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Solutions to Exercise 4.1 (Page No 63) of NCERT Class 7 Math 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 = -6 | ||
| (x) | m = 0 | ||
| (xi) | m = 6 |
Answer:
The completed table is shown below:
| S. No. | Equation | Value | Say, whether 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) | m = -6 | No | |
| (x) | m = 0 | No | |
| (xi) | m = 6 | Yes |
The calculations are shown below:
(i) The equation is x + 3 = 0.
LHS = x + 3
= 3 + 3 (Putting x = 3)
= 6
RHS = 0
We can see LHS ≠ RHS (since 6 ≠ 0) and therefore the equation is not satisfied.
(ii) The equation is x + 3 = 0.
LHS = x + 3
= 0 + 3 (Putting x = 0)
= 3
RHS = 0
We can see LHS ≠ RHS (since 3 ≠ 0) and therefore the equation is not satisfied.
(iii) The equation is x + 3 = 0.
LHS = x + 3
= -3 + 3 (Putting x = -3)
= 0
RHS = 0
We can see LHS = RHS (both = 0) and therefore the equation is satisfied.
(iv) The equation is x – 7 = 1.
LHS = x – 7
= 7 – 7 (Putting x = 7)
= 0
RHS = 1
We can see LHS ≠ RHS (since 0 ≠ 1) and therefore the equation is not satisfied.
(v) The equation is x – 7 = 1.
LHS = x – 7
= 8 – 7 (Putting x = 8)
= 1
RHS = 1
We can see LHS = RHS (both = 1) and therefore the equation is satisfied.
(vi) The equation is 5x = 25.
LHS = 5x
= 5 × 0 (Putting x = 0)
= 0
RHS = 25
We can see LHS ≠ RHS (since 0 ≠ 25) and therefore the equation is not satisfied.
(vii) The equation is 5x = 25.
LHS = 5x
= 5 × 5 (Putting x = 5)
= 25
RHS = 25
We can see LHS = RHS (both = 25) and therefore the equation is satisfied.
(viii) The equation is 5x = 25.
LHS = 5x
= 5 × (-5) (Putting x = -5)
= -25
RHS = 25
We can see LHS ≠ RHS (since -25 ≠ 25) and therefore the equation is not satisfied.
(ix) The equation is m/3 = 2.
LHS = m/3
= -6/3 (Putting m = -6)
= -2
RHS = 2
We can see LHS ≠ RHS (since -2 ≠ 2) and therefore the equation is not satisfied.
(x) The equation is m/3 = 2.
LHS = m/3
= 0/3 (Putting x = 0)
= 0
RHS = 2
We can see LHS ≠ RHS (since 0 ≠ 2) and therefore the equation is not satisfied.
(xi) The equation is m/3 = 2.
LHS = m/3
= 6/3 (Putting x = 6)
= 2
RHS = 2
We can see LHS = RHS (since 2 = 2) and therefore 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)
Answer:
The equation is n + 5 = 19.
LHS = n + 5
= 1 + 5 (Putting n = 1)
= 6
RHS = 19
We can see LHS ≠ RHS and therefore, n = 1 is not a solution to the given equation.
(b) 7n + 5 = 19 (n = – 2)
Answer:
The equation is 7n + 5 = 19.
LHS = 7n + 5
= 7 × (-2) + 5 (Putting n = -2)
= -9
RHS = 19
We can see LHS ≠ RHS and therefore, n = -2 is not a solution to the given equation.
(c) 7n + 5 = 19 (n = 2)
Answer:
The equation is 7n + 5 = 19.
LHS = 7n + 5
= 7 × (2) + 5 (Putting n = 2)
= 19
RHS = 19
We can see LHS = RHS and therefore, n = 2 is a solution to the given equation.
(d) 4p – 3 = 13 (p = 1)
Answer:
The equation is 4p – 3 = 13.
LHS = 4p – 3
= 4 × 1 – 3 (Putting p = 1)
= 1
RHS = 13
We can see LHS ≠ RHS and therefore, p = 1 is not a solution to the given equation.
(e) 4p – 3 = 13 (p = – 4)
Answer:
The equation is 4p – 3 = 13.
LHS = 4p – 3
= 4 × (-4) – 3 (Putting p = -4)
= -19
RHS = 13
We can see LHS ≠ RHS and therefore, p = -4 is not a solution to the given equation.
(f) 4p – 3 = 13 (p = 0)
Answer:
The equation is 4p – 3 = 13.
LHS = 4p – 3
= 4 × (0) – 3 (Putting p = 0)
= -3
RHS = 13 We can see LHS ≠ RHS and therefore, p = -4 is not a solution to the given equation.
3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17
Answer:
Let us try putting p = 0:
LHS = 5p + 2
= 5 × (0) + 2
= 2
RHS = 17
We can see LHS ≠ RHS and therefore, p = 0 is not a solution to the given equation.
Let us try putting p = 1:
LHS = 5p + 2
= 5 × (1) + 2
= 7
RHS = 17
We can see LHS ≠ RHS and therefore, p = 1 is not a solution to the given equation.
Let us try putting p = 2:
LHS = 5p + 2
= 5 × (2) + 2
= 12
RHS = 17
We can see LHS ≠ RHS and therefore, p = 2 is not a solution to the given equation.
Let us try putting p = 3:
LHS = 5p + 2
= 5 × (3) + 2
= 17
RHS = 17
We can see LHS = RHS and therefore, p = 3 is a solution to the given equation.
(ii) 3m – 14 = 4
Answer:
Let us try putting m = 1:
LHS = 3m – 14
= 3 × (1) – 14
= 3 – 14
= -11
RHS = 4
We can see LHS ≠ RHS and therefore, m = 1 is not a solution to the given equation.
Let us try putting m = 2:
LHS = 3m – 14
= 3 × (2) – 14
= 6 – 14
= -8
RHS = 4
We can see LHS ≠ RHS and therefore, m = 2 is not a solution to the given equation.
Let us try putting m = 3:
LHS = 3m – 14
= 3 × (3) – 14
= 9 – 14
= -5
RHS = 4
We can see LHS ≠ RHS and therefore, m = 3 is not a solution to the given equation.
Let us try putting m = 4:
LHS = 3m – 14
= 3 × (4) – 14
= 12 – 14
= -2
RHS = 4
We can see LHS ≠ RHS and therefore, m = 4 is not a solution to the given equation.
Let us try putting m = 5:
LHS = 3m – 14
= 3 × (5) – 14
= 15 – 14
= 1
RHS = 4
We can see LHS ≠ RHS and therefore, m = 5 is not a solution to the given equation.
Let us try putting m = 6:
LHS = 3m – 14
= 3 × (6) – 14
= 18 – 14
= 4
RHS = 4 In this case, we can see that LHS = RHS and therefore, m = 6 is the required solution.
4. Write equations for the following statements.
(i) The sum of numbers x and 4 is 9.
Answer:
The above statement can be expressed as the following equation:
x + 4 = 9
(ii) 2 subtracted from y is 8.
Answer:
The above statement can be expressed as the following equation:
y – 2 = 8
(iii) Ten times a is 70.
Answer:
The above statement can be expressed as the following equation:
10a = 70
(iv) The number b divided by 5 gives 6.
Answer:
The above statement can be expressed as the following equation:
b/5 = 6
(v) Three-fourths of t is 15.
Answer:
The above statement can be expressed as the following equation:
(3/4)t = 15
or, 3t/4 = 15
(vi) Seven times m plus 7 gets you 77.
Answer:
The above statement can be expressed as the following equation:
7m + 7 = 77
(vii) One-fourth of a number x minus 4 gives 4.
Answer:
The above statement can be expressed as the following equation:
(1/4)x – 4 = 4
or x/4 – 4 = 4
(viii) If you take away 6 from 6 times y, you get 60.
Answer:
The above statement can be expressed as the following equation:
6y – 6 = 60
(ix) If you add 3 to one-third of z, you get 30.
Answer:
The above statement can be expressed as the following equation:
(1/3)z + 3 = 30
or, z/3 + 3 = 30
5. Write the following equations in statement forms.
(i) p + 4 = 15
Answer:
The sum of numbers p and 4 is 15.
(ii) m – 7 = 3
Answer:
7 subtracted from m is 3.
(iii) 2m = 7
Answer:
Two times m is 7.
(iv) m/5 = 3
Answer:
The number m divided by 5 gives 3.
(v) 3m/5 = 6
Answer:
Three-fifth of m is 6.
(vi) 3p + 4 = 25
Answer:
Three times p plus 4 gives 25.
(vii) 4p – 2 = 18
Answer:
If you take away 2 from 4 times p, you get 18.
(viii) p/2 + 2 = 8
Answer:
If you add 2 to half of p, you get 8.
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).
Answer:
Let Parmit has m marbles.
Five times Parmit’s marbles = 5m. Irfan has 7 marbles more than 5m. Therefore, Irfan has (5m + 7) marbles.
It is also given that Irfan has 37 marbles.
Therefore, the required equation is: 5m + 7 = 37.
(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.)
Answer:
Laxmi’s age is y years.
Three times Laxmi’s age is 3y years. Laxmi’s father’s age is 4 years more than3y years. Therefore, Laxmi’s father is (3y + 4) years.
It is also given that Laxmi’s father is 49 years old.
Therefore, the required equation is: 3y + 4 = 49.
(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.)
Answer:
The lowest marks is l.
Twice the lowest marks = 2l. The highest marks obtained by a student is (2l + 7).
It is also given that the highest score is 87.
Therefore, the required equation is: 2l + 7 = 87.
(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).
Answer:
The base angle is b degrees.
The vertex angle = 2b.
There are two base angles and one vertex angle. Therefore, the sum of the angles = b + b + 2b = 4b.
It is also given that the sum of angles of a triangle is 180 degrees.
Therefore, the required equation is: 4b = 1800.
Solutions to Exercise 4.2 (Page No 68) of NCERT Class 7 Math Chapter 4 Simple Equations –
1. Give first the step you will use to separate the variable and then solve the equation.
(a) x – 1 = 0
Answer:
To separate the variable,we should add 1 to both sides.
Therefore, New LHS = x – 1 + 1 = x and New RHS = 0 + 1 = 1.
Therefore, x = 1 is the required solution.
(b) x + 1 = 0
Answer:
To separate the variable,we should subtract 1 from both sides.
Therefore, New LHS = x + 1 – 1 = x and New RHS = 0 – 1 = -1.
Therefore, x = -1 is the required solution.
(c) x – 1 = 5
Answer:
To separate the variable,we should add 1 to both sides.
Therefore, New LHS = x – 1 + 1 = x and New RHS = 5 + 1 = 6.
Therefore, x = 6 is the required solution.
(d) x + 6 = 2
Answer:
To separate the variable,we should subtract 6 from both sides.
Therefore, New LHS = x + 6 – 6 = x and New RHS = 2 – 6 = -4.
Therefore, x = -4 is the required solution.
(e) y – 4 = – 7
Answer:
To separate the variable,we should add 4 to both sides.
Therefore, New LHS = y – 4 + 4 = y and New RHS = -7 + 4 = -3.
Therefore, y = -3 is the required solution.
(f) y – 4 = 4
Answer:
To separate the variable,we should add 4 to both sides.
Therefore, New LHS = y – 4 + 4 = y and New RHS = 4 + 4 = 8.
Therefore, y = 8 is the required solution.
(g) y + 4 = 4
Answer:
To separate the variable,we should subtract 4 from both sides.
Therefore, New LHS = y + 4 – 4 = y and New RHS = 4 – 4 = 0.
Therefore, y = 0 is the required solution.
(h) y + 4 = – 4
Answer:
To separate the variable,we should subtract 4 from both sides.
Therefore, New LHS = y + 4 – 4 = y and New RHS = -4 – 4 = -8. Therefore, y = -8 is the required solution.
2. Give first the step you will use to separate the variable and then solve the equation:
(a) 3l = 42
Answer:
To separate the variable, we shall divide both the sides by 3. This will give us just l on LHS.
New LHS = 3l/3 = (3 × l)/3 = l and New RHS = 42/3 = 14
Therefore, l = 14 is the required solution.
(b) b/2 = 6
Answer:
To separate the variable, we shall multiply both the sides by 2. This will give us just b on LHS.
New LHS = b/2 × 2 = (b × 2)/2 = b and New RHS = 6 × 2 = 12
Therefore, b = 12 is the required solution.
(c) p/7 = 4
Answer:
To separate the variable, we shall multiply both the sides by 7. This will give us just p on LHS.
New LHS = p/7 × 7 = (p × 7)/7 = p and New RHS = 4 × 7 = 28
Therefore, p = 28 is the required solution.
(d) 4x = 25
Answer:
To separate the variable, we shall divide both the sides by 4. This will give us just x on LHS.
New LHS = 4x/4 = (4 × x)/4 = x and New RHS = 25/4 = 6 1/4
Therefore, x = 6 1/4 is the required solution.
(e) 8y = 36
Answer:
To separate the variable, we shall divide both the sides by 8. This will give us just y on LHS.
New LHS = 8y/8 = (8 × y)/8 = y and New RHS = 36/8 = (36 ÷ 4)/(8 ÷ 4) = 9/2 = 4 1/2
Therefore, y = 4 1/2 is the required solution.
(f) z/3 = 5/4
To separate the variable, we shall multiply both the sides by 3. This will give us just z on LHS.
New LHS = z/3 × 3 = (z × 3)/3 = z and New RHS = 5/4 × 3 = (5 × 3)/4 = 15/4 = 3 3/4
Therefore, z = 3 3/4 is the required solution.
(g) a/5 = 7/15
Answer:
To separate the variable, we shall multiply both the sides by 5. This will give us just a on LHS.
New LHS = a/5 × 5 = (a × 5)/5 = a and New RHS = 7/15 × 5 = (7 × 5)/15 = 7/3 = 2 1/3
Therefore, a = 2 1/3 is the required solution.
(h) 20t = – 10
Answer:
To separate the variable, we shall divide both the sides by 20. This will give us just t on LHS.
New LHS = 20t/20 = (20 × t)/20 = t and New RHS = (-10)/20 = – 1/2
Therefore, t = – 1/2 is the required solution.
3. Give the steps you will use to separate the variable and then solve the equation.
(a) 3n – 2 = 46
Answer:
To separate the variable, we follow these steps:
Step 1:
We add 2 to both sides. Therefore, New LHS = 3n – 2 + 2 = 3n and New RHS = 46 + 2 = 48.
Therefore, we get 3n = 48.
Step 2:
To separate the variable, we shall divide both the sides by 3. This will give us just n on LHS.
New LHS = 3n/3 = (3 × n)/3 = n and New RHS = 48/3 = 16.
Therefore, the solution is n = 16.
(b) 5m + 7 = 17
Answer:
To separate the variable, we follow these steps:
Step 1:
We subtract 7 from both sides. Therefore, New LHS = 5m + 7 – 7 = 5m and New RHS = 17 – 7 = 10.
Therefore, we get 5m = 10.
Step 2:
To separate the variable, we shall divide both the sides by 5. This will give us just m on LHS.
New LHS = 5m/5 = (5 × m)/5 = m and New RHS = 10/5 = 2.
Therefore, the solution is m = 2.
(c) 20p/3 = 40
Answer:
To separate the variable, we follow these steps:
Step 1:
We multiply both sides by 3. Therefore, New LHS = 20p/3 × 3 = (20p × 3)/3 = 20p and New RHS = 40 × 3 = 120.
Therefore, we get 20p = 120.
Step 2:
To separate the variable, we shall divide both the sides by 20. This will give us just p on LHS.
New LHS = 20p/20 = (20 × p)/20 = p and New RHS = 120/20 = 6.
Therefore, the solution is p = 6.
(d) 3p/10 = 6
Answer:
To separate the variable, we follow these steps:
Step 1:
We multiply both sides by 10. Therefore, New LHS = 3p/10 × 10 = (3p × 10)/10 = 3p and New RHS = 6 × 10 = 60.
Therefore, we get 3p = 60.
Step 2:
To separate the variable, we shall divide both the sides by 3. This will give us just p on LHS.
New LHS = 3p/3 = p and New RHS = 60/3 = 20.
Therefore, the solution is p = 20.
4. Solve the following equations.
(a) 10p = 100
Answer:
To separate the variable, we shall divide both the sides by 10. This will give us just p on LHS.
New LHS = 10p/10 = p and New RHS = 100/10 = 10.
Therefore, the solution is p = 10.
(b) 10p + 10 = 100
Answer:
To separate the variable, we follow these steps:
Step 1:
To separate the variable,we should subtract 10 from both sides.
New LHS = 10p + 10 – 10 = 10pand New RHS = 100 – 10 = 90.
Therefore, we get 10p = 90.
Step 2:
To separate the variable, we shall divide both the sides by 10. This will give us just p on LHS.
New LHS = 10p/10 = p and New RHS = 90/10 = 9.
Therefore, the solution is p = 9.
(c) p/4 = 5
Answer:
To separate the variable, we multiply both sides by 4. Therefore, New LHS = p/4 × 4 = (p × 4)/4 = p and New RHS = 5 × 4 = 20.
Therefore, the solution is p = 20.
(d) (-p)/3 = 5
Answer:
To separate the variable, we multiply both sides by -3. Therefore, New LHS = (-p)/3 × (-3) = (-p ×(-3))/3 = p and New RHS = 5 × (-3) = -15.
Therefore, the solution is p = -15.
(e) 3p/4 = 6
Answer:
To separate the variable, we follow these steps:
Step 1:
We multiply both sides by 4. Therefore, New LHS = 3p/4 × 4 = (3p × 4)/4 = 3p and New RHS = 6 × 4 = 24.
Therefore, we get 3p = 24.
Step 2:
To separate the variable, we shall divide both the sides by 3. This will give us just p on LHS.
New LHS = 3p/3 = p and New RHS = 24/3 = 8.
Therefore, the solution is p = 8.
(f) 3s = – 9
Answer:
To separate the variable, we shall divide both the sides by 3. This will give us just s in LHS.
New LHS = 3s/3 = s and New RHS = (-9)/3 = -3.
Therefore, the solution is s = -3.
(g) 3s + 12 = 0
Answer:
To separate the variable, we follow these steps:
Step 1:
To separate the variable,we should subtract 12 from both sides.
New LHS = 3s + 12 – 12 = 3sand New RHS = 0 – 12 = -12.
Therefore, we get 3s= -12.
Step 2:
To separate the variable, we shall divide both the sides by 3. This will give us just s in LHS.
New LHS = 3s/3 = s and New RHS = (-12)/3 = -4.
Therefore, the solution is s = -4.
(h) 3s = 0
Answer:
To separate the variable, we shall divide both the sides by 3. This will give us just s in LHS.
New LHS = 3s/3 = s and New RHS = 0/3 = 0.
Therefore, the solution is s = 0.
(i) 2q = 6
Answer:
To separate the variable, we shall divide both the sides by 2. This will give us just q on LHS.
New LHS = 2q/2 = q and New RHS = 6/2 = 3.
Therefore, the solution is q = 3.
(j) 2q – 6 = 0
Answer:
To separate the variable, we follow these steps:
Step 1:
To separate the variable,we should add 6 to both sides.
New LHS = 2q – 6 + 6 = 2q and New RHS = 0 + 6 = 6.
Therefore, we get 2q= 6.
Step 2:
To separate the variable, we shall divide both the sides by 2. This will give us just s in LHS.
New LHS = 2q/2 = q and New RHS = 6/2 = 3.
Therefore, the solution is q = 3.
(k) 2q + 6 = 0
Answer:
To separate the variable, we follow these steps:
Step 1:
To separate the variable,we should subtract 6 from both sides.
New LHS = 2q + 6 – 6 = 2qand New RHS = 0 – 6 = -6.
Therefore, we get 2q= -6.
Step 2:
To separate the variable, we shall divide both the sides by 2. This will give us just q on LHS.
New LHS = 2q/2 = q and New RHS = (-6)/2 = -3.
Therefore, the solution is q = -3.
(l) 2q + 6 = 12
Answer:
To separate the variable, we follow these steps:
Step 1:
To separate the variable,we should subtract 6 from both sides.
New LHS = 2q + 6 – 6 = 2qand New RHS = 12 – 6 = 6.
Therefore, we get 2q= 6.
Step 2:
To separate the variable, we shall divide both the sides by 2. This will give us just q on LHS.
New LHS = 2q/2 = q and New RHS = 6/2 = 3.
Therefore, the solution is q = 3.
Solutions to Exercise 4.3 (Page No 71) of NCERT Class 7 Math Chapter 4 Simple Equations –
1. Set up equations and solve them to find the unknown numbers in the following cases.
(a) Add 4 to eight times a number; you get 60.
Answer:
Let the unknown number be x.
When you add 4 to eight times the number, you get (8x + 4) which is = 60 (given).
Therefore,
8x + 4 = 60
or, 8x = 60 – 4 (On transposing 4 becomes -4)
or, 8x = 56
We divide both sides by 8 to get,
8x/8 = 56/8
or, x = 7 (This is the unknown number)
(b) One-fifth of a number minus 4 gives 3.
Answer:
Let the unknown number be x.
One-fifth of the number minus 4 = (1/5)x – 4 which is = 3 (given).
Therefore,
(1/5)x – 4 = 3
or, x/5 – 4 = 3
or, x/5 = 3 + 4 (On transposing -4 becomes 4)
or, x/5 = 7
We multiply both sides by 5 we get,
x/5 × 5 = 7 × 5
or, (x × 5)/5 = 35
or, x = 35 (This is the unknown number)
(c) If I take three-fourths of a number and add 3 to it, I get 21.
Answer:
Let the unknown number be x.
When you take three-fourths of the number and add 3 to it, we get (3/4)x + 3 which is = 21 (given).
Therefore,
(3/4)x + 3 = 21
or, 3x/4 + 3 = 21
or, 3x/4 = 21 – 3 (On transposing +3 becomes -3)
or, 3x/4 = 18
We multiply both sides by 4 we get,
3x/4 × 4 = 18 × 4
or, 3x = 72
We divide both sides by 3 to get,
3x/3 = 72/3
or, x = 24 (This is the unknown number)
or, x = 24 (This is the unknown number)
(d) When I subtracted 11 from twice a number, the result was 15.
Answer:
Let the unknown number be x.
When I subtracted 11 from twice the number, we get (2x – 11) which is = 15.
Therefore,
2x – 11 = 15
or, 2x = 15 + 11 (On transposing -11 becomes +11)
or, 2x = 26
We divide both sides by 2 to get,
2x/2 = 26/2
or, x = 13 (This is the unknown number)
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Answer:
Let the unknown number of notebooks be x.
When Munna subtracts thrice the number of notebooks he has from 50, he gets (50 – 3x) notebooks which is = 8.
Therefore,
50 – 3x = 8
or, – 3x = 8 – 50 (On transposing 50 becomes -50)
or, – 3x = – 42
We divide both sides by -3 to get,
(-3x)/(-3) = (-42)/(-3)
or, x = 14 (This is the unknown number of notebooks)
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
Answer:
Let the unknown number x.
When she adds 19 to it and divides the sum by 5, she gets (x + 19)/5 which is = 8.
Therefore,
(x + 19)/5 = 8
We multiply both sides by 5 we get,
((x + 19)/5) × 5 = 8 × 5
or, x + 19 = 40
or, x = 40 – 19 (On transposing 19 becomes -19)
or, x = 21 (This is the unknown number)
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
Answer:
Let the unknown number x.
When Anwar takes away 7 from 5/2 of the number, he gets ((5/2)x – 7) which is = 23.
Therefore,
(5/2)x – 7 = 23
or, 5x/2 = 23 + 7 (On transposing -7 becomes +7)
or, 5x/2 = 30
We multiply both sides by 2 we get,
(5x/2) × 2 = 30 × 2
or, (5x × 2 )/2 = 60
or, 5x = 60
We divide both sides by 5 to get,
5x/5 = 60/5
or, x = 12 (This is the unknown number)
2. Solve the following:
(a) 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. What is the lowest score?
Answer:
Let the lowest marks be x.
Therefore, the highest marks is (2x + 7) which is = 87 (given).
Therefore,
2x + 7 = 87
or, 2x = 87 – 7 (On transposing +7 becomes -7)
or, 2x = 80
We divide both sides by 2 to get,
2x/2 = 80/2
or, x = 40 (This is the lowest score)
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°.
What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
Answer:
Let each base angle be = b0.
There are two base angles and one vertex angle.
The sum of three angles of a triangle is 180°.
Therefore,
b + b + 40 = 180
or, 2b + 40 = 180
or, 2b = 180 – 40 (On transposing +40 becomes -40)
or, 2b = 140
We divide both sides by 2 to get,
2b/2 = 140/2
or, b = 700 (This is the value of each base angle)
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Answer:
Let Rahul’s score be x runs.
Therefore, Sachin’s score = 2x runs.
It is given that their runs fell two short of a double century.
Therefore,
2x + x = 200 – 2
or, 3x = 198
We divide both sides by 3 to get,
3x/3 = 198/3
or, x = 66 Therefore, Rahul’s score is 66 runs and Sachin’s score is 2x = 2 × 66 = 132 runs.
3. Solve the following:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has.
Irfan has 37 marbles. How many marbles does Parmit have?
Answer:
Let the number of marbles that Parmit has be x.
It is given that Irfan has 7 marbles more than five times the marbles Parmit has. So, Irfan has (5x + 7) marbles which is = 37 (given).
Therefore,
5x + 7 = 37
or, 5x = 37 – 7 (On transposing +7 becomes -7)
or, 5x = 30
We divide both sides by 5 to get,
5x/5 = 30/5
or, x = 6
Therefore, Parmit has 6 marbles.
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age.
What is Laxmi’s age?
Answer:
Let Laxmi’s age be x years.
It is given that Laxmi’s father is 4 years older than three times Laxmi’s age. So, Laxmi’s father’s age is (3x + 4) years which is = 49 years (given).
Therefore,
3x + 4 = 49
or, 3x = 49 – 4 (On transposing +4 becomes -4)
or, 3x = 45
We divide both sides by 3 to get,
3x/3 = 45/3
or, x = 15
Therefore, Laxmi’s age is 15 years.
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
Answer:
Let the number of fruit trees be x.
So, the number of non-fruit trees is (3x + 2) which is = 77 (given).
Therefore,
3x + 2 = 77
or, 3x = 77 – 2 (On transposing +2 becomes -2)
or, 3x = 75
We divide both sides by 3 to get,
3x/3 = 75/3
or, x = 25 Therefore, the number of fruit trees planted = 25.
4. Solve the following riddle.
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Answer:
Let the number be x.
When you take seven of them, that is multiply the number by 7, and add a 50 you get (7x + 50).
After obtaining the number, you will still need forty to reach 300 (given).
Therefore,
(7x + 50) + 40 = 300
or, 7x + 90 = 300
or, 7x = 300 – 90 (On transposing +90 becomes -90)
or, 7x = 210
We divide both sides by 7 to get,
7x/7 = 210/7
or, x = 30 (This is the required number)
Important Questions from Previous NCERT Textbook:
Exercise 4.3 Page No: 89 (Old Textbook):
1. Solve the following equations.
(a) 2y + 5/2 = 37/2
Answer:
Transposing 5/2 from LHS to RHS we get,
2y = 37/2 – 5/2 (On transposing + 5/2 becomes – 5/2)
or, 2y = (37 – 5 )/2
or, 2y = 32/2
or, 2y = 16
To separate the variable, we shall divide both the sides by 2. This will give us just yon LHS.
New LHS = 2y/2 = y and New RHS = 16/2 = 8.
Therefore, the solution is y = 8.
(b) 5t + 28 = 10
Answer:
Transposing 28 from LHS to RHS we get,
5t =10 – 28 (On transposing +28 becomes -28)
or, 5t =-18
To separate the variable, we shall divide both the sides by 5. This will give us just t in LHS.
New LHS = 5t/5 = t and New RHS = (-18)/5
Therefore, the solution is t = (-18)/5
(c) a/5 + 3 = 2
Answer:
Transposing 3 from LHS to RHS we get,
a/5 = 2 – 3 (On transposing +3 becomes -3)
or, a/5 = -1
To separate the variable, we shall multiply both the sides by 5. This will give us just aon LHS.
New LHS = a/5 × 5 = (a × 5)/5 = a and New RHS = -1 × 5
Therefore, the solution is a = -5.
(d) q/4 + 7 = 5
Answer:
Transposing 7 from LHS to RHS we get,
q/4 = 5 – 7 (On transposing +7 becomes -7)
or, q/4 = -2
To separate the variable, we shall multiply both the sides by 4. This will give us just q on LHS.
New LHS = q/4 × 4 = (q × 4)/4 = q and New RHS = -2 × 4
Therefore, the solution is q = -8.
(e) (5/2)x = -5
Answer:
To separate the variable, we follow these steps:
Step 1:
We multiply both sides by 2. Therefore, New LHS = (5/2)x × 2 = (5x × 2)/2 = 5x and New RHS = -5 × 2 = -10.
Therefore, we get 5x = -10.
Step 2:
To separate the variable, we shall divide both the sides by 5. This will give us just x on LHS.
New LHS = 5x/5 = x and New RHS = (-10)/5 = -2.
Therefore, the solution is x = -2.
(f) (5/2)x = 25/4
Answer:
To separate the variable, we follow these steps:
Step 1:
We multiply both sides by 2. Therefore, New LHS = (5/2)x × 2 = (5x × 2)/2 = 5x and New RHS = 25/4 × 2 = 25/2.
Therefore, we get 5x = 25/2.
Step 2:
To separate the variable, we shall divide both the sides by 5. This will give us just x on LHS.
New LHS = 5x/5 = x and New RHS = 25/2 ÷ 5 = 25/2 × 1/5 = (25 × 1)/(2 × 5) = 5/2
Therefore, the solution is x = 5/2
(g) 7m + 19/2 = 13
Answer:
Transposing 19/2 from LHS to RHS we get,
7m = 13 – 19/2 (On transposing +19/2 becomes -19/2)
or, 7m = 7/2
To separate the variable, we shall divide both the sides by 7. This will give us just m on LHS.
New LHS = 7m/7 = m and New RHS = 7/2 ÷ 7 = 7/2 × 1/7 = (7 × 1)/(2 × 7) = 1/2
Therefore, the solution is x = 1/2.
(h) 6z + 10 = – 2
Answer:
Transposing 10from LHS to RHS we get,
6z = -2 – 10 (On transposing +10 becomes -10)
or, 6z = -12
To separate the variable, we shall divide both the sides by 6. This will give us just z on LHS.
New LHS = 6z/6 = z and New RHS = (-12)/6 = -2
Therefore, the solution is z = -2.
(i) (3/2)l = 2/3
Answer:
To separate the variable, we follow these steps:
Step 1:
We multiply both sides by 2. Therefore, New LHS = (3/2)l × 2 = (3l × 2)/2 = 3l and New RHS = 2/3 × 2 = 4/3.
Therefore, we get 3l = 4/3.
Step 2:
To separate the variable, we shall divide both the sides by 3. This will give us just l on LHS.
New LHS = 3l/3 = l and New RHS = 4/3 ÷ 3 = 4/3 × 1/3 = (4 × 1)/(3 × 3) = 4/9
Therefore, the solution is x = 4/9
(j) 2b/3 – 5 = 3
Answer:
Transposing -5 from LHS to RHS we get,
2b/3 = 3 + 5 (On transposing -5 becomes +5)
or, 2b/3 = 8
We multiply both sides by 3. Therefore, New LHS = (2b/3) × 3 = (2b × 3)/3 = 2b and New RHS = 8 × 3 = 24.
Therefore, we get 2b = 24.
To separate the variable, we shall divide both the sides by 2. This will give us just b on LHS.
New LHS = 2b/2 = b and New RHS = 24/2 = 12
Therefore, the solution is b = 12.
2. Solve the following equations.
(a) 2(x + 4) = 12
Answer:
Let us divide both sides by 2. This will remove the brackets in the LHS. We get,
(2(x + 4))/2 = 12/2
or, x + 4 = 6
or, x = 6 – 4 (transposing 4 to RHS)
or, x = 2 (required solution)
(b) 3(n – 5) = 21
Answer:
Let us divide both sides by 3. This will remove the brackets in the LHS. We get,
(3(n – 5) )/3 = 21/3
or, n – 5 = 21/3
or, n – 5= 7
or, n = 7 + 5 (transposing -5 to RHS)
or, n = 12 (required solution)
(c) 3(n – 5) = – 21
Answer:
Let us divide both sides by 3. This will remove the brackets in the LHS. We get,
(3(n – 5) )/3 = (-21)/3
or, n – 5= -7
or, n = -7 + 5 (transposing -5 to RHS)
or, n = -2 (required solution)
(d) – 4(2 + x) = 8
Answer:
Let us divide both sides by -4. This will remove the brackets in the LHS. We get,
(– 4(2 + x) )/(-4) = 8/(-4)
or, 2 + x = -2
or, x = -2 – 2 (transposing 2 to RHS)
or, x = -4
(e) 4(2 – x) = 8
Answer:
Let us divide both sides by 4. This will remove the brackets in the LHS. We get,
(4(2 – x))/4 = 8/4
or, 2 – x = 2
or, –x = 2 – 2 (transposing 2 to RHS)
or, –x = 0
Multiplying both sides by (-1) we get,
–x × (-1) = 0 × (-1)
or, x = 0
3. Solve the following equations.
(a) 4 = 5(p – 2)
Answer:
An equation remains the same when the expressions on the left and the right are interchanged.
For convenience we interchange the expressions and write,
5(p – 2) = 4
Let us divide both sides by 5. This will remove the brackets in the LHS. We get,
(5(p – 2))/5 = 4/5
or, p – 2 = 4/5
Transposing -2 from LHS to RHS we get,
p = 4/5 + 2 (On transposing -2 becomes +2)
or, p = 14/5
(b) – 4 = 5(p – 2)
Answer:
An equation remains the same when the expressions on the left and the right are interchanged.
For convenience we interchange the expressions and write,
5(p – 2) = -4
Let us divide both sides by 5. This will remove the brackets in the LHS. We get,
(5(p-2))/5 = (-4)/5
Transposing -2 from LHS to RHS we get,
p = -4/5 + 2 (On transposing -2 becomes +2)
or, p = 6/5
(c) 16 = 4 + 3(t + 2)
Answer:
An equation remains the same when the expressions on the left and the right are interchanged.
For convenience we interchange the expressions and write,
4 + 3(t + 2) = 16
Transposing 4from LHS to RHS we get,
3(t + 2) = 16 – 4 (On transposing 4 becomes -4)
or, 3(t + 2) = 12
Let us divide both sides by 3. This will remove the brackets in the LHS. We get,
(3(t + 2))/3 = 12/3
or, t + 2 = 4
Transposing 2from LHS to RHS we get,
t= 4 – 2 (On transposing 2 becomes -2)
or, t = 2
(d) 4 + 5(p – 1) = 34
Answer:
Transposing 4from LHS to RHS we get,
5(p – 1) =34 – 4 (On transposing 4 becomes -4)
or 5(p – 1) = 30
Let us divide both sides by 3. This will remove the brackets in the LHS. We get,
(5(p – 1))/5 = 30/5
or, p – 1 = 6
Transposing -1from LHS to RHS we get,
p = 6 + 1 (On transposing -1 becomes 1)
or, p = 7
(e) 0 = 16 + 4(m – 6)
Answer:
An equation remains the same when the expressions on the left and the right are interchanged.
For convenience we interchange the expressions and write,
16 + 4(m – 6) = 0
Transposing 16from LHS to RHS we get,
4(m – 6) = -16 (On transposing 16 becomes -16)
Let us divide both sides by 4. This will remove the brackets in the LHS. We get,
(4(m – 6) )/4 = (-16)/4
or, m – 6 = -4
Transposing -6from LHS to RHS we get,
m = -4 + 6 (On transposing -6 becomes 6)
or, m = 2
4. (a) Construct 3 equations starting with x = 2
Answer:
We start with x = 2.
Multiply both sides by 5 we get,
5x = 5 × 2
or, 5x = 10 (This is equation 1)
Next we add 6 to both sides of equation 1 and get,
5x + 6 = 10 + 6
or, 5x + 6 = 16 (This is equation 2)
Next we divide equation 2 by 16 and get,
5x/16 + 6/16 = 16/16
or, 5x/16 + (6 ÷ 2)/(16 ÷ 2) = 1/1
or, 5x/16 + 3/8 = 1 (This is equation 3)
(b) Construct 3 equations starting with x = – 2
Answer:
We start with x = -2.
We divide both sides by 3 to get,
x/3 = (-2)/3
or, x/3 = – 2/3 (This is equation 1)
Next we add 1 to both sides to get,
x/3 + 1 = – 2/3 + 1
or, x/3 + 1 = 1/3 (This is equation 2)
Next, we multiply both sides by 4 to get,
4 × x/3 + 4 × 1 = 4 × 1/3
or, (4 × x)/3 + 4 × 1 = (4 × 1)/3
or, 4x/3 + 4 = 4/3 (This is equation 3)
Extra Questions to Complement Solutions to NCERT Class 7 Mathematics Chapter 4 Simple Equations:
Very Short Answer Type Questions:
1. What is wrong with the equation p + 3 = p – 2?
Answer:
The LHS and the RHS are not equal. So the equation cannot exist.
2. Which value of x satisfies the equation 2x = 1?
Answer:
x = 1/2 satisfies the equation 2x = 1.
3. If k – 8 = 5, then what is the value of 8(k – 8)?
Answer:
8(k – 8) = 8 × 5 = 40
4. What is the operation needed to solve the equation x + 7 = 2 called?
Answer:
The operation needed to solve the equation is called transposition.
5. What is the solution of 4x = 0?
Answer:
The solution of the equation is x = 0.
6. You are given that x/2 = 1/3 . By what number do you have to multiply the equation to get 1/6 on the RHS?
Answer:
You have to multiply the equation by 1/2.
7. The solution to equation p + 1 = 1/2 lies between which two integers?
Answer:
p + 1 = 1/2 or, p = 1/2 – 1 = – 1/2 . Therefore, the solution p = – 1/2 lies between integers 0 and – 1.
8. Can the equation p – 1 = 5 be written as 5 = p – 1?
Answer:
Yes it can be written both ways.
9. Right now Ram’s age is x years. What will be his age 10 years from now?
Answer:
10 years from now Ram’s age will be (x + 10) years.
10. If 1/5 – x = 0, what is the value of x?
Answer:
The value of x = 1/5 .
Multiple Choice Questions (MCQ):
1. The solution of the equation 4x – 3 = 0 is:
(a) Proper fraction
(b) Improper fraction
(c) Whole Number
(d) The equation does not have a solution
Answer: (a) Proper fraction
4x – 3 = 0
or, 4x = 3 or, x = 3/4 which is a proper fraction.
2. The equations x + 1 = -1 and 2p = -4 have:
(a) Different solutions
(b) Same solution
(c) On equation can be solved, the other cannot be solved
(d) Solutions cannot be compared because the variables are different
Answer: (b) Same solution
x + 1 = -1
or, x + 1 – 1 = -1 – 1 (Subtracting 1 from both sides)
or, x = -2
2p = -4
or, 2p/2 = (-4)/2 (Dividing both sides by 2)
or, p = -2 Therefore, both equations have the same solution.
3. If 2p = 1 what is the value of 5p?
(a) 1/2
(b) 1
(c) 5/2
(d) 0
Answer: (c)
2p = 1
or, 2p/2 = 1/2(Dividing both sides by 2)
or, p = 1/2
or, 5 × p = 1/2 × 5 (Multiplying both sides by 5)
or, 5p = 5/2
4. The solution to an equation can be:
(a) Integer
(b) Fraction
(c) Decimal
(d) All of the above
Answer: (d) All of the above
5. The sum of two consecutive numbers is 15. What is the bigger number?
(a) 7
(b) 8
(c) 6
(d) 9
Answer:
Let the smaller number be x.
So the greater number is (x + 1).
Therefore,
x + (x + 1) = 15
or, 2x + 1 = 15
or, 2x + 1 – 1 = 15 – 1 (Subtracting 1 from both sides)
or, 2x = 14
or, 2x/2 = 14/2 (Dividing both sides by 2)
or, x = 7
x + 1 = 7 + 1 = 8
Therefore, the bigger number is 8.
Short and Long Answer Type Questions:
1. How are the equations 5p – 14 = 1 and p/2 + 1/2 = 2 related?
Answer:
5p – 14 = 1
or, 5p – 14 + 14 = 1 + 14 (Adding 14 to both sides)
or, 5p = 15
or, 5p/5 = 15/5
or, p = 3
p/2 + 1/2 = 2
or, p/2 + 1/2 – 1/2 = 2 – 1/2 (Subtracting 1/2 from both sides)
or, p/2 = (4 – 1)/2
or, p/2 = 3/2
or, p/2 × 2 = 3/2 × 2 (Multiplying both sides by 2)
or, p = 3
Therefore, we can see that both equations have the same solution p = 3.
2. If 3x = 1, then find the value of 6x + 2.
Answer:
3x = 1
or, x = 1/3 (Dividing both sides by 3)
6x + 2 = 6 × 1/3 + 2 = 2 + 2 = 4 (Answer)
3. You are given two equations: 6x = 1 and 3x = 1. The solution to which equation is greater?
Answer:
6x = 1
or, 6x/6 = 1/6 (Dividing both sides by 6)
or, x = 1/6
3x = 1
or, 3x/3 = 1/3 (Dividing both sides by 3)
or, x = 1/3
1/3 > 1/6 and hence the solution to equation 3x = 1 is greater.
4. 3/5th of a number is 12. What is the number?
Answer:
Let the number be x.
Therefore,
(3/5)x = 12
or, (3/5)x × 5 = 12 × 5 (Multiplying both sides by 5)
or, 3x = 60
or, 3x/3 = 60/3 (Dividing both sides by 3)
or, x = 20
Therefore, the number is 20.
5. You are given 4/5x = 1. Convert the equation into an equation which has 0 in the RHS.
Answer:
4/5x = 1
or, 4/5x × 5 = 1 × 5 (Multiplying both sides by 5)
or, 4x = 5
or, 4x – 5 = 5 – 5 (Subtracting 5 from both sides)
or, 4x – 5 = 0 (There is 0 in the RHS)
6. Find the solution of the equation ax + b + c = 0.
Answer:
ax + b + c = 0
or, ax + b + c – c = 0 – c (Subtracting c from both sides)
or, ax + b = -c
or, ax + b – b = -c – b (Subtracting b from both sides)
or, ax = -c – b
or, ax/a = (-c-b)/a (Dividing both sides by a)
or, x = (-c-b)/a (Answer)
7. If you add two consecutive even numbers together the sum is 26. What are the numbers?
Answer:
Let the smaller even number be x.
Therefore, the bigger even number is (x + 2).
From the problem statement we get:
x + (x + 2) = 26
or, 2x + 2 = 26
or, 2x + 2 – 2 = 26 – 2 (Subtracting 2 from both sides)
or, 2x = 24
or, 2x/2 = 24/2
or, x = 12
x + 2 = 12 + 2 = 14
Therefore, the two even numbers are 12 and 14.
8. The difference between two numbers is 20. One number is 1/3rd of the other. What are the two numbers?
Answer:
Let the bigger number be x.
So, the smaller number = (1/3)x.
Therefore,
x – x/3 = 20
or, (3x- x)/3 = 20
or, 2x/3 = 20
or, (2x/3) × 3 = 20 × 3 (Multiplying both sides by 3)
or, 2x = 60
or, 2x/2 = 60/2 (Dividing both sides by 2)
or, x = 30
or, x/3 = 30/3
or, x/3 = 10
Therefore, the two numbers are 30 and 10.
9. When a number is multiplied by 2 it becomes 16. What will it become if it is multiplied by 3?
Answer:
Let the number be x.
Number multiplied by 2 becomes 2x.
Therefore, we get:
2x = 16
We need to find 3x.
2x/2 = 16/2
or, x = 8
3x = 3 × 8 = 24 Therefore, when it is multiplied by 3 it becomes 24.
10. What is the relation between the equations p + 1 = 2 and 3p + 3 = 6?
Answer:
3p + 3 = 6
or, 3(p + 1) = 6
or, (3(p + 1))/3 = 6/3 (Dividing both sides by 3)
or, p + 1 = 2
Therefore, we can that the equations p + 1 = 2 and 3p + 3 = 6 are the same.
Fill in the Blanks:
(a) In an equation LHS is ________ to RHS.
(b) The solution to the equation 0 – x = 0 is ________.
(c) If the side of a square is x, the perimeter of the square is _________.
(d) The length of a rectangle is 5x and the breadth is 3x. If the length = 20, the breadth is _________.
(e) x = 8 ________ be a solution to the equation (x – 3)/3 = 2.
Answer:
(a) In an equation LHS is equal to RHS. (equal)
(b) The solution to the equation 0 – x = 0 is x = 0.
(c) If the side of a square is x, the perimeter of the square is 4x.
(d) The length of a rectangle is 5x and the breadth is 3x. If the length = 20, the breadth is 12.
(e) x = 8 cannot be a solution to the equation (x – 3)/3 = 2.
++++++++++++++
Frequently Asked Questions (FAQs) on NCERT Solutions to Class 7 Maths Chapter 4 Simple Equations:
Our solutions have been prepared by Indian and foreign educated engineers and scientists, who love teaching young students like you and understand your needs well. These solutions and extra material on simple equations are especially designed for beginners and will help you easily understand the concepts. Please take the time to go through each step carefully – algebra can be fun when approached in the right way!
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The main topics covered are:
4.1 – A Mind-Reading Game
4.2 – Setting Up of an Equation
4.3 – Review of What We Know
4.4 – What Equation Is
4.5 – More Equations
4.6 – Applications of Simple Equations to Practical Situations
Here are the number of problems by each chapter:
Exercise 4.1 – 6 questions
Exercise 4.2 – 4 questions
Exercise 4.3 – 4 questions
Yes indeed! You can download the free PDFs of the solutions (please look towards the top of the page) and use them for practice!
5. What is the difference between and equation and an inequality?
In an equation the LHS and RHS will be equal. In an inequality, the LHS and RHS will not be equal.
All equations do not have solutions. For example, the equation (x + 3) = 
has no solution, because 5 is not divisible by 0.
Yes, equations can have many solutions. You will encounter equations of this type in higher classes.
Yes, equations can have many variables. Once again, you will encounter equations of this type in higher classes.
Start by practising how to solve simple equations. Then learn how to set up equations and solve word problems by forming equations. Our solutions and extra material are an excellent place to start. We have covered the topic from all angles and have shown you all the problem-solving techniques you need to know. Solve the problems on your own while looking at our solutions material. This approach called ‘learning by doing’ helps in quick learning and retention of the material.
We have intentionally used the concepts and methods explained in the textbook, to help you understand how to solve problems using them. Therefore, we suggest that you closely follow Chapter 4 in the textbook, alongside looking at the solutions. Algebra requires hands-on practice – with enough practice you will develop a knack for it.
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