RD Sharma 2014 Solutions for Class 7 Math Chapter 8 Linear Equations In One Variable are provided here with simple step-by-step explanations. These solutions for Linear Equations In One Variable are extremely popular among class 7 students for Math Linear Equations In One Variable Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2014 Book of class 7 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RD Sharma 2014 Solutions. All RD Sharma 2014 Solutions for class 7 Math are prepared by experts and are 100% accurate.

#### Page No 8.12:

#### Question 5:

Solve each of the following equations and check your answes:

$\frac{x}{2}=0$

#### Answer:

$\frac{x}{2}$ = 0

Multiplying both sides by 2, we get

⇒ $\frac{x}{2}\times 2=0\times 2$

⇒ x = 0

__Verification:__

Substituting x= 0 in LHS, we get

LHS =$\frac{0}{2}$= 0 and RHS = 0

LHS = 0 and RHS = 0

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 1:

Solve each of the following equations and check your answers:

*x* − 3 = 5

#### Answer:

x − 3 = 5

Adding 3 to both sides, we get

⇒ *x* − 3 + 3 = 5 + 3

⇒ x = 8

__Verification:__

Substituting x= 8 in LHS, we get

LHS = *x* − 3 and RHS = 5

LHS = 8 − 3 = 5 and RHS = 5

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Answer:

x + 9 = 13

Subtracting 9 from both sides, we get

=> x + 9 $-$ 9 = 13 $-$ 9

=> x = 4

Verification:

Substituting x = 4 on LHS, we get

LHS = 4 + 9 = 13 = RHS

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 3:

Solve each of the following equations and check your answers:

$x-\frac{3}{5}=\frac{7}{5}$

#### Answer:

*x* − $\frac{3}{5}$ = $\frac{7}{5}$

Adding $\frac{3}{5}$ to both sides, we get

*=> x* − $\frac{3}{5}$ + $\frac{3}{5}$ = $\frac{7}{5}$ + $\frac{3}{5}$

=> x = $\frac{7+3}{5}$

=> x = $\frac{10}{5}$

⇒ x = 2

__Verification:__

Substituting x = 2 in LHS, we get

LHS = 2 − $\frac{3}{5}$=$\frac{10-3}{5}=\frac{7}{5}$, and RHS = $\frac{7}{5}$

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 4:

Solve each of the following equations and check your answers:

3*x* = 0

#### Answer:

*3x* = 0

Dividing both sides by 3, we get

⇒ $\frac{3x}{3}$ = $\frac{0}{3}$

⇒ x = 0

__Verification:__

Substituting x = 0 in LHS = 3x, we get

LHS = 3 $\times $ 0 = 0 and RHS = 0

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 6:

Solve each of the following equations and check your answers:

$x-\frac{1}{3}=\frac{2}{3}$

#### Answer:

*x* − $\frac{1}{3}$ = $\frac{2}{3}$

⇒ Adding $\frac{1}{3}$ to both sides, we get

⇒ *x* − $\frac{1}{3}$ +$\frac{1}{3}$ = $\frac{2}{3}$ + $\frac{1}{3}$

=> x = $\frac{2+1}{3}$

=> x = $\frac{3}{3}$

⇒ x = 1

__Verification:__

Substituting x= 1 in LHS, we get

LHS = 1 −$$$\frac{1}{3}$ =$\frac{3-1}{3}$$=\frac{2}{3}$, and RHS = $\frac{2}{3}$

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 7:

Solve each of the following equations and check your answers:

$x+\frac{1}{2}=\frac{7}{2}$

#### Answer:

*x* + $\frac{1}{2}$ = $\frac{7}{2}$

⇒ Subtracting $\frac{1}{2}$ from both sides, we get

⇒ *x* + $\frac{1}{2}$ − $\frac{1}{2}$ = $\frac{7}{2}$ − $\frac{1}{2}$

$\Rightarrow \text{x=}\frac{7-1}{2}=\frac{6}{2}$

⇒ x = 3

__Verification:__

Substituting x = 3 in LHS, we get

LHS = 3 + $\frac{1}{2}$=$\frac{6+1}{2}=\frac{7}{2}$, and RHS = $\frac{7}{2}$

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 8:

Solve each of the following equations and check your answers:

10 − *y* = 6

#### Answer:

10 − y = 6

Subtracting 10 from both sides, we get

⇒ 10 − y − 10 = 6 − 10

⇒ −y = −4.

⇒ Multiplying both sides by −1, we get

⇒ −y $\times $−1 = −4 $\times $−1

⇒ y = 4

__Verification:__

Substituting y = 4 in LHS, we get

LHS = 10 − y = 10$-$4 = 6 and RHS = 6

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 9:

Solve each of the following equations and check your answers:

7 + 4*y* = −5

#### Answer:

7 + 4*y* = −5

Subtracting 7 from both sides, we get

⇒ 7 + 4*y* − 7 = −5 − 7

⇒ 4*y* = −12

Dividing both sides by 4, we get

⇒ $\text{y}$= $\frac{-12}{4}$

⇒ *y* = −3

Verification :

Substituting y = −3 in LHS, we get

LHS = 7 + 4*y* = 7 + 4(−3) = 7 − 12 = −5, and RHS = −5

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 10:

Solve each of the following equations and check your answers:

$\frac{4}{5}-x=\frac{3}{5}$

#### Answer:

$\frac{4}{5}$ − x = $\frac{3}{5}$

Subtracting$$$\frac{4}{5}$ from both sides, we get

⇒ $\frac{4}{5}$ − x − $\frac{4}{5}$= $\frac{3}{5}$ − $\frac{4}{5}$

⇒ −x = $\frac{3-4}{5}$

$\Rightarrow -\text{x}=\frac{-1}{5}$

Multiplying both sides by -1, we get

⇒ −x $\times $ (−1) = −$\frac{1}{5}$ $\times $ (−1)

⇒ x = $\frac{1}{5}$

__Verification:__

Substituting x= $\frac{1}{5}$ in LHS, we get

LHS = $\frac{4}{5}$ − $\frac{1}{5}$=$\frac{4-1}{5}=\frac{3}{5}$, and RHS = $\frac{3}{5}$

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 11:

Solve each of the following equations and check your answers:

$2y-\frac{1}{2}=-\frac{1}{3}$

#### Answer:

Adding $\frac{1}{2}$$$to both sides, we get

⇒ 2y − $\frac{1}{2}$$+$$\frac{1}{2}$$$$=-\frac{1}{3}$ + $\frac{1}{2}$$$

⇒ 2y = $\frac{-2+3}{6}$

⇒ 2y = $\frac{1}{6}$$$

Dividing both sides by 2, we get

⇒ $\frac{2y}{2}=\frac{1}{6\times 2}$

⇒ y = $\frac{1}{12}$

__Verification:__

Substituting y = $\frac{1}{12}$ in LHS, we get

LHS = $2\times \left(\frac{1}{12}\right)-\frac{1}{2}=\frac{1}{6}-\frac{1}{2}$ =$\frac{1-3}{6}=\frac{-2}{6}$ =$-\frac{1}{3}$, and RHS = $-\frac{1}{3}$

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 12:

Solve each of the following equations and check your answers:

$14=\frac{7x}{10}-8$

#### Answer:

$14=$$\frac{7\text{x}}{10}$

Adding 8 to both sides, we get

⇒ 14 + 8 = $\frac{7x}{10}$ − 8 + 8

⇒ 22 = $\frac{7x}{10}$

Multiplying both sides by 10, we get

⇒ 22 $\times 10$ = $\frac{7x}{10}$$\times 10$

⇒ 220 = 7x

Dividing both sides by 7, we get

⇒ $\frac{220}{7}=\frac{7x}{7}$

⇒ x = $\frac{220}{7}$

__Verification____:__

Substituting x = $\frac{220}{7}$ in RHS, we get

LHS =$$ 14, and RHS = $\frac{7\left({\displaystyle \frac{220}{7}}\right)}{10}-8$ = $\frac{220}{10}-8=22-8=14$

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 13:

Solve each of the following equations and check your answers:

3 (*x* + 2) = 15

#### Answer:

3 (*x* + 2) = 15

Dividing both sides by 3, we get

⇒ $$$\frac{\mathrm{3\; (x\; +\; 2)}}{3}=\frac{15}{3}$

⇒ (*x* + 2) = 5

Subtracting 2 from both sides, we get

⇒ *x* + 2 $-$ 2 = 5 $-$ 2

⇒ *x = 3*

__Verification:__

Substituting x = 3 in LHS, we get

LHS = 3 (*x* + 2)= 3 (3+2) = 3$\times $5 = 15, and RHS = 15

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 14:

Solve each of the following equations and check your answers:

$\frac{x}{4}=\frac{7}{8}$

#### Answer:

$\frac{x}{4}=\frac{7}{8}$

Multiplying both sides by 4, we get

⇒ $\frac{x}{4}\times 4=\frac{7}{8}\times 4$

⇒ x = $\frac{7}{2}$

__Verification:__

Substituting x = $\frac{7}{2}$ in LHS, we get

LHS = $\frac{7}{2\times 4}$ = $\frac{7}{8}$, and RHS = $\frac{7}{8}$

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 15:

Solve each of the following equations and check your answers:

$\frac{1}{3}-2x=0$

#### Answer:

$\frac{1}{3}-2x=0$

Subtracting $\frac{1}{3}$ from both sides, we get

⇒ $\frac{1}{3}-2x-\frac{1}{3}$ = 0 $-$$\frac{1}{3}$

⇒ $-2x=-\frac{1}{3}$

Multiplying both sides by $-$1, we get

⇒ $-2x\times (-1)=-\frac{1}{3}\times (-1)$

⇒ 2x = $\frac{1}{3}$

Dividing both sides by 2, we get

⇒ $\frac{2x}{2}=\frac{1}{3\times 2}$

⇒ x = $\frac{1}{6}$

__Verification:__

Substituting x = $\frac{1}{6}$ in LHS, we get

LHS = $\frac{1}{3}-\left(2\times \frac{1}{6}\right)=\frac{1}{3}-\frac{1}{3}=0$, and RHS = 0

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 16:

Solve each of the following equations and check your answers:

3(*x* + 6) = 24

#### Answer:

3(*x* + 6) = 24

Dividing both sides by 3, we get

⇒ $$$\frac{\mathrm{3\; (x\; +\; 6)}}{3}=\frac{24}{3}$

⇒ (*x* + 6) = 8

Subtracting 6 from both sides, we get

⇒ *x* + 6 $-$ 6 = 8 $-$ 6

⇒ *x = *2

__Verification:__

Substituting x = 2 in LHS, we get

LHS = 3 (*x* + 6) = 3 (2 + 6) = 3$\times $8 = 24, and RHS = 24

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 17:

Solve each of the following equations and check your answers:

3(*x* + 2) − 2(*x* − 1) = 7

#### Answer:

3(*x* + 2) − 2(*x* − 1) = 7

On expanding the brackets, we get

⇒ $\left(3\times x\right)+\left(3\times 2\right)-\left(2\times x\right)+\left(2\times 1\right)=7$

⇒ 3x + 6 $-$ 2x + 2 = 7

⇒ 3x $-$ 2x + 6 + 2 = 7

⇒ x + 8 = 7

Subtracting 8 from both sides, we get

⇒ *x* + 8 $-$8 = 7 $-$ 8

⇒ *x = *$-$1

__Verification:__

Substituting x = $-$1 in LHS, we get

LHS = 3 (*x* + 2) $-$2(x $-$1), and RHS = 7

LHS = 3 ($-$1 + 2) $-$2($-$1$-$1) = (3$\times $1) $-$ (2$\times $$-$2) = 3 + 4 = 7, and RHS = 7

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 18:

Solve each of the following equations and check your answers:

8(2*x* − 5) − 6(3*x* − 7) = 1

#### Answer:

8(2*x* − 5) − 6(3*x* − 7) = 1

On expanding the brackets, we get

⇒ (8$\times $2x) $-$ (8 $\times $ 5) $-$ (6 $\times $ 3x) + ($-$6)$\times $($-$7) = 1

⇒ 16x $-$ 40 $-$ 18x + 42 = 1

⇒ 16x $-$ 18x + 42 $-$ 40 = 1

⇒ $-$2x + 2 = 1

Subtracting 2 from both sides, we get

⇒ $-$2x + 2 $-$ 2 = 1 $-$ 2

⇒ $-$2x = $-$1

Multiplying both sides by $-$1, we get

⇒ $-$2x $\times $($-$1) = $-$1$\times $($-$1)

⇒ 2x = 1

Dividing both sides by 2, we get

⇒ $\frac{2x}{2}=\frac{1}{2}$

⇒ x = $\frac{1}{2}$

Verification:

Substituting x = $\frac{1}{2}$ in LHS, we get

= 8(2$\times $$\frac{1}{2}$ $-$ 5) $-$ 6(3$\times $$\frac{1}{2}$ − 7)

= 8(1 − 5) − 6($\frac{3}{2}$ − 7)

= 8$\times $(−4) − (6 $\times $$\frac{3}{2}$) + (6 $\times $7)

= −32 − 9 + 42 = − 41 + 42 = 1 = RHS

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 19:

Solve each of the following equations and check your answers:

6(1 − 4*x*) + 7(2 + 5*x*) = 53

#### Answer:

6(1 − 4*x*) + 7(2 + 5*x*) = 53

On expanding the brackets, we get

⇒ (6$\times $1) $-$ (6 $\times $ 4x) + (7 $\times $2) + (7$\times $5x) = 53

⇒ 6 $-$ 24x + 14 + 35x = 53

⇒ 6 + 14 + 35x $-$ 24x = 53

⇒ 20 + 11x = 53

Subtracting 20 from both sides, we get

⇒ 20 + 11x $-$ 20 = 53 $-$ 20

⇒ 11x = 33

⇒ Dividing both sides by 11, we get

⇒ $\frac{11x}{11}=\frac{33}{11}$

⇒ x = 3

Verification:

Substituting x = 3 in LHS, we get

= 6(1 − 4$\times $3) + 7(2 + 5$\times $3)

= 6(1 − 12) + 7(2 + 15)

= 6(−11) + 7(17)

= −66 + 119 = 53 = RHS

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 20:

Solve each of the following equations and check your answers:

5(2 − 3*x*) − 17(2*x* − 5) = 16

#### Answer:

5(2 − 3*x*) − 17(2*x* − 5) = 16

On expanding the brackets, we get

⇒ (5$\times $2) $-$ (5 $\times $ 3x) − (17 $\times $2x ) + (17$\times $5) = 16

⇒ 10 $-$ 15x − 34x + 85 = 16

⇒ 10 + 85 − 34x $-$ 15x = 16

⇒ 95 - 49x = 16

Subtracting 95 from both sides, we get

⇒ - 49x + 95 $-$ 95 = 16 $-$ 95

⇒ - 49x = -79

Dividing both sides by -49, we get

⇒ $\frac{-49x}{-49}=\frac{-79}{-49}$

⇒ x = $\frac{79}{49}$

Verification:

Substituting x = $\frac{79}{49}$ in LHS, we get

= 5(2 − 3$\times $$\frac{79}{49}$) − 17(2$\times $$\frac{79}{49}$ − 5)

= (5$\times $2) $-$ (5 $\times $ 3$\times $$\frac{79}{49}$ ) − (17 $\times $2$\times $$\frac{79}{49}$ ) + (17$\times $5)

= 10 $-$ $\frac{1185}{49}$ − $\frac{2686}{49}$ + 85

= $\frac{490-1185-2686+4165}{49}$

= $\frac{784}{49}$

= 16

= RHS

So, LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 21:

Solve each of the following equations and check your answers:

$\frac{x-3}{5}-2=-1$

#### Answer:

$\frac{x-3}{5}-2=-1$

Adding 2 to both sides, we get

⇒ $\frac{x-3}{5}-2+2=-1+2$

⇒ $\frac{x-3}{5}=1$

Multiplying both sides by 5, we get

⇒ $\left(\frac{x-3}{5}\right)\times 5=1\times 5$

⇒ x $-$ 3 = 5

Adding 3 to both sides, we get

⇒ x $-$ 3 + 3 = 5 + 3

⇒ x = 8

Verification:

Substituting x = 8 in LHS, we get

= $\frac{8-3}{5}-2$

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

= 1 $-$ 2

= $-$1

= RHS

LHS = RHS

Hence, verified.

#### Page No 8.12:

#### Question 22:

Solve each of the following equations and check your answers:

5(*x* − 2) + 3(*x* + 1) = 25

#### Answer:

5(*x* − 2) + 3(*x* + 1) = 25

On expanding the brackets, we get

⇒ (5 $\times $ x) $-$ (5 $\times $ 2) + (3 $\times $ x) + (3$\times $1) = 25

⇒ 5x $-$ 10 + 3x + 3 = 25

⇒ 5x + 3x $-$ 10 + 3 = 25

⇒ 8x $-$ 7 = 25

Adding 7 to both sides, we get

⇒ 8x $-$ 7 + 7 = 25 +7

⇒ 8x = 32

Dividing both sides by 8, we get

⇒ $\frac{8x}{8}=\frac{32}{8}$

⇒ x = 4

Verification:

Substituting x = 4 in LHS, we get

= 5(4 − 2) + 3(4 + 1)

= 5(2) + 3(5)

= 10 + 15

= 25

= RHS

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 1:

Solve each of the following equations. Also, verify the result in each case.

6*x* + 5 = 2*x* + 17

#### Answer:

We have

⇒ 6*x* + 5 = 2*x* + 17

Transposing 2x to LHS and 5 to RHS, we get

⇒ 6x $-$ 2x = 17 $-$ 5

⇒ 4x = 12

Dividing both sides by 4, we get

$\Rightarrow \frac{4\text{x}}{4}=\frac{12}{4}$

⇒ x = 3

Verification:

Substituting x =3 in the given equation, we get

6$\times $3 + 5 = 2$\times $3 + 17

18 + 5 = 6 + 17

23 = 23

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 2:

Solve each of the following equations. Also, verify the result in each case.

2(5*x* − 3) − 3(2*x* − 1) = 9

#### Answer:

We have

⇒2(5*x* − 3) − 3(2*x* − 1) = 9

Expanding the brackets, we get

⇒ $2\times 5x-2\times 3-3\times 2x+3\times 1=9$

⇒ 10x − 6 − 6x + 3 = 9

⇒ 10x − 6x − 6 + 3 = 9

⇒ 4x − 3 = 9

Adding 3 to both sides, we get

⇒ 4x − 3 + 3= 9 + 3

⇒ 4x = 12

Dividing both sides by 4, we get

⇒ $\frac{4x}{4}=\frac{12}{4}$

⇒ Thus, x = 3.

Verification:

Substituting x =3 in LHS, we get

=2(5$\times $3 − 3) − 3(2$\times $3 − 1)

=2$\times $12 − 3 $\times $ 5

=24 − 15

= 9

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 3:

Solve each of the following equations. Also, verify the result in each case.

$\frac{x}{2}=\frac{x}{3}+1$

#### Answer:

$\frac{x}{2}=\frac{x}{3}+1$

Transposing $\frac{x}{3}$ to LHS, we get

$\Rightarrow \frac{x}{2}-\frac{x}{3}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x-2x}{6}=1\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{x}{6}=1$

Multiplying both sides by 6, we get

$\Rightarrow \frac{x}{6}\times 6=1\times 6$

$\Rightarrow $ x = 6

Verification:

Substituting x = 6 in the given equation, we get

$\frac{6}{2}=\frac{6}{3}+1$

3 = 2 + 1

3 = 3

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 4:

Solve each of the following equations. Also, verify the result in each case.

$\frac{x}{2}+\frac{3}{2}=\frac{2x}{5}-1$

#### Answer:

$\frac{x}{2}+\frac{3}{2}=\frac{2x}{5}-1$

Transposing $\frac{2x}{5}$ to LHS and $\frac{3}{2}$ to RHS, we get

=> $\frac{x}{2}-\frac{2x}{5}=-1-\frac{3}{2}$

=> $\frac{5x-4x}{10}=\frac{-2-3}{2}$

=> $\frac{x}{10}=\frac{-5}{2}$

Multiplying both sides by 10, we get

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

=> x = $-25$

Verification:

Substituting x = $-25$ in the given equation, we get

$\frac{-25}{2}+\frac{3}{2}=\frac{2\times \left(-25\right)}{5}-1$

$\frac{-22}{2}=-10-1$

$-11=-11$

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 5:

Solve each of the following equations. Also, verify the result in each case.

$\frac{3}{4}(x-1)=x-3$

#### Answer:

$\frac{3}{4}\left(x-1\right)=x-3$

On expanding the brackets on both sides, we get

=>$\frac{3}{4}x-\frac{3}{4}=x-3$

Transposing $\frac{3}{4}x$ to RHS and 3 to LHS, we get

=> $3-\frac{3}{4}=x-\frac{3}{4}x$

=> $\frac{12-3}{4}=\frac{4x-3x}{4}$

=> $\frac{9}{4}=\frac{x}{4}$

Multiplying both sides by 4, we get

=> x = 9

Verification:

Substituting x = 9 on both sides, we get

$\frac{3}{4}\left(9-1\right)=9-3$

$\frac{3}{4}\times 8=6$

6 = 6

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 6:

Solve each of the following equations. Also, verify the result in each case.

3(*x* − 3) = 5(2*x* + 1)

#### Answer:

6. 3(*x* − 3) = 5(2*x* + 1)

On expanding the brackets on both sides, we get

=> $3\times x-3\times 3=5\times 2x+5\times 1$

=> 3x $-$ 9 = 10x + 5

Transposing 10x to LHS and 9 to RHS, we get

=> 3x $-$ 10x = 9 + 5

=> $-$7x = 14

Dividing both sides by 7, we get

=> $\frac{-7\mathrm{x}}{7}=\frac{14}{7}$

=> x = $-$2

Verification:

Substituting x = $-$2 on both sides, we get

$3\left(-2-3\right)=5\left(2\left(-2\right)+1\right)$

$3\left(-5\right)=5\left(-3\right)$

$-$15 = $-$15

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 7:

Solve each of the following equations. Also, verify the result in each case.

3*x* − 2 (2*x* − 5) = 2(*x* + 3) − 8

#### Answer:

3*x* − 2 (2*x* − 5) = 2(*x* + 3) − 8

On expanding the brackets on both sides, we get

=> $3x-2\times 2x+2\times 5=2\times x+2\times 3-8$

=> $3x-4x+10=2x+6-8$

=> $-x+10=2x-2$

Transposing x to RHS and 2 to LHS, we get

=> 10 + 2 = 2x + x

=> 3x = 12

Dividing both sides by 3, we get

=> $\frac{3x}{3}=\frac{12}{3}$

=> x = 4

Verification:

Substituting x = 4 on both sides, we get

$3\left(4\right)-2\left(2\left(4\right)-5\right)=2\left(4+3\right)-8$

12$-$2 (8 $-$ 5) = 14$-$8

12 $-$ 6 = 6

6 = 6

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 8:

Solve each of the following equations. Also, verify the result in each case.

$x-\frac{x}{4}-\frac{1}{2}=3+\frac{x}{4}$

#### Answer:

$x-\frac{x}{4}-\frac{1}{2}=3+\frac{x}{4}$

Transposing $\frac{x}{4}$ to LHS and $-\frac{1}{2}$ to RHS, we get

=> $x-\frac{x}{4}-\frac{x}{4}=3+\frac{1}{2}$

=> $\frac{4x-x-x}{4}=\frac{6+1}{2}$

=> $\frac{2x}{4}=\frac{7}{2}$

Multiplying both sides by 4, we get

=> $\frac{2x}{4}\times 4=\frac{7}{2}\times 4$

=> 2x = 14

Dividing both sides by 2, we get

=> $\frac{2x}{2}=\frac{14}{2}$

=> x = 7

Verification:

Substituting x = 7 on both sides, we get

$7-\frac{7}{4}-\frac{1}{2}=3+\frac{7}{4}$

$\frac{28-7-2}{4}=\frac{12+7}{4}$

$\frac{19}{4}=\frac{19}{4}$

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 9:

Solve each of the following equations. Also, verify the result in each case.

$\frac{6x-2}{9}+\frac{3x+5}{18}=\frac{1}{3}$

#### Answer:

$\frac{6x-2}{9}+\frac{3x+5}{18}=\frac{1}{3}$

=> $\frac{6x\times 2-2\times 2+3x+5}{18}=\frac{1}{3}$

=> $\frac{12x-4+3x+5}{18}=\frac{1}{3}$

=> $\frac{15x+1}{18}=\frac{1}{3}$

Multiplying both sides by 18, we get

=> $\frac{15x+1}{18}\times \left(18\right)=\frac{1}{3}\times \left(18\right)$

=> 15x + 1 = 6

Transposing 1 to RHS, we get

=> 15x = 6$-$1

=> 15x = 5

Dividing both sides by 15, we get

=> $\frac{15x}{15}=\frac{5}{15}$

=> x = $\frac{1}{3}$

Verification:

Substituting x = $\frac{1}{3}$ on both sides, we get

$\frac{6\left({\displaystyle \frac{1}{3}}\right)-2}{9}+\frac{3\left({\displaystyle \frac{1}{3}}\right)+5}{18}=\frac{1}{3}$

$\frac{2-2}{9}+\frac{1+5}{18}=\frac{1}{3}$

$0+\frac{6}{18}=\frac{1}{3}$

$\frac{1}{3}=\frac{1}{3}$

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 10:

Solve each of the following equations. Also, verify the result in each case.

$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

#### Answer:

$m-\frac{m-1}{2}=1-\frac{m-2}{3}$

=> $\frac{2m-m-(-1)}{2}=\frac{3-m-(-2)}{3}$

=> $\frac{m+1}{2}=\frac{3-m+2}{3}$

=> $\frac{m+1}{2}=\frac{5-m}{3}$

=> $\frac{m}{2}+\frac{1}{2}=\frac{5}{3}-\frac{m}{3}$

Transposing m/3 to LHS and 1/2 to RHS, we get

=> $\frac{m}{2}+\frac{m}{3}=\frac{5}{3}-\frac{1}{2}$

=> $\frac{3m+2m}{6}=\frac{10-3}{6}$

Multiplying both sides by 6, we get

=> $\frac{5m}{6}\times 6=\frac{7}{6}\times 6$

=> 5m = 7

Dividing both sides by 5, we get

=> $\frac{5m}{5}=\frac{7}{5}$

=> m = $\frac{7}{5}$

Verification:

Substituting m =$\frac{7}{5}$ on both sides, we get

$\frac{7}{5}-\frac{{\displaystyle \frac{7}{5}}-1}{2}=1-\frac{{\displaystyle \frac{7}{5}}-2}{3}$

$\frac{7}{5}-\frac{{\displaystyle \frac{7-5}{5}}}{2}=1-\frac{{\displaystyle \frac{7-10}{5}}}{3}$

$\frac{7}{5}-\frac{2}{5\times 2}=1-\frac{-3}{5\times 3}$

$\frac{7}{5}-\frac{1}{5}=1+\frac{1}{5}$

$\frac{6}{5}=\frac{6}{5}$

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 11:

Solve each of the following equations. Also, verify the result in each case.

$\frac{(5x-1)}{3}-\frac{(2x-2)}{3}=1$

#### Answer:

$\frac{\left(5x-1\right)}{3}-\frac{\left(2x-2\right)}{3}=1$

$\Rightarrow \frac{5x-1-2x-(-2)}{3}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5x-1-2x+2}{3}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5x-2x+2-1}{3}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x+1}{3}=1$

Multiplying both sides by 3, we get

$\Rightarrow \left(\frac{3x+1}{3}\right)\times 3=1\times 3$

=> 3x + 1 = 3

Subtracting 1 from both sides, we get

=> 3x + 1 $-$ 1 = 3 $-$ 1

=> 3x = 2

Dividing both sides by 3, we get

=> $\frac{3x}{3}=\frac{2}{3}$

=> x = $\frac{2}{3}$

Verification:

Substituting x =$\frac{2}{3}$ in LHS, we get

$=\frac{5\left({\displaystyle \frac{2}{3}}\right)-1}{3}-\frac{2\left({\displaystyle \frac{2}{3}}\right)-2}{3}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle \frac{10}{3}}-1}{3}-\frac{{\displaystyle \frac{4}{3}}-2}{3}=\frac{{\displaystyle \frac{10-3}{3}}}{3}-\frac{{\displaystyle \frac{4-6}{3}}}{3}=\frac{7}{3\times 3}-\left({\displaystyle \frac{-2}{3\times 3}}\right)=\frac{7}{9}+\frac{2}{9}=\frac{9}{9}=1=RHS$

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 12:

Solve each of the following equations. Also, verify the result in each case.

$0.6x+\frac{4}{5}=0.28x+1.16$

#### Answer:

$0.6x+\frac{4}{5}=0.28x+1.16$

Transposing 0.28x to LHS and 4/5 to RHS, we get

=> 0.6x $-$ 0.28x = 1.16 $-$ $\frac{4}{5}$

=> 0.32x = 1.16 $-$ 0.8

=> 0.32x = 0.36

Dividing both sides by 0.32, we get

=> $\frac{0.32x}{0.32}=\frac{0.36}{0.32}$

=> x = $\frac{9}{8}$

Verification:

Substituting x = $\frac{9}{8}$ on both sides, we get

$0.6\left(\frac{9}{8}\right)+\frac{4}{5}=0.28\left(\frac{9}{8}\right)+1.16\phantom{\rule{0ex}{0ex}}\frac{5.4}{8}+\frac{4}{5}=\frac{2.52}{8}+1.16\phantom{\rule{0ex}{0ex}}0.675+0.8=0.315+1.16\phantom{\rule{0ex}{0ex}}1.475=1.475$

LHS = RHS

Hence, verified.

#### Page No 8.19:

#### Question 13:

Solve ech of the following question. Also, verify the result in each case.

$0.5x+\frac{x}{3}=0.25x+7$

#### Answer:

$0.5x+\frac{x}{3}=0.25x+7\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{05}{10}\text{x}+\frac{\text{x}}{3}=\frac{25\text{x}}{100}+7\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{2}+\frac{x}{3}=\frac{x}{4}+7$

Transposing x/4 to LHS, we get

$\Rightarrow \frac{x}{2}+\frac{x}{3}-\frac{x}{4}=7$

$\Rightarrow \frac{6x+4x-3x}{12}=7\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7x}{12}=7$

Multiplying both sides by 12, we get

=> $\frac{7x}{12}\times 12=7\times 12$

=> 7x = 84

Dividing both sides by 7, we get

=> $\frac{7x}{7}=\frac{84}{7}$

=> x = 12

Verification:

Substituting x = 12 on both sides, we get

$0.5\left(12\right)+\frac{12}{3}=0.25\left(12\right)+7$

6 + 4 = 3 + 7

10 = 10

LHS =RHS

Hence, verified.

#### Page No 8.26:

#### Question 1:

If 5 is subtracted from three times a number, the result is 16. Find the number.

#### Answer:

Let the required number be 'x'. Then, 5 subtracted from 3 times x = 3x $-$ 5.

⇒ 3x $-$ 5 = 16

Adding 5 to both sides, we get

⇒ 3x $-$ 5 + 5 = 16 + 5

⇒ 3x= 21

Dividing both sides by 3, we get

⇒ $\frac{3x}{3}=\frac{21}{3}$

⇒ x =7

Thus, the required number is 7.

#### Page No 8.26:

#### Question 2:

Find the number which when multiplied by 7 is increased by 78.

#### Answer:

Let the required number be 'x'. Thus, when multiplied by 7, it gives 7x, and x increases by 78.

⇒ 7x = x + 78

Transposing x to LHS, we get

⇒ 7x $-$ x = 78

⇒ 6x = 78

Dividing both sides by 6, we get

⇒ $\frac{6x}{6}=\frac{78}{6}$

⇒ x =13

Thus, the required number is 13.

#### Page No 8.26:

#### Question 3:

Find three consecutive natural numbers such that the sum of the first and second is 15 more than the third.

#### Answer:

Let the first number be 'x'. Hence, the second number = x + 1 and the third number = x + 2.

⇒ Sum of first and second numbers = (x) + (x + 1).

ATQ:

⇒ (x) + (x + 1) = 15 + (x + 2)

⇒ 2x + 1 = 17 + x

Transposing x to LHS and 1 to RHS, we get

⇒ 2x $-$ x = 17 $-$ 1

⇒ x = 16

So, first number = x = 16

Second number = x + 1 = 16 + 1 = 17

Third number = x + 2 = 16 + 2 = 18

Thus, the required consecutive natural numbers are 16, 17 and 18.

#### Page No 8.26:

#### Question 4:

The difference between two numbers is 7. Six times the smaller plus the larger is 77. Find the numbers.

#### Answer:

Let the smaller number be 'x'. So, the larger number = x + 7.

ATQ:

⇒ 6x + (x + 7) = 77

⇒ 6x + x + 7 = 77

⇒ 7x + 7 = 77

Subtracting 7 from both sides, we get

⇒ 7x + 7 $-$ 7 = 77 $-$ 7

⇒ 7x = 70

Dividing both sides by 7, we get

⇒ $\frac{7x}{7}=\frac{70}{7}$

x = 10

Thus, the smaller number = x = 10, and the larger number = x + 7 = 10 + 7 = 17.

The two required numbers are 10 and 17.

#### Page No 8.26:

#### Question 5:

A man says, "I am thinking of a number. When I divide it by 3 and then add 5, my answer is twice the number I thought of". Find the number.

#### Answer:

Let the number thought of by the man be 'x'.

So, ATQ:

⇒ $\frac{x}{3}+5=2x\phantom{\rule{0ex}{0ex}}$

Transposing x/3 to RHS, we get

⇒ 5 = $2x-\frac{x}{3}$

⇒ 5 = $\frac{6x-x}{3}$

⇒ 5 = $\frac{5x}{3}$

Multiplying both sides by 3, we get

⇒ $5\times 3=\frac{5x}{3}\times 3$

⇒ 15 = 5x

Dividing both sides by 5, we get

⇒ $\frac{15}{5}=\frac{5x}{5}$

⇒ x = 3

Thus, the number thought of by the man is 3.

#### Page No 8.26:

#### Question 6:

If a number is tripled and the result is increased by 5, we get 50. Find the number.

#### Answer:

Let the required number be 'x'.

So, ATQ:

⇒ 3x + 5 = 50

Subtracting 5 from both sides, we get

⇒ 3x + 5 $-$ 5 = 50 $-$ 5

⇒ 3x = 45

Dividing both sides by 3, we get

⇒ $\frac{3x}{3}=\frac{45}{3}$

⇒ x = 15

Thus, the required number is 15.

#### Page No 8.26:

#### Question 7:

Shikha is 3 years younger to her brother Ravish. If the sum of their ages is 37 years, what are their present ages?

#### Answer:

Let the present age of Shikha = 'x' years.

So, the present age of Shikha's brother Ravish = (x + 3) years.

So, sum of their ages = x + (x+ 3)

⇒ x + ( x + 3 ) = 37

⇒ 2x + 3 = 37

Subtracting 3 from both sides, we get

⇒ 2x + 3 $-$ 3 = 37 $-$ 3

⇒ 2x = 34

Dividing both sides by 2, we get

⇒ $\frac{2x}{2}=\frac{34}{2}$

⇒ x = 17

So, the present age of Shikha = 17 years, and the present age of Ravish = x+ 3 = 17 + 3 = 20 years.

#### Page No 8.26:

#### Question 8:

Mrs. Jain is 27 years older than her daughter Nilu. After 8 years she will be twice as old as Nilu. Find their present ages.

#### Answer:

Let the present age of Nilu = 'x' years.

Therefore, the present age of Nilu's mother, Mrs. Jain = (x + 27) years.

So, after 8 years,

Nilu's age = (x + 8), and Mrs. Jain's age = (x + 27 + 8) = (x + 35) years

⇒ x + 35 = 2(x + 8)

Expanding the brackets, we get

⇒ x + 35 = 2x + 16

Transposing x to RHS and 16 to LHS, we get

⇒ 35 $-$ 16 = 2x $-$ x

⇒ x = 19

So, the present age of Nilu = x = 19 years, and the present age of Nilu's mother = x+ 27 = 19 + 27 = 46 years.

#### Page No 8.26:

#### Question 9:

A man is 4 times as old as his son. After 16 years, he will be only twice as old as his son. Find the their present ages.

#### Answer:

Let the present age of the son = 'x' years.

Therefore, the present age of his father = '4x' years.

So, after 16 years,

Son's age = (x + 16) and father's age = (4x + 16) years

ATQ:

⇒ 4x + 16 = 2(x + 16)

⇒ 4x + 16 = 2x + 32

Transposing 2x to LHS and 16 to RHS, we get

⇒ 4x - 2x = 32 - 16

⇒ 2x = 16

Dividing both sides by 2, we get

⇒ $\frac{2x}{2}=\frac{16}{2}$

⇒ x = 8

So, the present age of the son = x = 8 years, and the present age of the father = 4x = 4(8) = 32 years.

#### Page No 8.26:

#### Question 10:

The difference in age between a girl and her younger sister is 4 years. The younger sister in turn is 4 years older than her brother. The sum of the ages of the younger sister and her brother is 16. How old are the three children?

#### Answer:

Let the age of the girl = 'x' years.

So, the age of her younger sister = (x $-$ 4) years.

Thus, the age of the brother = (x $-$ 4 $-$ 4) years = (x $-$ 8) years.

ATQ:

⇒ (x $-$ 4) + (x $-$ 8) = 16

⇒ x + x $-$ 4 $-$ 8 = 16

⇒ 2x $-$ 12 = 16

Adding 12 to both sides, we get

⇒ 2x $-$ 12 + 12 = 16 + 12

⇒ 2x = 28

Dividing both sides by 2, we get

⇒ $\frac{2x}{2}=\frac{28}{2}$

⇒ x = 14

Thus, the age of the girl = x = 14 years, the age of the younger sister = x $-$ 4 = 14 $-$ 4 = 10 years,

and the age of the younger brother = x $-$ 8 = 14 $-$ 8 = 6 years.

#### Page No 8.26:

#### Question 11:

One day, during their vacation at a beach resort, Shella found twice as many sea shells as Anita and Anita found 5 shells more than sandy. Together sandy and Shella found 16 sea shells. How many did each of them find?

#### Answer:

Let the number of sea shells found by Sandy = 'x'.

So, the number of sea shells found by Anita = (x + 5).

The number of sea shells found by Shella = 2 (x + 5 ).

According to the question,

⇒ x + 2 (x + 5) = 16

⇒ x + 2x + 10 = 16

⇒ 3x + 10 = 16

Subtracting 10 from both sides, we get

⇒ 3x + 10 $-$ 10 = 16 $-$ 10

⇒ 3x = 6

Dividing both sides by 3, we get

⇒ $\frac{3x}{3}=\frac{6}{3}$

⇒ x = 2

Thus, the number of sea shells found by Sandy = x = 2, the number of sea shells found by Anita = x + 5 = 2 + 5 = 7,

and the number of sea shells found by Shella = 2(x + 5) = 2(2 + 5) = 2(7 ) = 14.

#### Page No 8.26:

#### Question 12:

Andy has twice as many marbles as Pandy, and Sandy has half as many has Andy and Pandy put together. Andy has 110 marbles which is 115 marbles less than Sandy. How many does each of them have?

#### Answer:

Let the number of marbles with Pandy = 'x'.

So, the number of marbles with Andy = '2x'.

Thus, the number of marbles with Sandy = $\frac{1}{2}\left(x+2x\right)$ = $\frac{3x}{2}$.

According to the question,

$\frac{3x}{2}$ $-$ 115 = 110

Adding 115 to both sides, we get

$\frac{3x}{2}$ $-$ 115 + 115 = 110 + 115

$\frac{3x}{2}$ = 225

Multiplying both sides by 2, we get

$\frac{3x}{2}\times 2=225\times 2$

3x = 450

Dividing both sides by 3, we get

$\frac{3x}{3}=\frac{450}{3}$

x = 150

So, Pandy has 150 marbles, Andy has 2x = 2(150) = 300 marbles, and Sandy has $\frac{3x}{2}$ = $\frac{3\times 150}{2}$ = 225 marbles.

#### Page No 8.26:

#### Question 13:

A bag contains 25 paise and 50 paise coins whose total value is Rs 30. If the number of 25 paise coins is four times that of 50 paise coins, find the number of each type of coins.

#### Answer:

Let the number of 50 paise coins = 'x'.

So, the money value contribution of 50 paise coins = 0.5x.

The number of 25 paise coins = '4x'.

The money value contribution of 25 paise coins = 0.25(4x) = x.

According to the question,

⇒ 0.5x + x = 30

⇒ 1.5x = 30

Dividing both sides by 1.5, we get

⇒ $\frac{1.5x}{1.5}=\frac{30}{1.5}$

⇒ x = 20

Thus, the number of 50 paise coins = 'x' = 20, and the number of 25 paise coins = '4x' = 4 (20) = 80.

#### Page No 8.26:

#### Question 14:

The length of a rectangular field is twice its breadth. If the perimeter of the field is 228 metres, find the dimensions of the field.

#### Answer:

Let the breadth of the rectangle = 'x' metres.

According to the question,

Length of the rectangle = '2x' metres

Perimeter of a rectangle = 2 (length + breadth)

So, 2 (2x + x) = 228

=> 2 (3x) = 228

=> 6x = 228

Dividing both sides by 6, we get

=> $\frac{6x}{6}=\frac{228}{6}$

=> x = 38

So, the breadth of the rectangle = x = 38 metres, and the length of the rectangle = 2x = 2(38) = 76 metres.

#### Page No 8.26:

#### Question 15:

There are only 25 paise coins in a purse. The value of money in the purse is Rs 17.50. Find the number of coins in the purse.

#### Answer:

Let the number of 25-paise coins in the purse be 'x'.

So, the value of money in the purse = 0.25x.

But 0.25x = 17.5.

Dividing both sides by 0.25, we get

=> $\frac{0.25x}{0.25}=\frac{17.5}{0.25}$

=> x = 70

Thus, the number of 25-paise coins in the purse = 70.

#### Page No 8.26:

#### Question 16:

In a hostel mess, 50 kg rice are consumed everyday. If each student gets 400 gm of rice per day, find the number of students who take meals in the hostel mess.

#### Answer:

Let the number of students in the hostel be 'x'.

Quantity of rice consumed by each student = 400 gm.

So, daily rice consumption in the hostel mess = 400(x).

But, daily rice consumption = 50 kg = 50 $\times $ 1000 = 50000 gm [since 1 kg = 1000 gm].

According to the question,

400x = 50000

Dividing both sides by 400, we get

=> $\frac{400x}{400}=\frac{50000}{400}$

=> x = 125

Thus, 125 students have their meals in the hostel mess.

#### Page No 8.6:

#### Question 1:

Verify by substitution that:

(i) *x* = 4 is the root of 3*x* − 5 = 7

(ii) *x* = 3 is the root of 5 + 3*x* = 14

(iii) *x* = 2 is the root of 3*x* − 2 = 8*x* − 12

(iv) *x* = 4 is the root of $\frac{3x}{2}=6$

(v) *y* = 2 is the root of *y* − 3 = 2*y* − 5

(vi) *x* = 8 is the root of $\frac{1}{2}x+7=11$

#### Answer:

(i) *x* = 4 is the root of 3*x* − 5 = 7.

Now, substituting x = 4 in place of 'x' in the given equation 3x − 5 = 7,

3(4) − 5 = 7

12 − 5 = 7

7 = 7

LHS = RHS

Hence, *x* = 4 is the root of 3*x* − 5 = 7.

(ii) *x* = 3 is the root of 5 + 3*x* = 14.

Now, substituting x = 3 in place of 'x' in the given equation 5 + 3*x* = 14,

5 + 3(3) = 14

5 + 9 = 14

14 = 14

LHS = RHS

Hence, *x* = 3 is the root of 5 + 3*x* = 14.

(iii) *x* = 2 is the root of 3*x* − 2 = 8*x* − 12.

Now, substituting x = 2 in place of 'x' in the given equation 3*x* − 2 = 8*x* − 12,

3(2) − 2 = 8(2) − 12

6 − 2 = 16 − 12

4 = 4

LHS = RHS

Hence, *x* = 2 is the root of 3*x* − 2 = 8*x* − 12.

(iv) x = 4 is the root of $\frac{3x}{2}=6$.

Now, substituting x = 4 in place of 'x' in the given equation $\frac{3x}{2}=6$,

$\frac{3\times 4}{2}=6\phantom{\rule{0ex}{0ex}}\frac{12}{2}=6\phantom{\rule{0ex}{0ex}}6=6$

LHS = RHS

Hence, x = 4 is the root of $\frac{3x}{2}=6$.

(v) *y* = 2 is the root of *y* − 3 = 2*y* − 5.

Now, substituting y = 2 in place of 'y' in the given equation *y* − 3 = 2*y* − 5,

2 − 3 = 2(2) − 5

−1 = 4 − 5

−1 = −1

LHS = RHS

Hence, *y *= 2 is the root of *y* − 3 = 2*y* − 5.

(vi) *x* = 8 is the root of $\frac{1}{2}x+7=11$.

Now, substituting x = 8 in place of 'x' in the given equation $\frac{1}{2}x+7=11$,

$\frac{1}{2}\times 8+7=11$

4 + 7 = 11

11 = 11

LHS = RHS

Hence, *x* = 8 is the root of $\frac{1}{2}x+7=11$.

#### Page No 8.6:

#### Question 2:

Solve each of the following equations by trial-and-error method:

(i) *x* + 3 = 12

(ii) *x* − 7 = 10

(iii) 4*x* = 28

(iv) $\frac{x}{2}+7=11$

(v) 2*x* + 4 = 3*x*

(vi) $\frac{x}{4}=12$

(vii) $\frac{15}{x}=3$

(viii) $\frac{x}{18}=20$

#### Answer:

(i) *x* + 3 = 12

Here, LHS = x + 3 and RHS = 12.

x | LHS | RHS | Is LHS = RHS? |

1 | 1+3=4 | 12 | No |

2 | 2+3=5 | 12 | No |

3 | 3+3=6 | 12 | No |

4 | 4+3=7 | 12 | No |

5 | 5+3=8 | 12 | No |

6 | 6+3=9 | 12 | No |

7 | 7+3=10 | 12 | No |

8 | 8+3=11 | 12 | No |

9 | 9+3=12 | 12 | Yes |

Hence, x = 9 is the solution to this equation.

(ii)

*x*− 7 = 10

Here, LHS = x −7 and RHS =10.

x | LHS | RHS | Is LHS = RHS? |

9 | 9−7 = 2 | 10 | No |

10 | 10−7 = 3 | 10 | No |

11 | 11−7=4 | 10 | No |

12 | 12−7=5 | 10 | No |

13 | 13−7=6 | 10 | No |

14 | 14−7=7 | 10 | No |

15 | 15−7=8 | 10 | No |

16 | 16−7=9 | 10 | No |

17 | 17−7=10 | 10 | Yes |

Hence, x = 17 is the solution to this equation.

(iii) 4

*x*= 28

Here, LHS = 4x and RHS = 28.

x | LHS | RHS | Is LHS = RHS? |

1 | 4$\times $1=4 | 28 | No |

2 | 4$\times $2=8 | 28 | No |

3 | 4$\times $3=12 | 28 | No |

4 | 4$\times $4=16 | 28 | No |

5 | 4$\times $5=20 | 28 | No |

6 | 4$\times $6=24 | 28 | No |

7 | 4$\times $7=28 | 28 | Yes |

Hence, x = 7 is the solution to this equation.

(iv) $\frac{x}{2}+7=11$

Here, LHS = $\frac{x}{2}+7$ and RHS = 11.

Since RHS is a natural number, $\frac{x}{2}$ must also be a natural number, so we must substitute values of x that are multiples of 2.

x | LHS | RHS | Is LHS = RHS? |

2 | $\frac{2}{2}$+7=8 | 11 | No |

4 | $\frac{4}{2}$+7=9 | 11 | No |

6 | $\frac{6}{2}$+7=10 | 11 | No |

8 | $\frac{8}{2}$+7=11 | 11 | Yes |

Hence, x = 8 is the solution to this equation.

(v) 2

*x*+ 4 = 3

*x*

*Here, LHS = 2*

*x*+ 4 and RHS = 3x.

x | LHS | RHS | Is LHS = RHS? |

1 | 2(1)+4=6 | 3(1)=3 | No |

2 | 2(2)+4=8 | 3(2)=6 | No |

3 | 2(3)+4=10 | 3(3)=9 | No |

4 | 2(4)+4=12 | 3(4)=12 | Yes |

Hence, x = 4 is the solution to this equation.

(vi) $\frac{x}{4}$ = 12

Here, LHS =$\frac{x}{4}$ and RHS = 12.

Since RHS is a natural number, $\frac{x}{4}$ must also be a natural number, so we must substitute values of x that are multiples of 4.

x | LHS | RHS | Is LHS = RHS? |

16 | $\frac{16}{4}$=4 | 12 | No |

20 | $\frac{20}{4}$=5 | 12 | No |

24 | $\frac{24}{4}$=6 | 12 | No |

28 | $\frac{28}{4}$=7 | 12 | No |

32 | $\frac{32}{4}$=8 | 12 | No |

36 | $\frac{36}{4}$=9 | 12 | No |

40 | $\frac{40}{4}$=10 | 12 | No |

44 | $\frac{44}{4}$=11 | 12 | No |

48 | $\frac{48}{4}$=12 | 12 | Yes |

Hence, x = 48 is the solution to this equation.

(vii) $\frac{15}{x}$ = 3

Since RHS is a natural number, $\frac{15}{x}$ must also be a natural number, so we must substitute values of x that are factors of 15.

x | LHS | RHS | Is LHS = RHS? |

1 | $\frac{15}{1}$=15 | 3 | No |

3 | $\frac{15}{3}$=5 | 3 | No |

5 | $\frac{15}{5}$=3 | 3 | Yes |

Hence, x = 5 is the solution to this equation.

(viii) $\frac{x}{18}$= 20

Here, LHS =$\frac{x}{18}$ and RHS = 20.

Since RHS is a natural number, $\frac{x}{18}$ must also be a natural number, so we must substitute values of x that are multiples of 18.

x | LHS | RHS | Is LHS = RHS? |

324 | $\frac{324}{18}$=18 | 20 | No |

342 | $\frac{342}{18}$=19 | 20 | No |

360 | $\frac{360}{18}$=20 | 20 | Yes |

Therefore, if x = 360, LHS = RHS.

Hence, x = 360 is the solution to this equation.

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