**NCERT Solutions for Class 7 Maths**

Chapter 4 Simple Equations

Exercise 4.3

Chapter 4 Simple Equations

Exercise 4.3

1. Solve the following equations:

(a) $2 y+\frac{5}{2}=\frac{37}{2}$

(b) $5 t+28=10$

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

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

(e) $\frac{5}{2} x=-5$

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

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

(h) $6 z+10=$

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

(j) $\frac{2 b}{3}-5=3$

**Answer**

☛ Here, to solve given equations, separate the variable on LHS by applying suitable operation and apply same operation on RHS, also to get required solution

(a) We have, $2 y+\frac{5}{2}=\frac{37}{2}$

On transposing $\left(+\frac{5}{2}\right)$ from LHS to RHS, we get

$2 y=\frac{37}{2}-\frac{5}{2}$

$\Rightarrow$ $2 y=\frac{1}{2}(37-5)$

[taking $\frac{1}{2}$ as common]

$\Rightarrow$ $2 y=\frac{32}{2}=16$

On dividing both sides by 2, we get

$\frac{2 y}{2}=\frac{16}{2} \Rightarrow y=8 $

Hence, $y=8$ is a solution of the given equation.

(b) We have, 5 t+28 = 10

On transposing (+28) from LHS to RHS, we get

5 t = 10-28 $\Rightarrow$ 5 t = -18

On dividing both sides by 5 , we get

$\frac{5 t}{5}=\frac{-18}{5} \Rightarrow t=-\frac{18}{5}$

Hence, $t=-\frac{18}{5}$ is a solution of the given equation.

(c) We have, $\frac{a}{5}+3=2$

On transposing $(+3)$ from LHS to RHS, we get

$\frac{a}{5}=2-3$ $\Rightarrow$ $\frac{a}{5}=-1$

On multiplying both sides by 5, we get

$\frac{a}{5} \times 5=-1 \times 5$ $\Rightarrow$ a=-5

Hence, $a = -5$ is a solution of the given equation.

(d) We have, $\quad \frac{q}{4}+7=5$

On transposing $(+7)$ from LHS to RHS, we get

$\frac{q}{4}=5-7 \Rightarrow \frac{q}{4}=-2$

On multiplying both sides by 4 , we get

$\frac{q}{4} \times 4=-2 \times 4 \Rightarrow q=-8$

Hence, $q=-8$ is a solution of the given equation.

(e) We have, $\frac{5}{2} x=-10$

On multiplying both sides by 2 , we get

$\frac{5 x}{2} \times 2=-10 \times 2 \Rightarrow 5 x=-20$

On dividing both sides by 5 , we get

$\frac{5 x}{5}=-\frac{20}{5} \Rightarrow x=-4$

Hence, $x=-4$ is a solution of the given equation.

(f) We have, $\frac{5}{2} x=\frac{25}{4}$

On multiplying both sides by 2 , we get

$\frac{5 x}{2} \times 2=\frac{25}{4} \times 2 \Rightarrow 5 x=\frac{25}{2}$

On dividing both sides by 5, we get

$\frac{5 x}{5}=\frac{25}{2 \times 5} \Rightarrow x=\frac{5}{2}$

Hence, $x=\frac{5}{2}$ is a solution of the given equation.

(g) We have, $7 m+\frac{19}{2}=13$

On transposing $\left(+\frac{19}{2}\right)$ from LHS to RHS, we get

$7 m=13-\frac{19}{2} \Rightarrow 7 m=\frac{13}{1}-\frac{19}{2}$ $7 m=\frac{13 \times 2-19 \times 1}{2} \quad[\because$ LCM of 2 and $1=2]$ $7 m=\frac{26-19}{2} \Rightarrow 7 m=\frac{7}{2}$

$7 m=\frac{26-19}{2} \Rightarrow 7 m=\frac{7}{2}$

On dividing both sides by 7, we get

$\frac{7 m}{7}=\frac{7}{2 \times 7} \Rightarrow m=\frac{1}{2}$

(h) We have, $6 z+10=-2$

Transposing $(+10)$ from LHS to RHS, we get

$ 6 z=-2-10 \Rightarrow 6 z=-12 $

On dividing both sides by 6 , we get

$ \frac{6 z}{6}=\frac{-12}{6} \quad \Rightarrow \quad z=-2 $

Hence, $z=-2$ is a solution of the given equation.

(i) We have,

$ \frac{3 l}{2}=\frac{2}{3}$

On multiplying both sides by 2 , we get $\frac{3 l}{2} \times 2=\frac{2}{3} \times 2 \Rightarrow 3 t=\frac{4}{3}$

On dividing both sides by 3 , we get $\frac{3 l}{3}=\frac{4}{3 \times 3} \Rightarrow l=\frac{4}{9}$

Hence, $l=\frac{4}{9}$ is a solution of the given equation.

(j) We have, $\frac{2 b}{3}-5=3$

On transposing (-5) from LHS to RHS, we get

$\frac{2 b}{3}=3+5 \Rightarrow \frac{2 b}{3}=8$

$\mathrm{On}$ multiplying both sides by 3, we get

$\frac{2 b}{3} \times 3=8 \times 3 \Rightarrow 2 b=24$

Again, dividing both sides by 2, we get

$\frac{2 b}{2}=\frac{24}{2}$

$\Rightarrow \quad b=12$

Hence, $b=12$ is a solution of the given equation.

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

**Answer**

(a) 2(x + 4) = 12

Let us divide both the side by 2,

= (2(x + 4))/2 = 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,

= (3(n – 5))/3 = 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,

= (3(n – 5))/3 = – 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,

= (-4(2 + x))/ (-4) = 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,

= (4(2 – x))/ 4 = 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)

**(a) 4 = 5(p – 2) **

Let us divide both the side by 5,

= 4/5 = (5(p – 2))/5

= 4/5 = p -2

By transposing – 2 from RHS to LHS it becomes 2

= (4/5) + 2 = p

= (4 + 10)/ 5 = p

= p = 14/5

**(b) – 4 = 5(p – 2)
**Let us divide both the side by 5,

= – 4/5 = (5(p – 2))/5

= – 4/5 = p -2

By transposing – 2 from RHS to LHS it becomes 2

= – (4/5) + 2 = p

= (- 4 + 10)/ 5 = p

= p = 6/5

**(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,

= 12/3 = (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,

= (5(p – 1))/ 5 = 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,

= – 16/4 = (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

**Answer**

**(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

= (6x/6) – (4/6) = (8/6)

= x – (4/6) = (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

= (5x/2) – (3/2) = (-13/2) … [equation 3]

## Leave a Reply