RS Aggarwal Solutions for Class 7 Maths Exercise 7A Chapter 7 Linear Equations in One Variable

RS Aggarwal Solutions for Class 7 Maths Exercise 7A Chapter 7 Linear Equations in One Variable, an equation is a statement of equality which contains one or more unknown quantities or variables. An equation involving only a linear polynomial is a linear equation. Students of Class 7 are suggested to solve RS Aggarwal Class 7 Solutions Chapter 7 to strengthen the fundamentals and be able to solve questions that are usually asked in the examination. By practising RS Aggarwal Solutions for Class 7 Maths Chapter 7 Linear Equations in One Variable, students can get good marks in Maths.

Download the PDF of RS Aggarwal Solutions For Class 7 Maths Chapter 7 Linear Equations in One Variable – Exercise 7A

 

rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a
rs aggarwal solution class 7 maths chapter 7 ex 7a

 

Access answers to Maths RS Aggarwal Solutions for Class 7 Chapter 7 – Linear Equations in One Variable Exercise 7A

Solve the following equations. Check your result in each case.

  1. 3x – 5 = 0

    Solution:-

    The value of the variable which makes the equation a true statement is called the root of the equation.

    Now, find the value of x by considering the given equation,

    = 3x – 5 = 0

    Transposing – 5 to RHS it becomes 5

    = 3x = 5

    Multiplying both side by (1/3)

    = 3x × (1/3) = 5 × (1/3)

    = x = (5/3)

    By substituting (5/3) in the place of x in given equation, we get

    = 3 × (5/3) – 5 = 0

    = 5 – 5 = 0

    = 0= 0

    ∴ LHS = RHS

    Hence, the result is verified.

  2. 8x – 3 = 9 – 2x

    Solution:-

    Now, find the value of x by considering the given equation,

    =8x – 3 = 9 – 2x

    Transposing – 3 to RHS and it becomes 3 and -2x to LHS it becomes 2x

    = 8x + 2x = 9 + 3

    = 10x = 11

    Multiplying both side by (1/10)

    = 10x × (1/10) = 12 × (1/10)

    = x = (12/10)

    = x = (6/5)

    By substituting (6/5) in the place of x in given equation, we get

    LHS,

    = 8 × (6/5) – 3

    = (48/5) – 3

    = (48-15/5)

    = (33/5)

    RHS,

    = 9 – 2 × (6/5)

    = 9 – (12/5)

    = (45 – 12)/5

    = (33/5)

    By comparing LHS and RHS

    = (33/5) = (33/5)

    ∴ LHS = RHS

    Hence, the result is verified.

  3. 7 – 5x = 5 – 7x

    Solution:-

    Now, find the value of x by considering the given equation,

    =7 – 5x = 5 – 7x

    Transposing 7 to RHS and it becomes -7 and -7x to LHS it becomes 7x

    = 7x – 5x = 5 – 7

    = 2x = -2

    Multiplying both side by (1/3)

    = 2x × (1/2) = -2 × (1/2)

    = x = (-2/2)

    = x = -1

    By substituting -1 in the place of x in given equation, we get

    LHS,

    = 7 – 5 × (-1)

    = 7 – (-5)

    = 7 + 5

    = 12

    RHS,

    = 5 – 7 × (-1)

    = 5 – (-7)

    = 5 + 7

    = 12

    By comparing LHS and RHS

    = 12 = 12

    ∴ LHS = RHS

    Hence, the result is verified.

  4. 3 + 2x = 1 – x

    Solution:-

    Now, find the value of x by considering the given equation,

    =3 + 2x = 1 – x

    Transposing 3 to RHS and it becomes -3 and -x to LHS it becomes x

    = 2x + x = 1 – 3

    = 3x = -2

    Multiplying both side by (1/3)

    = 3x × (1/3) = -2 × (1/3)

    = x = (-2/3)

    By substituting -2/3 in the place of x in given equation, we get

    LHS,

    = 3 + 2 × (-2/3)

    = 3 + (-4/3)

    = 3 – (4/3)

    = (9 – 4) / 3

    = 5/3

    RHS,

    = 1 – (-2/3)

    = 1 + (2/3)

    = (3 + 2)/ 3

    = 5/3

    By comparing LHS and RHS

    = 5/3 = 5/3

    ∴ LHS = RHS

    Hence, the result is verified.

  5. 2 (x – 2) + 3 (4x – 1) = 0

    Solution:-

    The above question can be written as,

    = 2x – 4 + 12x – 3 = 0

    Now, find the value of x by considering the above equation,

    = 2x + 12x – 4 – 3 = 0

    = 14x – 7 = 0

    Transposing -7 to RHS and it becomes 7

    = 14x = 7

    Multiplying both side by (1/14)

    = 14x × (1/14) = 7 × (1/14)

    = x = (7/14)

    = x = 1/2

    By substituting 1/2 in the place of x in given equation, we get

    LHS,

    = (2 × (1/2)) – 4 + (12 × (1/2)) – 3

    = 1– 4 + 6 – 3

    = 7 – 7

    = 0

    By comparing LHS and RHS

    = 0 = 0

    ∴ LHS = RHS

    Hence, the result is verified.

  6. 5 (2x – 3) – 3 (3x – 7) = 5

    Solution:-

    The above question can be written as,

    = 10x – 15 – 9x + 21 = 5

    Now, find the value of x by considering the given equation,

    = 10x – 15 – 9x + 21 = 5

    = x + 6 = 5

    Transposing 6 to RHS and it becomes -6

    = x = 5 – 6

    = x = -1

    By substituting -1 in the place of x in given equation, we get

    LHS,

    = 10 (-1) -15 -9(-1) + 21

    = -10 – 15 + 9 + 21

    = -25 + 30

    = 5

    By comparing LHS and RHS

    = 5 = 5

    ∴ LHS = RHS

    Hence, the result is verified.

  7. 2x – (1/3) = (1/5) – x

    Solution:-

    Now, find the value of x by considering the given equation,

    =2x – (1/3) = (1/5) – x

    Transposing – (1/3) to RHS and it becomes (1/3) and – x to LHS it becomes x

    = 2x + x = (1/5) + (1/3)

    = 3x = (3 + 5) / 15

    = 3x = 8/15

    Multiplying both side by (1/3)

    = 3x × (1/3) = (8/15) × (1/3)

    = x = (8/45)

    By substituting (8/45) in the place of x in given equation, we get

    LHS,

    = 2 × (8/45) – (1/3)

    = (16/45) – (1/3)

    = (16 – 15) / 45

    = (1/45)

    RHS

    = (1/5) – (8/45)

    = (9-8) / 45

    = (1/45)

    By comparing LHS and RHS

    = (1/45) = (1/45)

    ∴ LHS = RHS

    Hence, the result is verified.

  8. (1/2)x- 3 = 5 + (1/3)x

    Solution:-

    Now, find the value of x by considering the given equation,

    = (1/2)x- 3 = 5 + (1/3)x

    Transposing -3 to RHS and it becomes 3 and (1/3)x to LHS it becomes (-1/3)x

    = (1/2)x – (1/3)x = 5 + 3

    = ((3-2)/6)x = 8

    = (1/6)x = 8

    Multiplying both side by 6

    = 6 × (1/6)x = 8 × 6

    = x = 48

    By substituting 48 in the place of x in given equation, we get

    LHS,

    = (1/2) × (48) – 3

    = (48/2) – 3

    = 24 – 3

    = 21

    RHS,

    = 5 + (1/3) × (48)

    = 5 + (48/3)

    = 5 + 16

    = 21

    By comparing LHS and RHS

    = 21 = 21

    ∴ LHS = RHS

    Hence, the result is verified.

  9. (x/2) + (x/4) = (1/8)

    Solution:-

    Now, find the value of x by considering the given equation,

    = (x/2) + (x/4) = (1/8)

    = ((2 + 1) / 4) = (1/8)

    = (3/4)x = (1/8)

    Multiplying both side by (4/3)

    = (3/4)x × (4/3) = (1/8) × (4/3)

    = x = (1/6)

    By substituting (1/6) in the place of x in given equation, we get

    LHS,

    = (3/4)x

    = (3/4) × (1/6)

    = (1/8)

    RHS,

    = (1/8)

    By comparing LHS and RHS

    = (1/8) = (1/8)

    ∴ LHS = RHS

    Hence, the result is verified.

  10. 3x + 2(x + 2) = 20 – (2x – 5)

    Solution:-

    The above question can be written as,

    = 3x + 2x + 4 = 20 – 2x + 5

    Now, find the value of x by considering the above equation,

    = 5x + 4 = 20 – 2x + 5

    Transposing 4 to RHS and it becomes -4 and -2x to LHS it becomes 2x

    = 5x + 2x = 20 – 4 + 5

    = 7x = 25 – 4

    = 7x = 21

    Multiplying both side by (1/7)

    = 7x × (1/7) = 21 × (1/7)

    = x = 21/7

    = x = 3

    By substituting 3 in the place of x in given equation, we get

    LHS,

    = 5x + 4

    = 5 × (3) + 4

    = 15 + 4

    = 19

    RHS,

    = 20 – 2x + 5

    = 20 – 2 × (3) + 5

    = 20 – 6 + 5

    = 25 – 6

    = 19

    By comparing LHS and RHS

    = 19 = 19

    ∴ LHS = RHS

    Hence, the result is verified.

  11. 13( y – 4) – 3(y-9) – 5(y + 4) = 0

    Solution:-

    The above question can be written as,

    = 13y – 52 – 3y + 27 – 5y – 20 = 0

    Now, find the value of y by considering the above equation,

    = 13y – 3y – 5y – 52 + 27 – 20 = 0

    = 13y – 8y – 72 + 27 = 0

    = 5y – 45 = 0

    Transposing -45 to RHS and it becomes 45

    = 5y = 45

    Multiplying both side by (1/5)

    = 5y × (1/5) = 45 × (1/5)

    = y = 9

    By substituting 9 in the place of y in given equation, we get

    LHS,

    = 5y – 45

    = 5 × (9) – 45

    = 45 – 45

    = 0

    RHS,

    = 0

    By comparing LHS and RHS

    = 0 = 0

    ∴ LHS = RHS

    Hence, the result is verified.

  12. (2m + 5) / 3 = (3m – 10)

    Solution:-

    The above question can be written as,

    = 3 × [(2m + 5)/ 3] = 3 × (3m -10) … [Multiplying both side by 3]

    = (2m + 5) = 9m – 30

    Transposing 2m to RHS and it becomes -2m and -30 to LHS it becomes 30

    = 30 + 5 = 9m -2m

    = 35 = 7m

    Multiplying both side by (1/7),

    = 7m × (1/7) = 35 × (1/7)

    = m = (35/7)

    = m = 5

    By substituting 5 in the place of m in given equation, we get

    LHS,

    = 2m + 5

    = 2 × (5) + 5

    = 10 + 5

    = 15

    RHS,

    = 9m – 30

    = 9 × (5) – 30

    = 45 – 30

    = 15

    By comparing LHS and RHS

    = 15 = 15

    ∴ LHS = RHS

    Hence, the result is verified.

  13. 6(3x + 2) – 5(6x – 1) = 3(x – 8) – 5(7x – 6) + 9x

    Solution:-

    The above question can be written as,

    = 18x + 12 – 30x + 5 = 3x – 24 – 35x + 30 + 9x

    Now, find the value of x by considering the above equation,

    Transposing 12, 5 to RHS and it becomes -12, -5 and -35x, 9x, 3x to LHS it becomes 35x, -9x, -3x

    = 18x – 30x + 35x – 9x – 3x = -24 + 30 – 12 – 5

    = 53x – 42x = 30 – 41

    = 11x = -11

    Multiplying both side by (1/11),

    = 11x × (1/11) = -11 × (1/11)

    = x = (-11/11)

    = x = -1

    By substituting 5 in the place of x in given equation, we get

    LHS,

    = 18x + 12 – 30x + 5

    = 18 (-1) + 12 – 30 (-1) + 5

    = -18 + 12 + 30 + 5

    = -18 + 47

    = 29

    RHS,

    = 3x – 24 – 35x + 30 + 9x

    = 3 (-1) – 24 – 35 (-1) + 30 + 9 (-1)

    = -3 – 24 + 35 + 30 – 9

    = -36 + 65

    = 29

    By comparing LHS and RHS

    = 15 = 15

    ∴ LHS = RHS

    Hence, the result is verified.

  14. t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t -4)

    Solution:-

    The above question can be written as,

    = t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12

    Now, find the value of t by considering the above equation,

    Transposing -5, -5 to RHS and it becomes 5, 5 and 8t, -3t to LHS it becomes -8t, 3t

    = t – 2t + 10t – 8t + 3t = 6 + 12 + 5 + 5

    = 14t – 10t = 28

    = 4t = 28

    Multiplying both side by (1/4),

    = 4t × (1/4) = 28 × (1/4)

    = t = 7

    By substituting 7 in the place of t in given equation, we get

    LHS,

    = t – 2t – 5 – 5 + 10t

    = 7 – 2× (7) – 5 – 5 + 10(7)

    = 7 – 14 – 5 – 5 + 70

    = 77 – 24

    = 53

    RHS,

    = 6 + 8t – 3t + 12

    = 6 + 8 × (7) – 3 × (7) + 12

    = 6 + 56 – 21 + 12

    = 74 – 21

    = 53

    By comparing LHS and RHS

    = 15 = 15

    ∴ LHS = RHS

    Hence, the result is verified.

  15. (2/3)x = (3/8)x + (7/12)

    Solution:-

    Find the value of x by considering the above equation,

    Transposing (3/8)x to RHS and it becomes -(3/8)x.

    = (2/3)x – (3/8)x = (7/12)

    = [(16 – 9) / 24]x = (7/12)

    = (7/24) x = (7/12)

    Multiplying both side by (24/7),

    = (7/24)x × (24/7) = (7/12) × (24/7)

    = x = 2

    By substituting 2 in the place of x in given equation, we get

    LHS,

    = (2/3)x

    = (2/3) × (2)

    = (4/3)

    RHS,

    = (3/8)x + (7/12)

    = (3/8) × (2) + (7/12)

    = (3/4) + (7/12)

    = (9 + 7) / (12)

    = (16/12)

    = (4/3)

    By comparing LHS and RHS

    = (4/3) = (4/3)

    ∴ LHS = RHS

    Hence, the result is verified.

  16. [(3x – 1)/ 5] – (x/7) = 3

    Solution:-

    The above question can be written as,

    = [(7 × (3x -1)) – (5 × (x))/35] = 3

    = ((21x – 7) – (5x))/35 = 3

    = (21x – 7 – 5x)/35 = 3

    = (16x – 7)/35 = 3 … [Multiplying both side by 35]

    = [(16x – 7)/35] × 35 = 3 × 35

    = 16x – 7 = 105

    = 16x = 105 + 7

    = 16x = 112 … [Multiplying both side by 1/16]

    = 16x × (1/16) = 112 × (1/16)

    = x = 7

    By substituting 7 in the place of x in given equation, we get

    LHS,

    = (16x – 7)/35

    = ((16 × 7) – 7)/35

    = (112 – 7)/ 35

    = (105/35)

    = 3

    RHS,

    = 3

    By comparing LHS and RHS

    = (4/3) = (4/3)

    ∴ LHS = RHS

    Hence, the result is verified.


Access other exercises of RS Aggarwal Solutions For Class 7 Chapter 7 – Linear Equations in One Variable

Exercise 7B Solutions

Exercise 7C Solutions