NCERT Class 7 Maths Simple Equations
The chapter 4 begins with an introduction to Simple Equations by explaining the basic concept of setting up of an equation and recalling the concepts learnt earlier classes such as variables.Then the concept of equation and equality is discussed in detail.Now, the procedure to solve an equation and the steps involved in it is explained elaborately.Now, the reverse process i.e. to obtain the equation from the solution is explained.In the last section of the chapter, applications of simple equations to practical situations is dealt in detail.
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Exercise 4.1
Chapter 4 Ex.4.1 Question 1
Complete the last column of the table.
S. No 
Equation 
Value 
Say, whether the equation 
(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) 
\(\begin{align}\frac{m}{3} = 2\end{align}\) 
\(m= 6\) 

(x) 
\(\begin{align}\frac{m}{3} = 2\end{align}\) 
\(m= 0\) 

(xi) 
\(\begin{align}\frac{m}{3} = 2\end{align}\) 
\(m= 6\) 

Solution
What is Known?
Equation and the value of the variable.
What is unknown?
Whether the given value is a solution of the equation or not.
Reasoning:
Put the value of the variable in the given equation. If LHS is equal to the RHS then the equation is satisfied and if it is not that means this given value is not a solution of the equation.
Steps:
(i) \(\begin{align}x + 3 = 0,x = 3\end{align}\)
L.H.S \(=\) \(x+ 3\), R.H.S \(= 0\)
By putting, \(x= 3\)
L.H.S \( =3 + 3= 6 ≠ \)R.H.S
Therefore, “No”, the equation is not satisfied.
(ii) \(\begin{align}x+ 3 = 0,x= 0\end{align}\)
L.H.S \(=\) \(x+ 3\), R.H.S\(= 0\)
By putting, \(x= 0;\)
L.H.S \( =0+3= 3 ≠\)R.H.S
Therefore, “No”, the equation is not satisfied.
(iii) \(\begin{align}x+ 3 = 0, x= 3\end{align}\)
L.H.S \(=\) \(x+ 3\), R.H.S \(= 0\)
By putting, \(x= 3\)
L.H.S \(= 3 + 3 = 0 =\) R.H.S
Therefore, “Yes”, the equation is satisfied.
(iv) \(\begin{align}x– 7 = 1, x= 7\end{align}\)
L.H.S \(=\) \(x\,– 7\), R.H.S \(= 1\)
By putting, \(x= 7\);
L.H.S \(= 7 \,– 7 = 0 ≠\) R.H.S
Therefore, “No”, the equation is not satisfied.
(v) \(\begin{align}x– 7 = 1, x= 8\end{align}\)
L.H.S \(=\) \(x– 7\), R.H.S \(= 1\)
By putting, \(x= 8;\)
L.H.S \(= 8 \,– 7 = 1 =\) R.H.S
Therefore, “Yes”, the equation is satisfied.
(vi) \(\begin{align}5x= 25, x= 0,\end{align}\)
L.H.S \(=\) \(5x\), R.H.S\( = 25\)
By putting, \(x= 0;\)
L.H.S \(=\) \(5 \times 0=0 ≠ \)R.H.S
Therefore, “No”, the equation is not satisfied.
(vii) \(\begin{align}5x= 25, x= 5,\end{align}\)
L.H.S \(= 5x\), R.H.S \(= 25\)
By putting, \(x= 5;\)
L.H.S \(=\) \(5 \times 5 = 25\)= R.H.S
Therefore, “Yes”, the equation is satisfied.
(viii) \(\begin{align}5x = 25, x= – 5,\end{align}\)
L.H.S \(=\) \(5x\), R.H.S \(= 25\)
By putting, \(x= – 5;\)
L.H.S \(=\) \(5 \times \left( {  5} \right) =  25≠\)R.H.S
Therefore, “No”, the equation is not satisfied.
(ix) \(\begin{align}\frac{m}{3} = 2, m= – 6,\end{align}\)
L.H.S \(=\) \(\begin{align}\frac{m}{3}\end{align}\), R.H.S \(= 2\)
By putting, \(m\)\(= – 6;\)
L.H.S \(=\) \(\begin{align}\frac{{  6}}{3}=−2≠\end{align}\) R.H.S
Therefore, “No”, the equation is not satisfied.
(x) \(\begin{align}\frac{m}{3} = 2, m=0\end{align}\)
L.H.S \(=\) \(\begin{align}\frac{m}{3}\end{align}\), R.H.S \(= 2\)
By putting, \(m= 0;\)
L.H.S \(=\) \(\begin{align}\frac{0}{3}=0≠\end{align}\)R.H.S
Therefore, “No”, the equation is not satisfied.
(xi) \(\begin{align}\frac{m}{3} = 2 m= 6,\end{align}\)
L.H.S \(=\) \(\begin{align}\frac{m}{3}\end{align}\), R.H.S \(= 2\)
By putting, \(m\) \(= 6;\)
L.H.S \(=\) \(\begin{align}\frac{6}{3}= 2 =\end{align}\) R.H.S
Therefore, “Yes”, the equation is satisfied
S. No 
Equation 
Value 
Say, weather the equation is satisfied. (Yes/No) 
(i) 
\(x + 3 = 0\) 
\(x= 3\) 
No 
(ii) 
\(x + 3 = 0\) 
\(x= 0\) 
No 
(iii) 
\(x + 3 = 0\) 
\(x= 3\) 
Yes 
(iv) 
\(x  7 = 1\) 
\(x= 7\) 
No 
(v) 
\(x  7 = 1\) 
\(x= 8\) 
Yes 
(vi) 
\(5x= 25\) 
\(x= 0\) 
No 
(vii) 
\(5x= 25\) 
\(x= 5\) 
Yes 
(viii) 
\(5x= 25\) 
\(x= 5\) 
No 
(ix) 
\(\begin{align}\frac{m}{3} = 2\end{align}\) 
\(m= 6\) 
No 
(x) 
\(\begin{align}\frac{m}{3} = 2\end{align}\) 
\(m= 0\) 
No 
(xi) 
\(\begin{align}\frac{m}{3} = 2\end{align}\) 
\(m= 6\) 
Yes 
Chapter 4 Ex.4.1 Question 2
Check whether the value given in the brackets is a solution to the given equation or not:
(a) \(n + 5 = 19\left( {n = 1} \right)\)
(b) \(7n + 5 = 19\left( {n = 2} \right)\)
(c) \(7n + 5 = 19\left( {n = 2} \right)\)
(d) \(4p3 = 13\left( {p = 1} \right)\)
(e) \(4p3 = 13\left( {p = 4} \right)\)
(f) \(4p3 = 13\left( {p = 0} \right)\)
Solution
What is unknown?
Whether the given value is a solution of the equation or not.
What is Known?
Equation and the value of the variable.
Reasoning:
Put the value of the given variable in the equation. If LHS is equal to the RHS then the equation is satisfied and if it is not that means this given value is not a solution of the equation
Steps:
a) Here, \(n+ 5\) is L.H.S, \(19\) is R.H.S and \(n= 1\) (given)
L.H.S = \(n+ 5,\)
By putting, \(n= 1,\)
L.H.S \(= 1 + 5 = 6 ≠\) R.H.S
L.H.S \(≠\) R.H.S, so \(n= 1\) is not a solution of the equation.
b) Here, \(7n+ 5\) is L.H.S, \(19\) is R.H.S and \(n= – 2\) (given)
L.H.S = \(7n+ 5\),
By putting, \(n= \,– 2\),
L.H.S\( =7 \times (−2) + 5= −9 ≠\) R.H.S
As, L.H.S \(≠\) R.H.S, so \(n= \,– 2\) is not a solution of the equation.
c) Here, \(7n+ 5\) is L.H.S, \(19\) is R.H.S and \(n= 2\) (given)
L.H.S = \(7n+ 5\),
By putting, \(n= 2\),
L.H.S \(= 7 \times (2) + 5 = 19 = \)R.H.S
As, L.H.S = R.H.S, so \(n= 2\) is a solution of the equation.
d) Here, \(4p\,– 3\) is L.H.S, \(13\) is R.H.S and \(p= 1\) (given)
L.H.S = \(4p\,– 3,\)
By putting, \(p= 1\),
L.H.S \(=4 \times (1)\, – 3 = 1 ≠ \) R.H.S
As, L.H.S \(≠\) R.H.S, so \(p= 1\) is not a solution of the equation.
e) Here, \(4p– 3\) is L. H.S, \(13\) is R.H.S and \(p= 4\) (given)
L.H.S \(=4p– 3\),
By putting, \(p= 1\),
L.H.S \(=4 \times (4) – 3 = 19 ≠\) R.H.S
As, L.H.S \(≠\) R.H.S, so \(p= 1\) is not a solution.
f) Here, \(4p– 3\) is L. H.S, \(13\) is R.H.S and \(p= 0\) (given data)
L.H.S \(=\) \(4p\,– 3\),
By putting, \(p= 0\),
L.H.S \(=4 \times(0)–3=−3≠ \)R.H.S
As, L.H.S\(≠\)R.H.S, so \(p= 0\) is not a solution of the equation.
Chapter 4 Ex.4.1 Question 3
Solve the equation given below by trial and error method:
(i) \(\begin{align}5p + 2 = 17\end{align}\)
(ii) \(\begin{align}3m{\rm{1}}4 = 4\end{align}\)
Solution
What is Known?
Equations
What is unknown?
Solution of the equation or the value of the variable.
Reasoning:
Put the different values of the variable in the given equation. If LHS is equal to the RHS then the equation is satisfied and if it is not that means this variable is not a solution of the equation.
Steps:
(i) \(5p + 2 = 17\)
\(5p + 2=\) L.H.S
By putting, \(p= 0\),
\(5 \times 0+2=2≠17\)
By putting, \(p=1\),
\(5 \times (1)+2=7≠17\)
By putting, \(p= 2\),
\(5 \times (2) + 2 = 12 ≠\) R.H.S
By putting, \(p= 3\),
\( 5\times (3) + 2 = 17 =\) R.H.S
Therefore, \(p= 3\) is a solution of the equation.
(ii) \(3m{\rm{1}}4 = 4\)
\(3m{\rm{1}}4=\) L.H.S
By putting, \(m= 5\),
\(3 \times (5)–14=1≠6\)
By putting, \(m= 6\),
\(3 \times (6)–14 =4 =\)R.H.S.
Therefore, \(m= 6\) is a solution of the equation.
Chapter 4 Ex.4.1 Question 4
Write equations for the following statements:
(i) The sum of numbers \(x\) and is
(ii) subtracted from \(y\) is
(iii) Ten times \(a\) is
(iv) The number \(b\) divided by gives
(v) Three fourth of \(t\) is
(vi) Seven times \(m\) plus gets you
(vii) One fourth of a number \(x\) minus gives
(viii) If you take away from times \(y\), you get
(ix) If you add to one third of , you get
Solution
What is Known?
Statements of the equations.
What is unknown?
Equations for the given statement.
Reasoning:
This question is very simple. Read the statement carefully and frame the equation in steps.
Steps:
(i) \(x~+4=9\)
(ii) \(y2 = 8\)
(iii) \(10a = 70\)
(iv) \(\begin{align}\frac{b}{5} = 6\end{align}\)
(v)\(\begin{align} \frac{3}{4}t = 15\end{align}\)
(vi) \(7m + 7 = 77\)
(vii) \(\begin{align}\frac{1}{4}x4 = 4\end{align}\)
(viii) \(6y6 = 60\)
(ix)\(\begin{align}\frac{1}{3}z + 3 = 30\end{align}\)
Chapter 4 Ex.4.1 Question 5
Write the following equations in statement forms:
(i) \(p + 4 = 15\)
(ii) \(m7=3\)
(iii) \(2m=7\)
(iv)\(\begin{align}\frac{m}{5} = 3\end{align}\)
(v) \(\begin{align}\frac{3m}{5}=6~\end{align}\)
(vi) \(3p+4=25~\)
(vii) \(4p2=18~\)
(viii) \(\begin{align}\frac{p}{2} + 2{\rm{ = }}8\end{align}\)
Solution
What is Known?
Equations
What is unknown?
Statements of the given equations.
Reasoning:
To solve this question, look at the equation carefully and decide the variable and numbers. Then write the suitable statement using the variables and numbers.
Steps:
(i) The sum of \(p\) and \(4\) is \(15.\)
(ii) 7 subtracted from \(m\) is \(3.\)
(iii) Two times \(m\) is \(7.\)
(iv) Onefifth of \(m\) is \(3.\)
(v) Threefifth of \(m\) is \(6.\)
(vi) When \(4\) is added to three times of a number \(p,\) it gives \(25.\)
(vii) When \(2\) is subtracted from four times of a number \(p\), gives \(18.\)
(viii) When \(2\) is added to half of \(p\) gives \(8.\)
Chapter 4 Ex.4.1 Question 6
Set up an equation in the following cases:
(i) Irfan says that he has \(7\) marbles more than five times the marbles Parmit has. Irfan has \(37\) marbles. (Take \(‘m’\) to be the number of Parmit’s marbles.)
(ii) Laxmi’s father is \(49\) years old. He is \( 4\) years older than three times Laxmi’s age. (Take Laxmi’s age to be \(y\) years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus \(7.\) The highest score is \(87.\) (Take the lowest score to be \(l\).)
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be \(b\) in degrees. Remember that the sum of angles of a triangle is \(180\) degrees).
Solution
What is unknown?
Equations for the different statements.
What is Known?
Statements of the equations.
Reasoning:
Read the statement carefully and frame the equation.
Steps:
(i) Let permit has \(m\) number of marbles
Number of marbles Irfan has \(=5m + 7\)
Total number of marbles Irfan has \(37\)
So, \(5m + 7=37\)
(ii) Let the age of Laxmi be \(y\) years
Laxmi’s father is four years older than three times Laxmi’s age \(=3y + 4\)
Age of Laxmi’s father is \(49\) years
According to question,
\(3y + 4= 49\)
(iii) Let the lowest marks obtained by the student be \(l\)
Highest marks obtained by the student be \(2l + 7\)
And the highest score is \(87\)
Therefore, According to question
\(2l + 7= 87\)
(iv) Let the base angle of a triangle be \(b\)
Vertex angle of the triangle \(= 2b\)
According to question,
\[\begin{align}b + b + 2b = 180^\circ\\ 4b = 180^\circ\end{align}\]