NCERT Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.3

NCERT Solutions Class 7 Mathematics 
Chapter – 4 (Simple Equations)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 4 Simple Equations Exercise 4.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 4 Simple Equations

Exercise – 4.3 

1. Solve the following equations:
(a) 2y + \mathbf{\frac{5}{2}} = \mathbf{\frac{37}{2}}

(b) 5t + 28 = 10

(c) \mathbf{\frac{a}{5}} + 3 = 2

(d) \mathbf{\frac{q}{4}} + 7 = 5

(e) \mathbf{\frac{5}{2}}x = – 5

(f) \mathbf{\frac{5}{2}} x = \mathbf{\frac{25}{4}}

(g) 7m + \mathbf{\frac{19}{2}} = 13

(h) 6z + 10 = – 2

(i) \mathbf{\frac{3}{2}} l = \frac{2}{3}

(j) \mathbf{\frac{2b}{3}} – 5 = 3

Solution –
(a) 2y + \mathbf{\frac{5}{2}} = \mathbf{\frac{37}{2}}

By transposing \frac{5}{2} from LHS to RHS it becomes –\frac{5}{2}
Then,
⇒ 2y = \frac{37}{2} – \frac{5}{2} = \frac{37-5}{2} = \frac{32}{2}
Now,
Divide both side by 2,

⇒ \frac{2y}{2} = \frac{\frac{32}{2}}{2}

⇒ y = \frac{32}{2} × \frac{1}{2}

⇒ y = \frac{32}{4}

⇒ y = 8

(b) 5t + 28 = 10
By transposing 28 from LHS to RHS it becomes -28
⇒ 5t = 10 – 28
⇒ 5t = – 18
Now,
Divide both side by 5,
⇒ \frac{5t}{5} = – \frac{18}{5}

⇒ t = – \frac{18}{5}

(c) \mathbf{\frac{a}{5}} + 3 = 2
By transposing 3 from LHS to RHS it becomes -3
⇒ \frac{a}{5} = 2 – 3 = – 1
Now,
Multiply both side by 5,

⇒ \frac{a}{5} × 5 = -1 × 5
⇒ a = – 5

(d) \mathbf{\frac{q}{4}} + 7 = 5
By transposing 7 from LHS to RHS it becomes -7
⇒ \frac{q}{4} = 5 – 7

⇒ \frac{q}{4} = – 2
Now,
Multiply both side by 4,
⇒ \frac{q}{4} × 4= -2 × 4
⇒ q = -8

(e) \mathbf{\frac{5}{2}}x = – 5
First we have to multiply both the side by 2,
⇒ \frac{5x}{2} × 2 = – 5 × 2

⇒ 5x = – 10
Now,
We have to divide both the side by 5,
⇒ \frac{5x}{5} = \frac{-10}{5}
⇒ x = -2

(f) \mathbf{\frac{5}{2}} x = \mathbf{\frac{25}{4}}
First we have to multiply both the side by 2,
⇒ \frac{5x}{2} × 2 = \frac{25}{4} × 2

⇒ 5x = \frac{25}{2}
Now,
We have to divide both the side by 5,
⇒ \frac{5x}{2} = \frac{\frac{25}{2}}{5}

⇒ x = \frac{25}{2} × \frac{1}{5}

⇒ x = \frac{5}{2}

(g) 7m + \mathbf{\frac{19}{2}} = 13
By transposing \frac{19}{2} from LHS to RHS it becomes – \frac{19}{2}
Then,
⇒ 7m = 13 – \frac{19}{2}
⇒ 7m = \frac{(26-19)}{2}
⇒ 7m = \frac{7}{2}
Now,
Divide both side by 7,
⇒ \frac{7m}{7} = \frac{\frac{7}{2}}{7}

⇒ m = \frac{7}{2} × \frac{1}{7}

⇒ m = \frac{1}{2}

(h) 6z + 10 = – 2
By transposing 10 from LHS to RHS it becomes – 10
Then,
⇒ 6z = -2 – 10
⇒ 6z = – 12
Now,
Divide both side by 6,
⇒ \frac{6z}{6} = – \frac{12}{6}
⇒ z = – 2

(i) \mathbf{\frac{3}{2}} l = \frac{2}{3}
First we have to multiply both the side by 2,
⇒ \frac{3l}{2} × 2 = \frac{2}{3} × 2

⇒ 3l = \frac{4}{3}
Now,
We have to divide both the side by 3,
Then we get,
⇒ \frac{3l}{3} = \frac{\frac{4}{3}}{3}

⇒ l = \frac{4}{3} × \frac{1}{3}

⇒ x = \frac{4}{9}

(j) \mathbf{\frac{2b}{3}} – 5 = 3
By transposing -5 from LHS to RHS it becomes 5
Then,
⇒ \frac{2b}{3} = 3 + 5

⇒ \frac{2b}{3} = 8
Now,
Multiply both side by 3,

⇒ \frac{2b}{3} × 3 = 8 × 3

⇒ 2b = 24
Divide both side by 2,
⇒ \frac{2b}{2} = \frac{24}{2}

⇒ b = 12

2. Solve the following equations:
(a) 2 (x + 4) = 12
(b) 3 (n – 5) = 21
(c) 3 (n – 5) = – 21
(d) – 4 (2 + x) = 8
(e) 4 (2 – x) = 8

Solution –
(a) 2 (x + 4) = 12
Let us divide both the side by 2,
⇒ \frac{(2(x + 4))}{2} = \frac{12}{2}
⇒ x + 4 = 6
By transposing 4 from LHS to RHS it becomes -4
⇒ x = 6 – 4
⇒ x = 2

(b) 3 (n – 5) = 21
Let us divide both the side by 3,
⇒ \frac{3(n-5)}{3} = \frac{21}{3}
⇒ n – 5 = 7
By transposing -5 from LHS to RHS it becomes 5
⇒ n = 7 + 5
⇒ n = 12

(c) 3 (n – 5) = – 21
Let us divide both the side by 3,
⇒ \frac{3(n-5)}{3} = – \frac{21}{3}
⇒ n – 5 = -7
By transposing -5 from LHS to RHS it becomes 5
⇒ n = – 7 + 5
⇒ n = – 2

(d) – 4 (2 + x) = 8
Let us divide both the side by -4,
⇒ \frac{-4(2+x)}{-4} = \frac{8}{-4}
⇒ 2 + x = -2
By transposing 2 from LHS to RHS it becomes – 2
⇒ x = -2 – 2
⇒ x = – 4

(e) 4 (2 – x) = 8
Let us divide both the side by 4,
⇒ \frac{4(2-x)}{4} = \frac{8}{4}
⇒ 2 – x = 2
By transposing 2 from LHS to RHS it becomes – 2
⇒ – x = 2 – 2
⇒ – x = 0
⇒ x = 0

3. Solve the following equations:
(a) 4 = 5 (p – 2)
(b) – 4 = 5 (p – 2)
(c) 16 = 4 + 3 (t + 2)
(d) 4 + 5(p – 1) =34
(e) 0 = 16 + 4(m – 6)

Solution –
(a) 4 = 5 (p – 2)
Let us divide both the side by 5,
⇒ \frac{4}{5} = \frac{5(p-2)}{5}

⇒ \frac{4}{5} = p – 2
By transposing – 2 from RHS to LHS it becomes 2
⇒ \frac{4}{5} + 2 = p

⇒ \frac{(4+10)}{5} = p

⇒ p = \frac{14}{5} = 2\frac{4}{5}

(b) – 4 = 5 (p – 2)
Let us divide both the side by 5,
⇒ – \frac{4}{5} = \frac{5(p-2)}{5}

⇒ – \frac{4}{5} = p – 2
By transposing – 2 from RHS to LHS it becomes 2
⇒ – \frac{4}{5} + 2 = p

⇒ \frac{(-4+10)}{5} = p

⇒ p = \frac{6}{5} = 1\frac{1}{5}

(c) 16 = 4 + 3 (t + 2)
By transposing 4 from RHS to LHS it becomes – 4
⇒ 16 – 4 = 3 (t + 2)
⇒ 12 = 3 (t + 2)
Let us divide both the side by 3,
⇒ \frac{12}{3} = \frac{3(t+2)}{3}
⇒ 4 = t + 2
By transposing 2 from RHS to LHS it becomes – 2
⇒ 4 – 2 = t
⇒ t = 2

(d) 4 + 5(p – 1) =34
By transposing 4 from LHS to RHS it becomes – 4
⇒ 5 (p – 1) = 34 – 4
⇒ 5 (p – 1) = 30
Let us divide both the side by 5,
⇒ \frac{5(p-1)}{5} = \frac{30}{5}
⇒ p – 1 = 6
By transposing – 1 from RHS to LHS it becomes 1
⇒ p = 6 + 1
⇒ p = 7

(e) 0 = 16 + 4(m – 6)
By transposing 16 from RHS to LHS it becomes – 16
⇒ 0 – 16 = 4(m – 6)
⇒ – 16 = 4(m – 6)
Let us divide both the side by 4,
⇒ – \frac{16}{4} = \frac{4(m-6)}{4}
⇒ – 4 = m – 6
By transposing – 6 from RHS to LHS it becomes 6
⇒ – 4 + 6 = m
⇒ m = 2

4. (a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2

Solution –
(a) Construct 3 equations starting with x = 2
First equation is,
Multiply both side by 6
⇒ 6x = 12  —————- [equation 1]

Second equation is,
Subtracting 4 from both side,
⇒ 6x – 4 = 12 – 4
⇒ 6x – 4 = 8 —————-  [equation 2]

Third equation is,
Divide both side by 6
⇒ \frac{6x}{6} – \frac{4}{6} = \frac{8}{6}

⇒ x – \frac{4}{6} = \frac{8}{6}  —————-  [equation 3]

(b) Construct 3 equations starting with x = – 2
First equation is,
Multiply both side by 5
⇒ 5x = – 10 —————-  [equation 1]

Second equation is,
Subtracting 3 from both side,
⇒ 5x – 3 = – 10 – 3
⇒ 5x – 3 = – 13 —————-  [equation 2]

Third equation is,
Dividing both sides by 2
⇒ \frac{5x}{2} – \frac{3}{2} = \frac{-13}{2} —————-  [equation 3]

 

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