NCERT Solutions for Class 7 Mathematics: Chapter 4 – Simple Equations
Check NCERT Solutions for Class 7 Mathematics. With this article, you can access solutions of all the questions of CBSE Class 7 Maths subject.
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.
S. No. 
Equation 
Value 
Say, whether the Equation is Satisfied. (Yes/ No) 
(i) 
x + 3 = 0 
x = 3 

(ii) 
x + 3 = 0 
x = 0 

(iii) 
x + 3 = 0 
x = – 3 

(iv) 
x – 7 = 1 
x = 7 

(v) 
x – 7 = 1 
x = 8 

(vi) 
5x = 25 
x = 0 

(vii) 
5x = 25 
x = 5 

(viii) 
5x = 25 
x = – 5 

(ix) 
m/3 = 2 
m = – 6 

(x) 
m/3 = 2 
m = 0 

(xi) 
m/3 = 2 
m = 6 

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
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