x² + y²)
= x³ x x² + x³ x y² – y³ x x² – y³ x y²
= x⁵ + x³ y² – x² y³ – y⁵

Question 16.
Solution:
(x⁴ + y⁴) (x² – y²)
= x⁴ (x² – y²) + y⁴ (x² – y²)
= x⁴ x x² – x⁴ x y² + y⁴ x x² – y⁴ x y²
= x⁶ – x⁴ y² + x² y⁴ – y⁶

Question 17.
Solution:

Question 18.
Solution:
(x² – y²) (x + 2y)
= x² (x + 2y) – y² (x + 2y)
= x² x x + x² x 2y – y² x x – y² x 2y
= x³ + 2x² y – xy² – 2y³

Question 19.
Solution:
(2x + 3y – 5) (x + y)
= 2x (x + y) + 3y (x + y) – 5 (x + y)
= 2x x x + 2x x y + 3y x x + 3y x y – 5 x x – 5 x y
= 2x² + 2xy + 3xy + 3y² – 5x – 5y
= 2x² + 5xy + 3y² – 5x – 5y

Question 20.
Solution:
(3x + 2y – 4) (x – y)
= 3x (x – y) + 2y (x – y) – 4 (x – y)
= 3x x x – 3x x y + 2y x x – 2y x y – 4 x x – 4 x (-y)
= 3x² – 3xy + 2xy – 2y² – 4x + 4y
= 3x² – xy – 2y² – 4x + 4y

Question 21.
Solution:
(x² – 3x + 7) (2x + 3)
= x² (2x + 3) – 3x (2x + 3) + 7 (2x + 3)
= x² x 2x + x² x 3 – 3x x 2x – 3x x 3 + 7 x 2x + 7 x 3
= 2x³ + 3x² – 6x² – 9x + 14x + 21
= 2x³ – 3x² + 5x + 21

Question 22.
Solution:
(3x² + 5x – 9) (3x – 9)
= 3x² (3x – 9) + 5x (3x – 9) – 9 (3x – 9)
= 3x² x 3x – 3×2 x 9 + 5x x 3x + 5x x (-9) – 9 x 3x – 9 x (-9)
= 9x³ – 27x² + 15x² – 45x – 27x + 81
= 9x³ – 12x² – 72x + 81

Question 23.
Solution:
(9x² – x + 15) (x² – 3)
= 9x² (x² – 3) – x (x² – 3) + 15 (x² – 3)
= 9x² x x² – 9x² x 3 – x x x² + x x 3 + 15 x x² – 15 x 3
= 9x⁴ – 27x² – x³ + 3x + 15x² – 45
= 9x⁴ – x³ – 12x² + 3x – 45

Question 24.
Solution:
(x² + xy + y²) (x – y)
= x² (x – y) + xy (x – y) + y² (x – y)
= x² x x – x² x y + xy x x – xy x y + y² x x – y² x y
= x³ – x²y + x²y – xy² + xy² – y²
= x³ – y³

Question 25.
Solution:
(x² – xy + y²) (x + y)
= x² (x + y) – xy (x + y) + y² (x + y)
= x³ + x² y – x² y – xy² + xy² + y³
= x³ + y³

Question 26.
Solution:
(x² – 5x + 8) (x² + 2)
= x² (x² + 2) – 5x (x² + 2) + 8 (x² + 2)
= x² x x² + x² x 2 – 5x x x² – 5x x 2 + 8 x x² + 8 x 2
= x⁴ + 2x² – 5x³ – 10x + 8x² + 16
= x⁴ – 5x³ + 2x² + 8x² – 10x + 16
= x⁴ – 5x³ + 10x² – 10x + 16

Simplify:
Question 27.
Solution:
(3x + 4) (2x – 3) + (5x – 4) (x + 2)
= [3x (2x – 3) + 4 (2x – 3)] + [5x (x + 2) – 4 (x + 2)]
= [3x x 2x – 3x x 3 + 4 x 2x – 4 x 3] + [5x x x + 5x x 2 – 4 x x – 4 x 2]
= [6x² – 9x + 8x – 12] + [5x² + 10x – 4x – 8]
= (6x² – x – 12) + (5x² + 6x – 8)
= 6x² – x – 12 + 5x² + 6x – 8
= 6x² + 5x² – x + 6x – 12 – 8
= 11x² + 5x – 20

Question 28.
Solution:
(5x – 3) (x + 4) – (2x + 5) (3x – 4)
= [5x x x + 5x x 4 – 3 x x – 3 x 4] – [2x x 3x + 2x x (-4) + 5 x 3x + 5x (-4)]
= [5x² + 20x – 3x – 12] – [6x² – 8x + 15x – 20]
= (5x² + 17x – 12) – (6x² + 7x – 20)
= 5x² + 17x – 12 – 6x² – 7x + 20
= 5x² – 6x² + 17x – 7x – 12 + 20
= -x² + 10x + 8

Question 29.
Solution:
(9x – 7) (2x – 5) – (3x – 8) (5x – 3)
= [9x (2x – 5) -7 (2x – 5)] – [3x (5x – 3) -8 (5x – 3)]
= [9x x 2x – 9x x 5 – 7 x 2x – 7 x (-5)] – [3x x 5x – 3x x 3 – 8 x 5x – 8 x (-3)]
= [18x² – 45x – 14x + 35] – [15x² – 9x – 40x + 24]
= 18x² – 45x – 14x + 35 – 15x² + 9x + 40x – 24
= 18x² – 15x² – 45x – 14x + 9x + 40x + 35 – 24
= 3x² – 59x + 49x + 11
= 3x² – 10x + 11

Question 30.
Solution:
(2x + 5y) (3x + 4y) – (7x + 3y) (2x + y)
= [2x (3x + 4y) + 5y (3x + 4y)] – [7x (2x + y) + 3y (2x + y)]
= [2x x 3x + 2x x 4y + 5y x 3x + 5y x 4y] – [7x x 2x + 7x x y + 3y x 2x + 3y x y]
= [6x² + 8xy + 15xy + 20y²] – [14x² + 7xy + 6xy + 3y²]
= [6x² + 23xy + 20y²] – [14x² + 13xy + 3y^2]
= 6x² + 23xy + 20y² – 14x² – 13xy – 3y²
= 6x² – 14x² + 23xy – 13xy + 20y² – 3y²
= -8x² + 10xy + 11y²

Question 31.
Solution:
(3x² + 5x – 7) (x – 1) – (x² – 2x + 3) (x + 4)
= [3x² (x – 1) + 5x (x – 1) – 7 (x – 1)] – [x² (x + 4) – 2x (x + 4) + 3 (x + 4)]
= [3x² x x – 3x² x 1 + 5x x x – 5x x (-1) – 7 x x -7 x (-1)] -[x² x x + x² x 4 – 2x x x – 2x x 4 + 3 x x + 3 x 4]
= [3x³ – 3x² + 5x² – 5x – 7x + 7] – [x³ + 4x² – 2x² – 8x + 3x + 12]
= [3x³ + 2x² – 12x + 7] – [x³ + 2x² – 5x + 12]
= 3x³ + 2x² – 12x + 7 – x³ – 2x² + 5x – 12
= 3x³ – x³ + 2x² – 2x² – 12x + 5x – 12 + 7
= 2x³ – 7x – 5

Hope given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D 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 7 Solutions Chapter 6 Algebraic Expressions Ex 6C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D

Find each of the following products:
Question 1.
Solution:
4a(3a + 7b) = 4a x 3a + 4a x 7b = 12a² + 28ab

Question 2.
Solution:
5a(6a – 3b) = 5a x 6a – 5a x 3b = 30a² – 15ab

Question 3.
Solution:
8a² (2a + 5b) = 8a² x 2a + 8a² x 5b = 16a³ + 40a²b

Question 4.
Solution:
9x² (5x + 7) = 9x² x 5x + 9x² x 7 = 45x³ + 63x²

Question 5.
Solution:
ab(a² – b²) = ab x a² – ab x b² = a³b – ab³

Question 6.
Solution:
2x² (3x – 4x²) = 2x² x 3x – 2x² x 4x² = 6x³ – 8x⁴

Question 7.
Solution:

Question 8.
Solution:
-1 7x² (3x – 4) = -17x² x 3x – 17x² x (-4) = -51x³ + 68x²

Question 9.
Solution:

Question 10.
Solution:
-4x²y (3x² – 5y)
= -4x² y x 3x² – 4x² y x (-5y)
= -12x² y + 20x² y²

Question 11.
Solution:

Question 12.
Solution:
9t² (t + 7t³) = 9t² x t + 9t² x 7t³ = 9t³ + 63t⁵

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
24x² (1 – 2x)
= 24x² x 1 – 24x² x 2x
= 24x² – 48x³
If x = 2, then
24x² – 48x³
= 24(2)² – 48(2)³
= 24 x 4 – 48 x 8
= 96 – 384
= -288

Question 17.
Solution:

Question 18.
Solution:
s (s² – st) = s x s² – s x st = s³ – s²t
If s = 2, t = 3, then
s³ – s²t = (2)³ – (2)² x 3 = 8 – 4 x 3 = 8 – 12 = -4

Question 19.
Solution:
-3y (xy + y²) = -3y x xy + (-3y) x y² = -3xy² – 3y³
if x = 4, y = 5, then
-3xy² – 3y³
= -3(4)(5)² – 3(5)³ = -3 x 4 x 25 – 3 x 125 = -300 – 375 = -675

Simplify each of the following:
Question 20.
Solution:
a(b – c) + b(c – a) + c(a – b) = ab – ac + bc – ab + ac – bc = 0

Question 21.
Solution:
a(b – c) – b(c – a) – c(a – b) = ab – ac – bc + ab – ac + bc = 2ab – 2ac

Question 22.
Solution:
3x² + 2(x + 2) – 3x (2x + 1)
= 3x² + 2x + 4 – 6x² – 3x = 3x² – 6x² + 2x – 3x + 4 = -3x² – x + 4

Question 23.
Solution:
x (x + 4) + 3x (2x² – 1) + 4x² + 4
= x² + 4x + 6x³ – 3x + 4x² + 4
= 6x³ + x² + 4x² + 4x – 3x + 4
= 6x³ + 5x² + x + 4

Question 24.
Solution:
2x² + 3x (1 – 2x³) + x (x + 1)
= 2x² + 3x – 6x⁴ + x² + x
= – 6x⁴ + 2x² + x² + 3x + x
= – 6x⁴ + 3x² + 4x

Question 25.
Solution:
a²b (a – b²) + ab(4ab – 2a²) – a³b (1 – 2b)
= a³b – a²b³ – 2a³b² + 4a²b³ – 2a³b² – a³b + 2a³b²
= a³b – a³b + 2a³b² – a²b³ + 4a²6³
= 3a²b³

Question 26.
Solution:
4st (s – t) – 6s² (t – t²) – 3t² (2s² – s) + 2st(s – t)
= 4s²t – 4st² – 6s²t + 6s²t² – 6s²t² + 3st² + 2s²t – 2st²
= 4s²t – 6s²t + 2s²t – 4st² + 3st² – 2st² + 6s²t² – 6s²t²
= 6s²t – 6s²t – 6st² + 3st² + 6s²t² – 6s²t²
= – 3st²

Hope given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C 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 7 Solutions Chapter 6 Algebraic Expressions Ex 6B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D

Find the products:
Question 1.
Solution:
3 x 8 a²⁺⁴ = 24a⁶

Question 2.
Solution:
(-6x³) x (5x²) = -6 x 5x²⁺³ = -30x⁵

Question 3.
Solution:
(-4ab) x (-3a²bc)
= (-4) x (-3) a. a².b. b. c
= 12.a²⁺¹. b¹⁺¹.c
= 12a³b²c

Question 4.
Solution:
= (2a²b³) x (-3a³b)
= 2 x (-3) a². a³. b³. b. b
= -6a²⁺³.b³⁺¹
= -6a⁵.b⁴

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Find the products given below and in each case verify the result for a = 1, b = 2 and c = 3 :
Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B 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 7 Solutions Chapter 6 Algebraic Expressions Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A.

Other Exercises

• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6B
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6C
• RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6D

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:
Sum of (a + 3b – 4c), (4a – b + 9c) and (-2b + 3c – a)

Now subtract (2a – 3b + 4c) from 4a + 8c
= 4a + 8c – (2a – 3b + 4c)
= 4a + 8c – 2a + 3b – 4c
= 4a – 2a + 3b + 8c – 4c
= 2a + 3b + 4c

Question 4.
Solution:

Question 5.
Solution:
Sum of (8a – 6a² + 9) and (-10a – 8 + 8a²)
= 8a – 6a² + 9 + (-10a) – 8 + 8a²
= 8a – 10a – 6a² + 8a² + 9 – 8
= -2a + 2a² + 1
Now -3 – (-2a + 2a² + 1)
= (-3 + 2a – 2a² – 1)
= -4 + 2a – 2a²

Question 6.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 6 Algebraic Expressions Ex 6A 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.