Study MaterialsNCERT SolutionsNCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.3

NCERT Solutions for Class 8 Maths Chapter 2 solutions are important for CBSE Exam preparation as well as preparation for competitive exams like JEE

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    Solve the following equations and check your results.
    Ex 2.3 Class 8 Maths Question 1.
    3x = 2x + 18
    Solution:
    We have 3x = 2x + 18
    ⇒ 3x – 2x = 18 (Transposing 2x to LHS)
    ⇒ x = 18
    Hence, x = 18 is the required solution.
    Check: 3x = 2x + 18
    Putting x = 18, we have
    LHS = 3 × 18 = 54
    RHS = 2 × 18 + 18 = 36 + 18 = 54
    LHS = RHS
    Hence verified.

    Ex 2.3 Class 8 Maths Question 2.
    5t – 3 = 3t – 5
    Solution:
    We have 5t – 3 = 3t – 5
    ⇒ 5t – 3t – 3 = -5 (Transposing 3t to LHS)
    ⇒ 2t = -5 + 3 (Transposing -3 to RHS)
    ⇒ 2t = -2
    ⇒ t = -2 ÷ 2
    ⇒ t = -1
    Hence t = -1 is the required solution.
    Check: 5t – 3 = 3t – 5
    Putting t = -1, we have
    LHS = 5t – 3 = 5 × (-1)-3 = -5 – 3 = -8
    RHS = 3t – 5 = 3 × (-1) – 5 = -3 – 5 = -8
    LHS = RHS
    Hence verified.

    Ex 2.3 Class 8 Maths Question 3.
    5x + 9 = 5 + 3x
    Solution:
    We have 5x + 9 = 5 + 3x
    ⇒ 5x – 3x + 9 = 5 (Transposing 3x to LHS) => 2x + 9 = 5
    ⇒ 2x = 5 – 9 (Transposing 9 to RHS)
    ⇒ 2x = -4
    ⇒ x = -4 ÷ 2 = -2
    Hence x = -2 is the required solution.
    Check: 5x + 9 = 5 + 3x
    Putting x = -2, we have
    LHS = 5 × (-2) + 9 = -10 + 9 = -1
    RHS = 5 + 3 × (-2) = 5 – 6 = -1
    LHS = RHS
    Hence verified.

    Ex 2.3 Class 8 Maths Question 4.
    4z + 3 = 6 + 2z
    Solution:
    We have 4z + 3 = 6 + 2z
    ⇒ 4z – 2z + 3 = 6 (Transposing 2z to LHS)
    ⇒ 2z + 3 = 6
    ⇒ 2z = 6 – 3 (Transposing 3 to RHS)
    ⇒ 2z = 3
    ⇒ z = \(\frac { 3 }{ 2 }\)
    Hence z = \(\frac { 3 }{ 2 }\) is the required solution.
    Check: 4z + 3 = 6 + 2z
    Putting z = \(\frac { 3 }{ 2 }\), we have
    LHS = 4z + 3 = 4 × \(\frac { 3 }{ 2 }\) + 3 = 6 + 3 = 9
    RHS = 6 + 2z = 6 + 2 × \(\frac { 3 }{ 2 }\) = 6 + 3 = 9
    LHS = RHS
    Hence verified.

    Ex 2.3 Class 8 Maths Question 5.
    2x – 1 = 14 – x
    Solution:
    We have 2x – 1 = 14 – x
    ⇒ 2x + x = 14 + 1 (Transposing x to LHS and 1 to RHS)
    ⇒ 3x = 15
    ⇒ x = 15 ÷ 3 = 5
    Hence x = 5 is the required solution.
    Check: 2x – 1 = 14 – x
    Putting x = 5
    LHS we have 2x – 1 = 2 × 5 – 1 = 10 – 1 = 9
    RHS = 14 – x = 14 – 5 = 9
    LHS = RHS
    Hence verified.

    Ex 2.3 Class 8 Maths Question 6.
    8x + 4 = 3(x – 1) + 7
    Solution:
    We have 8x + 4 = 3(x – 1) + 7
    ⇒ 8x + 4 = 3x – 3 + 7 (Solving the bracket)
    ⇒ 8x + 4 = 3x + 4
    ⇒ 8x – 3x = 4 – 4 [Transposing 3x to LHS and 4 to RHS] ⇒ 5x = 0
    ⇒ x = 0 ÷ 5 [Transposing 5 to RHS] or x = 0
    Thus x = 0 is the required solution.
    Check: 8x + 4 = 3(x – 1) + 7
    Putting x = 0, we have
    8 × 0 + 4 = 3(0 – 1) + 7
    ⇒ 0 + 4 = -3 + 7
    ⇒ 4 = 4
    LHS = RHS
    Hence verified.

    Ex 2.3 Class 8 Maths Question 7.
    x = \(\frac { 4 }{ 5 }\) (x + 10)
    Solution:
    We have x = \(\frac { 4 }{ 5 }\) (x + 10)
    ⇒ 5 × x = 4(x + 10) (Transposing 5 to LHS)
    ⇒ 5x = 4x + 40 (Solving the bracket)
    ⇒ 5x – 4x = 40 (Transposing 4x to LHS)
    ⇒ x = 40
    Thus x = 40 is the required solution.
    Check: x = \(\frac { 4 }{ 5 }\) (x + 10)
    Putting x = 40, we have
    40 = \(\frac { 4 }{ 5 }\) (40 + 10)
    ⇒ 40 = \(\frac { 4 }{ 5 }\) × 50
    ⇒ 40 = 4 × 10
    ⇒ 40 = 40
    LHS = RHS
    Hence verified.

    Ex 2.3 Class 8 Maths Question 8.
    \(\frac { 2x }{ 3 }\) + 1 = \(\frac { 7x }{ 15 }\) + 3
    Solution:
    We have \(\frac { 2x }{ 3 }\) + 1 = \(\frac { 7x }{ 15 }\) + 3
    15(\(\frac { 2x }{ 3 }\) + 1) = 15(\(\frac { 7x }{ 15 }\) + 3)
    LCM of 3 and 15 is 15
    \(\frac { 2x }{ 3 }\) × 15 + 1 × 15 = \(\frac { 7x }{ 15 }\) × 15 + 3 × 15 [Multiplying both sides by 15] ⇒ 2x × 5 + 15 = 7x + 45
    ⇒ 10x + 15 = 7x + 45
    ⇒ 10x – 7x = 45 – 15 (Transposing 7x to LHS and 15 to RHS)
    ⇒ 3x = 30
    ⇒ x = 30 ÷ 3 = 10 (Transposing 3 to RHS)
    Thus the required solution is x = 10
    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Q8

    Ex 2.3 Class 8 Maths Question 9.
    2y + \(\frac { 5 }{ 3 }\) = \(\frac { 26 }{ 3 }\) – y
    Solution:
    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Q9
    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Q9.1

    Ex 2.3 Class 8 Maths Question 10.
    3m = 5m – \(\frac { 8 }{ 5 }\)
    Solution:
    We have
    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Q10
    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Q10.1

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-1

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-2

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-3

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-4

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-5

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-6

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-7

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-8

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-9

    NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 q-10

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