NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2.

- Simple Equations Class 7 Ex 4.1
- Simple Equations Class 7 Ex 4.3
- Simple Equations Class 7 Ex 4.4
- Simple Equations Class 7 MCQ

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Simple Equations |

Exercise |
Ex 4.2 |

Number of Questions Solved |
4 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2

**Question 1.**

Give first the step you will use to separate the variable and then solve the equation:

**(a)** x – 1 = 0

**(b)** x + 1 = 0

**(c)** x – 1 = 5

**(e)** y – 4 = – 7

**(f)** y – 4 = 4

**(g)** y + 4 = 4

**(h)** y + 4 = – 4

**Solution:
**

**(a)**The given equation is x – 1 = 0

Add 1 to both sides,

x – 1 + 1 = 0 + 1 ⇒ x = 1

It is the required solution.

**Check.**Put the solution x = 1 back into the equation.

L.H.S. = x – 1 = 1 = 1 – 0 = R.H.S.

The solution is thus checked for its correctness.

**(b)** The given equation is x + 1 = 0

Subtract 1 from both sides, x + 1 – 1 = 0 – 1 ⇒ x = – 1

It is the required solution.

**Check.** Put the solution x = – 1 back into the equation.

L.H.S. = x + 1 = (-1)+1

= 0 = R.H.S.

The solution is thus checked for its correctness.

**(c)** The given equation is

x – 1 = 5

Add 1 to both sides,

x + 1 – 1 = 5 + 1 ⇒ x = 6

It is the required solution

**Check.** Put the solution x = 6 back into the equation.

L.H.S. = x – 1 = 6 – 1 = 5 = R.H.S.

The solution is thus checked for its correctness.

**(d)** The given equation is x + 6 = 2

Subtract 6 from both sides, x + 6 – 6 = 2 – 6 ⇒ x = – 4

It is the required solution.

**Check.** Put the solution x = – 4 back into the equation.

L.H.S. = x + 6 = – 4 + 6 = 2 = R.H.S.

The solution is thus checked for its correctness.

**(e)** The given equation is y – 4 = – 7

Add 4 to both sides, y – 4 + 4 = – 7 + 4 ⇒ y = – 3

It is the required solution.

**Check.** Put the solution

L.H.S. = y – 4 = – 3 – 4 = – 7 = R.H.S.

The solution is thus checked for its correctness.

**(f)** The given equation is y – 4 = 4

Add 4 to both sides,

y – 4 + 4 = 4 + 4 ⇒ y = 8

It is the required solution.

**Check.** Put the solution y = 8 back into the equation.

L.H.S. = y – 4 = 8 – 4 = 4 = R.H.S.

The solution is thus checked for its correctness.

**(g)** The given equation is y + 4 = 4

Subtract 4 from both sides, y + 4 – 4 = 4 – 4 ⇒ y = 0

It is the required solution.

**Check.** Put the solution y = 0 back into the equation.

L.H.S. =y + 4 = 0 + 4 = 4 = R.H.S.

The solution is thus checked for its correctness.

**(h)** The given equation is y + 4 = – 4

Subtract 4 from both sides, y + 4 – 4 = – 4 – 4 ⇒ y = -8

It is the required solution.

**Check.** Put the solution y = – 8 back into the equation.

L.H.S. = y + 4 = -8 + 4 = – 4 = R.H.S.

The solution is thus checked for its correctness.

**Question 2.**

Give first the step you will use to separate the variable and then solve the equation:

**(a)** 3l = 42

**(b)** \(\frac { b }{ 2 } \) = 6

**(c)** \(\frac { p }{ 7 } \) = 4

**(d)** 4x = 25

**(e)** 8y = 36

**(f)** \(\frac { z }{ 3 } \) = \(\frac { 5 }{ 4 } \)

**(g)** \(\frac { a }{ 5 } \) = \(\frac { 7 }{ 15 } \)

**(h)** 20t = – 10

**Solution:
**

**Question 3.**

Give the steps you will use to separate the variable and then solve the equation :

**(a)** 3n – 2 = 46

**(b)** 5m + 7 = 17

**(c)** \(\frac { 20p }{ 3 } \) = 40

**(d)** \(\frac { 3p }{ 10 } \) = 6

**Solution:
**

**Question 4.**

Solve the following equations:

**Solution:
**

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