NCERT Solutions for Class 7 Mathematics: Chapter 4 – Simple Equations

Get NCERT Solutions for Class 7 Mathematics,  Chapter 4 – Simple Equations. This chapter is one of the most important chapters of CBSE Class 7 Mathematics subject. Students of Class 7 are advised to prepare this chapter well as concepts given in this chapter are required to solve questions of other chapters.

NCERT Solutions for Class 7 Mathematics  Chapter 4 – Simple Equations

1. Complete the last column of the table.

Answer 1

(i)

Given, x + 3 = 0

Now, L.H.S. = x + 3

Putting x = 3, we have,

L.H.S. = 3 + 3 = 6 ≠ R.H.S.

So, the equation is not satisfied.

(ii) 

Given, x + 3 = 0

Now, L.H.S. = x + 3

Putting x = 0, we have

L.H.S. = 0 + 3 = 3 ≠ R.H.S.

So, the equation is not satisfied.

(iii)

Given, x + 3 = 0

Now, L.H.S. = x + 3

Putting x = −3, we have

L.H.S. = − 3 + 3 = 0 = R.H.S.

So, the equation is satisfied.

(iv) 

Given, x − 7 = 1

Now, L.H.S. = x − 7

Putting x = 7,

We have, L.H.S. = 7 − 7 = 0 ≠ R.H.S.

So, the equation is not satisfied.

(v)

Given, x − 7 = 1

Now, L.H.S. = x − 7

Putting x = 8,

We have, L.H.S. = 8 − 7 = 1 = R.H.S.

So, the equation is satisfied.

(vi)

Now, 5x = 25

Now, L.H.S. = 5x

Putting x = 0,

We have, L.H.S. = 5 × 0 = 0 ≠ R.H.S.

So, the equation is not satisfied.

(vii)

Given, 5x = 25

Now, L.H.S. = 5x

Putting x = 5,

We have, L.H.S. = 5 × 5 = 25 = R.H.S.

So, the equation is satisfied.

(viii)

Given, 5x = 25

Now, L.H.S. = 5x

Putting x = −5,

We have, L.H.S. = 5 × (−5) = −25 ≠ R.H.S.

So, the equation is not satisfied.

(ix) 

Given, m/3 = 2

Now, L.H.S. = m/3

Putting m = −6,

We have, L. H. S. = -6/3 = -2 ≠ R.H.S.

So, the equation is not satisfied.

(x) 

Given, m/3 = 2

Now, L.H.S. = m/3

Putting m = 0,

We have, L.H.S. = 0/3 = 0 ≠ R.H.S.

So, the equation is not satisfied.

(xi) 

Given, m/3 = 2

Now, L.H.S. = m/3

By putting m = 6,

We have L.H.S. = 6/3 = 2 = R.H.S.

So, the equation is satisfied.

2. Check whether the value given in the brackets is a solution to the given equation or not:

(a) n + 5 = 19 (n = 1)

(b) 7n + 5 = 19 (n = – 2)

(c) 7n + 5 = 19 (n = 2)

(d) 4p – 3 = 13 (p = 1)

(e) 4p – 3 = 13 (p = – 4)

(f) 4p – 3 = 13 (p = 0)

Solution:

(a) Given, n + 5 = 19 (n = 1)

Putting n = 1 in L.H.S.,

We have, n + 5 = 1 + 5 = 6 ≠ 19

Now, L.H.S. ≠ R.H.S.,

So, n = 1 is not a solution of the given equation.

(b) Given, 7n + 5 = 19 (n = −2)

Putting n = −2 in L.H.S.,

We have, 7n + 5 = 7 × (−2) + 5 = −14 + 5 = −9 ≠ 19

Now, L.H.S. ≠ R.H.S.,

So, n = −2 is not a solution of the given equation.

(c) Given, 7n + 5 = 19 (n = 2)

Putting n = 2 in L.H.S.,

We have, 7n + 5 = 7 × (2) + 5 = 14 + 5 = 19 = R.H.S.

Now, L.H.S. = R.H.S.,

So, n = 2 is a solution of the given equation.

(d) Given, 4p − 3 = 13 (p = 1)

Putting p = 1 in L.H.S.,

We have, 4p − 3 = (4 × 1) − 3 = 1 ≠ 13

As L.H.S ≠ R.H.S.,

So, p = 1 is not a solution of the given equation, 4p − 3 = 13.

(e) Given, 4p − 3 = 13 (p = −4)

Putting p = −4 in L.H.S.,

We have, 4p − 3 = 4 × (−4) − 3 = − 16 − 3 = −19 ≠ 13

As L.H.S. ≠ R.H.S.,

So, p = −4 is not a solution of the given equation

(f) Given, 4p − 3 = 13 (p = 0)

Putting p = 0 in L.H.S.,

We have, 4p − 3 = (4 × 0) − 3 = −3 ≠ 13

As, L.H.S. ≠ R.H.S.,

So, p = 0 is not a solution of the given equation.

3. Solve the following equations by trial and error method:

(i) 5p + 2 = 17

(ii) 3m – 14 = 4

Solutions:

(i) Given, 5p + 2 = 17

Putting p = 2 in L.H.S.,

We have, (5 × 2) + 2 = 10 + 2 = 12 ≠ R.H.S.

Putting p = 3 in L.H.S.,

We have (5 × 3) + 2 = 17 = R.H.S.

At p = 3, L.H.S. = R.H.S, so  p = 3 is a solution of the given equation.

(ii) 3m − 14 = 4

Putting m = 6,

We have, (3 × 6) − 14 = 18 − 14 = 4 = R.H.S.

At, m= 6, L.H.S. = R.H.S., so, m = 6 is a solution of the given equation.

More questions will be available here shortly.

Also check:

NCERT Solutions for Class 7 Maths: Chapter 1 - Integers

NCERT Solutions for Class 7 Maths: Chapter 2 - Fractions & Decimals

NCERT Solutions for Class 7 Maths: Chapter 3 - Data Handling