RD Sharma solution class 7 chapter 8 Linear Equations In One Variables Exercise 8.3

Exercise 8.3

Page-8.19

Question 1:

Solve each of the following equations. Also, verify the result in each case.
6x + 5 = 2x + 17

Answer 1:

We have
⇒ 6x + 5 = 2x + 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.

Question 2:

Solve each of the following equations. Also, verify the result in each case.
2(5x − 3) − 3(2x − 1) = 9

Answer 2:

We have
⇒2(5x − 3) − 3(2x − 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.

Question 3:

Solve each of the following equations. Also, verify the result in each case.
$\frac{x}{2}=\frac{x}{3}+1$

Answer 3:

$\frac{x}{2} = \frac{x}{3} + 1$
Transposing $\frac{x}{3}$ to LHS, we get
$$\begin{gathered} \Rightarrow \frac{x}{2}-\frac{x}{3} = 1 \\ \Rightarrow \frac{3x-2x}{6} = 1 \end{gathered}$$ 
$\Rightarrow \frac{x}{6} = 1$      
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.

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 4:

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

Question 5:

Solve each of the following equations. Also, verify the result in each case.
$\frac{3}{4}(x-1)=x-3$

Answer 5:

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

Question 6:

Solve each of the following equations. Also, verify the result in each case.
3(x − 3) = 5(2x + 1)

Answer 6:

6. 3(x − 3) = 5(2x + 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.

Question 7:

Solve each of the following equations. Also, verify the result in each case.
3x − 2 (2x − 5) = 2(x + 3) − 8

Answer 7:

3x − 2 (2x − 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.

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 8:

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

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 9:

$\frac{6x - 2}{9} + \frac{3x + 5}{18} = \frac{1}{3}$
=> $\frac{6x\times 2 - 2\times 2 + 3x + 5}{18} = \frac{1}{3}$
=> $\frac{12x - 4 + 3x + 5}{18} = \frac{1}{3}$
=> $\frac{15x + 1}{18} = \frac{1}{3}$
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.

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 10:

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

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 11:

$\frac{\left(5x -1\right)}{3}- \frac{\left(2x - 2\right)}{3} = 1$
 $$\begin{gathered} \Rightarrow \frac{5x - 1 - 2x - (-2)}{3}= 1 \\ \Rightarrow \frac{5x - 1 - 2x + 2}{3} = 1 \\ \Rightarrow \frac{5x - 2x + 2 -1}{3 } = 1 \\ \Rightarrow \frac{3x + 1}{3} = 1 \end{gathered}$$
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
$$\begin{gathered} =\frac{5\left({\displaystyle \frac{2}{3}}\right) - 1}{3}-\frac{2\left({\displaystyle \frac{2}{3}}\right) - 2}{3} \\ =\frac{{\displaystyle \frac{10}{3}}-1}{3} - \frac{{\displaystyle \frac{4}{3}}- 2}{3}=\frac{{\displaystyle \frac{10-3}{3}}}{3}-\frac{{\displaystyle \frac{4-6}{3}}}{3}=\frac{7}{3\times 3}-\left({\displaystyle \frac{-2}{3\times 3}}\right)=\frac{7}{9} + \frac{2}{9}= \frac{9}{9}= 1 = RHS \end{gathered}$$
LHS = RHS
Hence, verified.

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 12:

$0.6x + \frac{4}{5} = 0.28x + 1.16$
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
$$\begin{gathered} 0.6\left(\frac{9}{8}\right) + \frac{4}{5} = 0.28\left(\frac{9}{8}\right) + 1.16 \\ \frac{5.4}{8} + \frac{4}{5} = \frac{2.52}{8} + 1.16 \\ 0.675 + 0.8 = 0.315 + 1.16 \\ 1.475 = 1.475 \end{gathered}$$

    LHS = RHS
Hence, verified.

Question 13:

Solve ech of the following question. Also, verify the result in each case.
$0.5x+\frac{x}{3}=0.25x+7$

Answer 13:

$$\begin{gathered} 0.5x + \frac{x}{3} = 0.25x + 7 \\ \Rightarrow \frac{05}{10}\text{x}+\frac{\text{x}}{3}=\frac{25\text{x}}{100}+7 \\ \Rightarrow \frac{x}{2} + \frac{x}{3} = \frac{x}{4} + 7 \end{gathered}$$
Transposing x/4 to LHS, we get

$\Rightarrow \frac{x}{2} + \frac{x}{3} -\frac{x}{4} = 7$                     
$$\begin{gathered} \Rightarrow \frac{6x + 4x - 3x}{12 }= 7 \\ \Rightarrow \frac{7x}{12} = 7 \end{gathered}$$
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.