RS Aggarwal 2019,2020 solution class 7 chapter 7 Linear Equation In One Variable Exercise 7A

Exercise 7A

Page-111

Question 1:

3x − 5 = 0

Answer 1:

$$\begin{gathered} 3x-5=0 \\ \Rightarrow 3x=5 (\mathrm{Transposing} -5 \mathrm{to} \mathrm{RHS}) \\ \Rightarrow x=\frac{5}{3} \\ \mathrm{CHECK}: \mathrm{By} \mathrm{substituting} x=\frac{5}{3} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{LHS} =3\left(\frac{5}{3}\right) - 5 = 5 - 5 = 0 \\ \mathrm{RHS} = 0 \\ \therefore \mathrm{LHS} = \mathrm{RHS} \\ \mathrm{Hence} \mathrm{checked}. \end{gathered}$$

Question 2:

8x − 3 = 9 − 2x

Answer 2:

8x − 3 = 9 − 2x
$\Rightarrow$8x + 2x = 9 + 3                   (By transposition)
$\Rightarrow$10x = 12
$$\begin{gathered} \Rightarrow x=\frac{12}{10}=\frac{6}{5} \\ \mathrm{CHECK}: \mathrm{By} \mathrm{substituting} x=\frac{6}{5} \mathrm{in} \mathrm{the} \mathrm{given} \mathrm{equation}, \mathrm{we} \mathrm{get}: \\ \mathrm{LHS}: 8\left(\frac{6}{5}\right)-3=\frac{48}{5}-3=\frac{48-15}{5}=\frac{33}{5} \\ \mathrm{RHS}: 9-2\left(\frac{6}{5}\right)=9-\frac{12}{5}=\frac{45-12}{5}=\frac{33}{5} \\ \therefore \mathrm{LHS}= \mathrm{RHS} \\ \mathrm{Hence} \mathrm{checked}. \end{gathered}$$

Question 3:

7 − 5x = 5 − 7x

Answer 3:

$\begin{array}{l}\text{We have:}\\ 7 - 5x= 5 - 7x\\ \\ \Rightarrow -5x + 7x = 5 - 7 \left[\text{transposing -7}x\text{ to LHS and 7 to RHS}\right]\\ \Rightarrow 2x = -2\\ \Rightarrow x = \frac{-{\sout{2}}^{-1}}{{\sout{2}}^1}\\ \Rightarrow x = -1\\ \text{Thus, }x \text{= }-\text{1 is a solution to the given equation.}\\ \text{CHECK: Substituting }x \text{= }-\text{1 in the given equation, we get:}\\ \text{LHS: }\text{= 7 }- 5x\\ \text{ = 7 }-5 \times (-1)\\ = 7+5\\ = 12\\ \text{RHS: }\\ \text{= 5 }- 7\text{x}\\ \text{=5 }- 7 \times (-1)\\ \text{= 5 + 7}\\ \text{=12}\\ \therefore \text{LHS = RHS}\\ \text{Hence, }x\text{ = }-1\text{ is a solution of the given equation.}\\ \end{array}$

Question 4:

3 + 2x = 1 − x

Answer 4:

$\begin{array}{l}\text{We have:}\\ \text{3}+\text{2}x=\text{ 1 – }x\\ \text{⇒ 2}x\text{ + }x\text{+ 3 – 1 = 0 (By transposition)}\\ \Rightarrow \text{ 3}x\text{ + 2 = 0}\\ \Rightarrow x\text{ = –}\frac{\text{2}}{3}\\ \text{CHECK: Substituting }x\text{=}-\frac{2}{3}\text{ in the given equation, we get:}\\ \\ \text{ LHS: 3+2}\mathit{\text{x}}\\ \text{ =3+2}\times (-\frac{2}{3})\\ \text{ =3 }-\frac{4}{3}\\ \text{ =}\frac{9-4}{3}\\ \text{ =}\frac{5}{3}\\ \text{RHS: 1}-x\\ \text{ =1}-\left(\frac{-2}{3}\right)\\ \text{ =1+}\frac{2}{3}\\ \text{ =}\frac{3+2}{3}\\ \text{ =}\frac{5}{3}\\ \therefore \mathrm{LHS}=\mathrm{RHS}\\ \text{Hence, }x\text{=}-\frac{2}{3}\text{ is a solution of the given equation}\text{. }\end{array}$

Question 5:

2(x − 2) +3(4x − 1) = 0

Answer 5:

$\begin{array}{l}\text{We have:}\\ 2(x-2)+3(4x-1)=0\\ \Rightarrow \text{ 2}x-4+12x-3\text{ =0}\\ \Rightarrow \text{ 14}x-\text{7 =0 }\\ \Rightarrow \text{ 14}x\text{ = }7 \text{(By transposition)}\\ \Rightarrow x=\frac{1}{2}\\ \\ \text{CHECK: Substituting }x\text{=}\frac{1}{2}\text{ in the given equation, we get:}\\ \\ \text{LHS: }2(x-2)+3(4x-1)\\ \text{ =2}x-4+12x-3\\ \text{ =2}\times \frac{1}{2}-4+12\times \frac{1}{2}-3\\ \text{ =1-4+6-3}\\ \text{ =}-7+7\\ \text{ =0}\\ \text{RHS: 0}\\ \text{∴ LHS= RHS }\\ \text{Hence, }x\text{=}\frac{1}{2} is a solution \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{equation}.\end{array}$

Question 6:

5(2x − 3) − 3(3x − 7) = 5

Answer 6:

$\begin{array}{l}\text{We have:}\\ 5(2x-3)-3(3x-7)=5\\ \Rightarrow \text{ 10}x-15-9x+21\text{ = 5}\\ \Rightarrow \text{ 10}x-\text{9}x\text{ =5+15}-21\text{ (By transposition)}\\ \Rightarrow x\text{ = 20}-21\\ \Rightarrow x=-1\\ \\ \text{CHECK: Substituting }x\text{=}-1\text{ in the given equation, we get:}\\ \\ \text{ LHS: }5(2x-3)-3(3x-7)\\ \text{ =10}x-15-9x+21\\ \text{ =10}\times (-1)-15-9\times (-1)+21\\ \text{ =}-10-15+9+21\\ \text{ =}-25+30\\ \text{ =5}\\ \text{ RHS: 5}\\ \therefore \mathrm{LHS}= \mathrm{RHS} \\ \text{Hence, }x\text{=}-1\text{ is a solution of the given equation}\text{.}\end{array}$

Question 7:

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

Answer 7:

$$\begin{gathered} \begin{array}{l}\\ \text{We have:}\\ 2x-\frac{1}{3}=\frac{1}{5}-x\\ \Rightarrow 2x+x=\frac{1}{5}+\frac{1}{3}\\ \Rightarrow 3x=\frac{3\times 1+5\times 1}{15}\\ \Rightarrow 3x=\frac{3+5}{15}\\ \Rightarrow 3x=\frac{8}{15}\\ \Rightarrow x=\frac{8}{15\times 3}\\ \Rightarrow x=\frac{8}{45}\\ \\ \text{CHECK: Substituting }x\text{=}\frac{8}{45}\text{ in the given equation, we get:}\\ \\ \text{ LHS: }2x-\frac{1}{3}\\ \text{ =2}\times \frac{8}{45}-\frac{1}{3}\\ \text{ =}\frac{16}{45}-\frac{1}{3}\\ \text{ =}\frac{16\times 1-15\times 1}{45}\\ \text{ =}\frac{16-15}{45}\\ \text{ =}\frac{1}{45}\\ \\ \\ \mathrm{RHS}: \frac{1}{5}-x\\ \text{ =}\frac{1}{5}-\frac{8}{45}\\ \text{ =}\frac{1\times 9-1\times 8}{45}\\ \text{ =}\frac{9-8}{45}\\ \text{ =}\frac{1}{45}\\ \text{ ∴ LHS=RHS }\end{array} \\ \mathrm{Hence}, x\text{=}\frac{8}{45}\text{ is a solution of the given equation}\text{. } \end{gathered}$$

Question 8:

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

Answer 8:

$\begin{array}{l}\\ \text{We have:}\\ \frac{1}{2}x-3=5+\frac{1}{3}x\\ \Rightarrow \frac{1}{2}x-\frac{1}{3}x=5+3(\text{transposing }\frac{1}{3}x\text{ to LHS and }-3\text{ to RHS})\\ \Rightarrow \left(\frac{1\times 3-1\times 2}{6}\right)x=8\\ \Rightarrow \left(\frac{3-2}{6}\right)x=8\\ \Rightarrow \frac{1}{6}x=8\\ \Rightarrow x=8 \times 6\\ \Rightarrow x=48\\ \\ \text{CHECK: Substituting }x\text{=48 in the given equation, we get:}\\ \\ \text{LHS: }\frac{1}{2}x-3\\ \text{ =}\frac{1}{{\sout{2}}^1}\times \sout{4}{\sout{8}}^{24}-3\\ \text{ =24}-3\\ \text{ =21}\\ \\ \text{RHS: }5+\frac{1}{3}x\\ \text{ =5+}\frac{1}{{\sout{3}}^1}\times \sout{4}{\sout{8}}^{16}\\ \text{ =5+16}\\ \text{ =21}\\ \text{ ∴ LHS=RHS }\\ \text{ Hence, }x\text{=48 is a solution of the given equation}\text{. }\end{array}$

Question 9:

$\frac{x}{2}+\frac{x}{4}=\frac{1}{8}$

Answer 9:

$\begin{array}{l}\frac{x}{2}+\frac{x}{4}=\frac{1}{8}\\ \Rightarrow \frac{x\times 2+x\times 1}{4}=\frac{1}{8}\\ \Rightarrow \frac{2x+x}{4}=\frac{1}{8}\\ \Rightarrow \frac{3x}{4}=\frac{1}{8}\\ \Rightarrow 3x=\frac{1}{{\sout{8}}^2}\times {\sout{4}}^1\\ \Rightarrow 3x=\frac{1}{2}\\ \Rightarrow x=\frac{1}{6}\\ \\ \text{CHECK: Substituting }x\text{=}\frac{1}{6}\text{ in the given equation, we get:}\\ \\ \text{ LHS: }\frac{x}{2}+\frac{x}{4}\\ \text{ =}\frac{x\times 2+x\times 1}{4}\\ \text{ =}\frac{2x+x}{4}\\ \text{ =}\frac{3x}{4}\\ \text{ =}\frac{{\sout{3}}^1}{4}\times \frac{1}{\sout{6^2}}\\ \text{ =}\frac{1}{8}\\ \text{RHS: }\frac{1}{8}\\ \therefore \mathrm{LHS}=\mathrm{RHS}\\ \text{ Hence, }x\text{=}\frac{1}{3}\text{ is a solution of the given equation}\text{. }\end{array}$

Question 10:

3x + 2(x + 2) = 20 − (2x − 5)

Answer 10:

$\begin{array}{l}\text{We have:}\\ 3x+2(x+2)=20-(2x-5)\\ \Rightarrow \text{ 3}x+2x+4\text{ = 20}-2x+5\\ \Rightarrow \text{ 3}x+2x+2x=20+5-4(\mathrm{T}\text{ransposing }-\text{2x to LHS and 4 to RHS})\\ \Rightarrow \text{ 7}x\text{ =21}\\ \Rightarrow x=\frac{\sout{2}{\sout{1}}^3}{{\sout{7}}^1}\\ \Rightarrow x=3\\ \\ \text{CHECK: Substituting }x\text{=3 in the given equation, we get:}\\ \\ \text{LHS=}3x+2(x+2)\\ \text{ =3}x+2x+4\\ \text{ =5}x+4\\ \text{ =5}\times \text{3+4}\\ \text{ =15+4}\\ \text{ =19}\\ \\ \\ \text{ RHS=}20-(2x-5)\\ \text{ =20}-2x+5\\ \text{ =25}-2\times 3\\ \text{ =25}-6\\ \text{ =19}\\ \therefore \mathrm{LHS}=\mathrm{RHS}\\ \text{Hence, }x\text{=3 is a solution of the given equation}\text{. }\end{array}$

Question 11:

13(y − 4) − 3(y − 9) − 5(y + 4) = 0

Answer 11:

$\begin{array}{l}\text{We have:}\\ 13(y-4)-3(y-9)-5(y+4)=0\\ \Rightarrow \text{ 13y}-52-3y+27-5y-20\text{ = 0}\\ \Rightarrow \text{13y}-3y-5y=52+20-27(\mathrm{T}\text{ransposing }-52,-20\text{ and 27 to RHS})\\ \Rightarrow \text{5y =45}\\ \Rightarrow y=\frac{\sout{4}{\sout{5}}^9}{{\sout{5}}^1}\\ \Rightarrow y=9\\ \\ \text{CHECK: Substituting }x\text{=9 in the given equation, we get:}\\ \\ \text{LHS=}13(y-4)-3(y-9)-5(y+4)\\ \text{ =13y}-52-3y+27-5y-20\\ \text{ =13y}-3y-5y-52+27-20\\ \text{ =5y}-45\\ \text{ =5}\times \text{9}-45\\ \text{ =45}-45\\ \text{ =0}\\ \text{RHS=0}\\ \text{∴ LHS=RHS }\\ \text{Hence, }x\text{=9 is a solution of the given equation}\text{. }\\ \end{array}$

Question 12:

$\frac{2m+5}{3}=3m-10$

Answer 12:

$\begin{array}{l}\\ \text{We have,}\\ \frac{2m+5}{3}=3m-10\\ \Rightarrow 2m+5=3(3m-10)\\ \Rightarrow 2m+5=9m-30\\ \Rightarrow 2m-9m=-30-5\left(\mathrm{T}\text{ransposing 9}\mathit{\text{m }}\text{to LHS and 5 to RHS}\right)\\ \Rightarrow -7m=-35\\ \Rightarrow m=\frac{-\sout{3}{\sout{5}}^5}{-{\sout{7}}^1}\\ \Rightarrow m=5\\ \\ \text{CHECK: Substituting }m=\text{ 5 in the given equation, we get:}\\ \\ \text{LHS=}\frac{2m+5}{3}\\ \text{ =}\frac{2\times 5+5}{3}\\ \text{ =}\frac{10+5}{3}\\ \text{ =}\frac{\sout{1}{\sout{5}}^5}{{\sout{3}}^1}\\ \text{ =5}\\ \\ \text{ RHS=}3m-10\\ \text{ =3}\times \text{5}-10\\ \text{ =15}-10\\ \text{ =5}\\ \text{ ∴ LHS=RHS }\\ \text{ Hence, }x\text{=5 is a solution of the given equation}\text{.}\end{array}$

Question 13:

6(3 x + 2) − 5(6x − 1) = 3(x − 8) − 5(7x − 6) + 9x

Answer 13:

$\begin{array}{l}\text{We have:}\\ 6(3x+2)-5(6x-1)=3(x-8)-5(7x-6)+9x\\ \text{⇒ 18}x\text{+1}2-30x+5\text{ =3x}-24-35x+30+9x\\ \Rightarrow \text{18}x-30x-3x+35x-9x=-24+30-12-5(\mathrm{T}\text{ransposing 3}x,9x \mathrm{and} -35x\text{ to LHS and 12 and 5 to RHS})\\ \Rightarrow \text{ 53}x-42x\text{ =30}-41\\ \Rightarrow \text{11}x=-11\\ \Rightarrow x=\frac{-\sout{1}{\sout{1}}^1}{\sout{1}{\sout{1}}^1}\\ \Rightarrow x=-1\\ \\ \text{CHECK: Substituting }x\text{=}-1\text{ in the given equation, we get:}\\ \\ \text{LHS=}6(3x+2)-5(6x-1)\\ \text{ =18}x\text{+1}2-30x+5\\ \text{ =}-12x+17\\ \text{ =}-12\times (-1)+17\\ \text{ =12+17}\\ \text{ =29}\\ \\ \text{RHS=}3(x-8)-5(7x-6)+9x\\ \text{ =3}x-24-35x+30+9x\\ \text{ =12}x-35x-24+30\\ \text{ =}-23x+6\\ \text{ =}-23\times (-1)+6\\ \text{ = 23+6}\\ \text{ =29 }\\ \text{ ∴ LHS=RHS }\\ \text{Hence, }x\text{=}-1\text{ is a solution of the given equation}\text{. }\\ \end{array}$

Question 14:

t − (2t + 5) − 5(1 − 2t) = 2(3 + 4t) −3(t − 4)

Answer 14:

$\begin{array}{l}\text{We have:}\\ t-(2t+5)-5(1-2t)=2(3+4t)-3(t-4)\\ \Rightarrow t-2t-5-5+10t\text{ =6+8t}-3t+12\\ \Rightarrow t-2t+10t-8t+3t=6+12+5+5(\mathrm{By} \text{transposition })\\ \Rightarrow \text{14}t-10t\text{ =28}\\ \Rightarrow \text{4}t=28\\ \Rightarrow x=\frac{\sout{2}{\sout{8}}^7}{{\sout{4}}^1}\\ \Rightarrow x=7\\ \\ \text{CHECK: Substituting }x\text{=7 in the given equation, we get:}\\ \\ \text{LHS=}t-(2t+5)-5(1-2t)\\ \text{ =}t-2t-5-5+10t\\ \text{ =11}t-2t-10\\ \text{ =9}t-10\\ \text{ =9}\times 7-10\\ \text{ =63}-10\\ \text{ =53}\\ \\ \text{ RHS=}2(3+4t)-3(t-4)\\ \text{ =6+8}t-3t+12\\ \text{ =5}t+18\\ \text{ =5}\times 7+18\\ \text{ =35}+18\\ \text{ = 53}\\ \text{ ∴ LHS=RHS }\\ \\ \text{Hence, }x\text{=7 is a solution of the given equation}\text{. }\\ \end{array}$

Question 15:

$\frac{2}{3}x=\frac{3}{8}x+\frac{7}{12}$

Answer 15:

$\begin{array}{l}\\ \text{We have:}\\ \frac{2}{3}x=\frac{3}{8}x+\frac{7}{12}\\ \Rightarrow \frac{2}{3}x-\frac{3}{8}x=\frac{7}{12}\left(\mathrm{T}\text{ransposing }\frac{3}{8}x\text{ to LHS}\right)\\ \Rightarrow \left(\frac{2\times 8-3\times 3}{24}\right)x=\frac{7}{12}\\ \Rightarrow \left(\frac{16-9}{24}\right)x=\frac{7}{12}\\ \Rightarrow \frac{7}{24}x=\frac{7}{12}\\ \Rightarrow x=\frac{{\sout{7}}^1}{\sout{1}{\sout{2}}^1}\times \frac{\sout{2}{\sout{4}}^2}{{\sout{7}}^1}\\ \Rightarrow x=2\\ \\ \text{CHECK: Substituting }x\text{=2 in the given equation, we get:}\\ \\ \text{LHS=}\frac{2}{3}x\\ \text{ =}\frac{2}{3}\times 2\\ \text{ =}\frac{4}{3}\\ \\ \text{RHS=}\frac{3}{8}x+\frac{7}{12}\\ \text{ =}\frac{3}{8}\times 2+\frac{7}{12}\\ \text{ =}\frac{6}{8}+\frac{7}{12}\\ \text{ =}\frac{6\times 3+7\times 2}{24}\\ \text{ =}\frac{18+14}{24}\\ \text{ =}\frac{\sout{3}{\sout{2}}^4}{\sout{2}{\sout{4}}^3}\\ =\frac{4}{3}\\ \begin{array}{l}\text{ ∴ LHS=RHS }\\ \text{ Hence, }x\text{=2 is a solution of the given equation}\text{. }\end{array}\end{array}$

Question 16:

$\frac{3x-1}{5}-\frac{x}{7}=3$

Answer 16:

$\begin{array}{l}\\ \text{We have:}\\ \frac{3x-1}{5}-\frac{x}{7}=3\\ \Rightarrow \frac{7(3x-1)-5\times x}{35}=3\\ \Rightarrow \left(\frac{21x-7-5x}{35}\right)=3\\ \Rightarrow \left(\frac{16x-7}{35}\right)=3\\ \Rightarrow 16x-7=3\times 35\left(\mathrm{T}\text{ransposing 35 to RHS}\right)\\ \Rightarrow 16x-7=105\\ \Rightarrow 16x=105+7\\ \Rightarrow 16x=112\\ \Rightarrow x=\frac{\sout{1}\sout{1}{\sout{2}}^7}{\sout{1}{\sout{6}}^1}\\ \Rightarrow x=7\\ \\ \text{CHECK: Substituting }x\text{=7 in the given equation, we get:}\\ \\ \text{ LHS=}\frac{3x-1}{5}-\frac{x}{7}\\ \text{ =}\frac{7(3x-1)-5\times x}{35}\\ \text{ =}\left(\frac{21x-7-5x}{35}\right)\\ \text{ =}\left(\frac{16x-7}{35}\right)\\ \text{ =}\left(\frac{16\times 7-7}{35}\right)\\ \text{ =}\frac{112-7}{35}\\ \text{ =}\frac{\sout{1}\sout{0}{\sout{5}}^3}{\sout{3}{\sout{5}}^1}\\ \text{ =3 }\\ \\ \text{ RHS=3}\\ \text{ ∴ LHS=RHS }\\ \text{ Hence, }x\text{=3 is a solution of the given equation}\text{. }\end{array}$

Question 17:

$2x-3=\frac{3}{10}(5x-12)$

Answer 17:

$\begin{array}{l}\\ \text{We have:}\\ 2x-3=\frac{3}{10}(5x-12)\\ \Rightarrow 10(2x-3)=3(5x-12)\\ \Rightarrow 20x-30=15x-36\\ \Rightarrow 20x-15x=-36+30(\mathrm{T}\text{ransposing 15}x\text{ to LHS and }-30\text{ to RHS})\\ \Rightarrow 5x=-6\\ \Rightarrow x=\frac{-6}{5}\\ \text{CHECK: Substituting }x\text{=}\frac{-6}{5}\text{ in the given equation, we get:}\\ \\ \text{LHS=}2x-3\\ \text{ =2}\times \left(\frac{-6}{5}\right)-3\\ \text{ =}\frac{-12}{5}-3\\ \text{ =}\frac{-12-(3\times 5)}{5}\\ \text{ =}\frac{-12-15}{5}\\ \text{ =}\frac{-27}{5}\\ \\ \text{ RHS=}\frac{3}{10}(5x-12)\\ \text{ =}\frac{3}{10}({\sout{5}}^1\times \frac{-6}{{\sout{5}}^1}-12)\\ \text{ =}\frac{3}{10}\times (-18)\\ \text{ =}\frac{3}{\sout{1}{\sout{0}}^5}\times -\sout{1}{\sout{8}}^9\\ \text{ =}\frac{-27}{5}\\ \text{∴ LHS=RHS }\\ \text{ Hence, }x\text{=}\frac{-6}{5}\text{ is a solution of the given equation}\text{. }\end{array}$

Question 18:

$\frac{y-1}{3}-\frac{y-2}{4}=1$

Answer 18:

$\begin{array}{l}\\ \text{We have:}\\ \frac{y-1}{3}-\frac{y-2}{4}=1\\ \Rightarrow \frac{4(y-1)-3(y-2)}{12}=1\\ \Rightarrow \left(\frac{4y-4-3y+6}{12}\right)=1\\ \Rightarrow \left(\frac{y+2}{12}\right)=1\\ \Rightarrow y+2=1\times 12\\ \Rightarrow y=12-2\\ \Rightarrow y=10\\ \text{CHECK: Substituting y=10 in the given equation, we get:}\\ \\ \text{LHS=}\frac{y-1}{3}-\frac{y-2}{4}\\ \text{ =}\frac{4(y-1)-3(y-2)}{12}\\ \text{ =}\left(\frac{y+2}{12}\right)\\ \text{ =}\left(\frac{10+2}{12}\right)\\ \text{ =}\frac{\sout{1}{\sout{2}}^1}{\sout{1}{\sout{2}}^1}\\ \text{ =1}\\ \\ \text{ RHS=1}\\ \text{ ∴ LHS=RHS }\\ \text{ Hence, y=10 is a solution of the given equation}\text{. }\end{array}$

Page-112

Question 19:

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

Answer 19:

$\begin{array}{l}\\ \text{We have:}\\ \frac{x-2}{4}+\frac{1}{3}=x-\frac{2x-1}{3}\\ \Rightarrow \frac{x-2}{4}+\frac{2x-1}{3}-x=-\frac{1}{3}(\mathrm{T}\text{ransposing -}\frac{2x-1}{3}\text{ to LHS and }\frac{1}{3}\text{to RHS) }\\ \Rightarrow \left(\frac{3(x-2)+4(2x-1)-12x}{12}\right)=-\frac{1}{3}\\ \Rightarrow \left(\frac{3x-6+8x-4-12x}{12}\right)=-\frac{1}{3}\\ \Rightarrow 11x-12x-10=-\frac{1}{{\sout{3}}^1}\times \sout{1}{\sout{2}}^4\\ \Rightarrow -x=-4+10\\ \Rightarrow -x=6\\ \Rightarrow x=-6\\ \text{CHECK: Substituting }x\text{=}-\text{6 in the given equation, we get:}\\ \\ \text{ LHS=}\frac{x-2}{4}+\frac{1}{3}\\ \text{ =}\frac{-6-2}{4}+\frac{1}{3}\\ \text{ =}-2+\frac{1}{3}\\ \text{ =}\frac{-5}{3}\\ \text{ RHS=}x-\frac{2x-1}{3}\\ \text{ =}-6-\frac{2\times (-6)-1}{3}\\ \text{ =}-6-\frac{(-13)}{3}\\ \text{ =}-6+\frac{13}{3}\\ \text{ =}\frac{-5}{3}\\ \text{∴ LHS=RHS }\\ \text{ Hence, y=10 is a solution of the given equation}\text{. }\end{array}$

Question 20:

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

Answer 20:

$\begin{array}{l}\\ \text{We have:}\\ \frac{2x-1}{3}-\frac{6x-2}{5}=\frac{1}{3}\\ \Rightarrow \frac{5(2x-1)-3(6x-2)}{15}=\frac{1}{3}\\ \Rightarrow \frac{10x-5-18x+6}{15}=\frac{1}{3}\\ \Rightarrow \frac{-8x+1}{15}=\frac{1}{3}\\ \Rightarrow -8x+1=\frac{1}{3}\times \text{15 }\\ \Rightarrow -8x=5-1\\ \Rightarrow -x=\frac{4}{8}\\ \Rightarrow x=-\frac{2}{4}=\frac{-1}{2}\\ \text{CHECK: Substituting }x\text{=}-\frac{1}{2}\text{ in the given equation, we get:}\\ \\ \text{LHS=}\frac{2x-1}{3}-\frac{6x-2}{5}\\ \text{ =}\frac{-8x+1}{15}\\ \text{ =}\frac{-8\times (-\frac{1}{2})+1}{15}\\ \text{ =}\frac{5}{15}\\ \text{ =}\frac{1}{3}\\ \text{RHS=}\frac{1}{3}\\ \begin{array}{l}\text{ ∴ LHS=RHS}\\ \text{ Hence, y=}-\frac{1}{2}\text{ is a solution of the given equation}\text{. }\end{array}\end{array}$

Question 21:

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

Answer 21:

$\begin{array}{l}\\ \text{We have:}\\ \frac{y+7}{3}=1+\frac{3y-2}{5}\\ \Rightarrow \frac{y+7}{3}=\frac{5\times 1+3y-2}{5}\\ \Rightarrow 5(y+7)=3(3+3y)\\ \Rightarrow 5y+35=9+9y\\ \Rightarrow 9y-5y=35-9\\ \Rightarrow 4y=26\\ \Rightarrow y=\frac{13}{2}\\ \\ \text{CHECK: Substituting }x\text{=}\frac{13}{2}\text{ in the given equation, we get:}\\ \\ \text{ LHS=}\frac{y+7}{3}\\ \text{ =}\frac{\frac{13}{2}+7}{3}\\ \text{=}\frac{1\times 13+2\times 7}{2}\\ \text{=}\frac{13+14}{6}\\ \text{=}\frac{27}{6}\\ \text{=}\frac{9}{2}\\ \\ \text{ RHS=1+}\frac{3\times \frac{13}{2}-2}{5}\\ =1+\frac{\frac{39-2\times 2}{2}}{5}\\ =1+\frac{35}{10}\\ \text{=}\frac{45}{10}\\ \text{ =}\frac{9}{2}\\ \text{∴ LHS=RHS }\\ \text{ Hence, y=}\frac{13}{2}\text{ is a solution of the given equation}\text{. }\end{array}$

Question 22:

$\frac{2}{7}(x-9)+\frac{x}{3}=3$

Answer 22:

$\begin{array}{l}\\ \text{We have:}\\ \Rightarrow \frac{2}{7}(x-9)+\frac{x}{3}=3\\ \Rightarrow \frac{2\times 3(x-9)+7x}{21}=3\\ \Rightarrow 6(x-9)+7x=3\times 21\\ \Rightarrow 6x-54+7x=63\\ \Rightarrow 13x=63+54\\ \Rightarrow 13x=117\\ \Rightarrow x=9\\ \\ \text{CHECK: Substituting }x\text{=9 in the given equation we get}\text{.}\\ \\ \text{LHS=}\frac{2}{7}(x-9)+\frac{x}{3}\\ \text{ =}\frac{2}{7}(9-9)+\frac{x}{3}\\ \text{=0+}\frac{9}{3}\\ \text{=}\frac{9}{3}\\ \text{=3}\\ \text{ RHS=3}\\ \therefore \mathrm{LHS}=\mathrm{RHS}\\ \text{Hence, }x\text{=9 is a solution of the given equation}\text{. }\end{array}$

Question 23:

$\frac{2x-3}{5}+\frac{x+3}{4}=\frac{4x+1}{7}$

Answer 23:

$\begin{array}{l}\\ \text{We have:}\\ \Rightarrow \frac{2x-3}{5}+\frac{x+3}{4}=\frac{4x+1}{7}\\ \Rightarrow \frac{4(2x-3)+5(x+3)}{20}=\frac{4x+1}{7}\\ \Rightarrow \frac{8x-12+5x+15}{20}=\frac{4x+1}{7}\\ \Rightarrow \frac{13x+3}{20}=\frac{4x+1}{7}\\ \Rightarrow 7(13x+3)=20(4x+1)\\ \Rightarrow 91x+21=80x+20\\ \Rightarrow 91x-80x=20-21\\ \Rightarrow 11x=-1\\ \Rightarrow x=\frac{-1}{11}\\ \\ \text{CHECK: Substituting }x\text{=}-\frac{1}{11}\text{ in the given equation, we get:}\\ \text{LHS:}\\ \text{LHS=}\frac{2x-3}{5}+\frac{x+3}{4}\\ \text{ =}\frac{2\times \frac{1}{11}-3}{5}+\frac{-\frac{1}{11}+3}{4}\\ \text{=}\frac{-2-33}{55}+\frac{33-1}{44}\\ \text{=}-\frac{35}{55}+\frac{32}{44}\\ \text{=}\frac{-140+160}{220}\\ \text{ =}\frac{20}{220}\text{ =}\frac{1}{11}\\ \\ \text{ RHS=}\frac{4x+1}{7}\\ =\frac{4\times (-\frac{1}{11})+1}{7}\\ =\frac{-4+11}{7\times 11}\\ =\frac{7}{77}\\ =\frac{1}{11}\\ \therefore \mathrm{LHS}=\mathrm{RHS}\\ \text{Hence, }x\text{=}\frac{-1}{11}\text{ is a solution of the given equation}\text{. }\end{array}$

Question 24:

$\frac{3}{4}(7x-1)-\left(2x-\frac{1-x}{2}\right)=x+\frac{3}{2}$

Answer 24:

$\begin{array}{l}\\ \text{We have:}\\ \frac{3}{4}(7x-1)-\left(2x-\frac{1-x}{2}\right)=x+\frac{3}{2}\\ \Rightarrow \frac{3}{4}(7x-1)-2x+\frac{1-x}{2}-x=\frac{3}{2}\\ \Rightarrow \frac{3\times 7}{4}x-\frac{3}{4}-2x+\frac{1}{2}-\frac{x}{2}-x=\frac{3}{2}\\ \Rightarrow \frac{21}{4}x-2x-\frac{x}{2}-x=\frac{3}{2}+\frac{3}{4}-\frac{1}{2} (\mathrm{By} \mathrm{transposition})\\ \Rightarrow \frac{21x-8x-2\times x-4x}{4}=1+\frac{3}{4}\\ \Rightarrow \frac{21x-14x}{4}=\frac{7}{4}\\ \Rightarrow \frac{7x}{4}=\frac{7}{4}\\ \Rightarrow x=1\\ \text{CHECK: Substituting }x\text{=1 in the given equation, we get:}\\ \\ \text{LHS=}\frac{3}{4}(7x-1)-\left(2x-\frac{1-x}{2}\right)\\ \text{ =}\frac{3}{4}(7\times 1-1)-\left(2\times 1-\frac{1-1}{2}\right)\\ \text{=}\frac{3}{4}\times 6-2\\ \text{=}\frac{9}{2}-2\\ \text{=}\frac{9-4}{2}\\ =\frac{5}{2}\\ \text{RHS=}x+\frac{3}{2}\\ =1+\frac{3}{2}\\ =\frac{2+3}{2}\\ =\frac{5}{2}\\ \begin{array}{l}\therefore \mathrm{LHS}=\mathrm{RHS}\\ \text{Hence, }x\text{=1 is a solution of the given equation}\text{. }\end{array}\end{array}$

Question 25:

$\frac{x+2}{6}-\left(\frac{11-x}{3}-\frac{1}{4}\right)=\frac{3x-4}{12}$

Answer 25:

$\begin{array}{l}\\ \mathrm{We} \mathrm{have}:\\ \frac{x+2}{6}-\left(\frac{11-x}{3}-\frac{1}{4}\right)=\frac{3x-4}{12}\\ \Rightarrow \frac{x+2}{6}-\left(\frac{11-x}{3}\right)+\frac{1}{4}=\frac{3x-4}{12}\\ \Rightarrow \frac{x+2}{6}-\left(\frac{11-x}{3}\right)-\frac{3x-4}{12}=-\frac{1}{4}\text{ (By transposition)}\\ \Rightarrow \frac{2(x+2)-4(11-x)-1(3x-4)}{12}=-\frac{1}{4}\\ \Rightarrow \frac{2x+4-44+4x-3x+4}{12}=-\frac{1}{4}\\ \Rightarrow 3x-36=-\frac{1}{4}\times \text{12 }\\ \Rightarrow 3x=-3+36\\ \Rightarrow x=\frac{33}{3}\\ \Rightarrow x=11\\ \text{CHECK: Substituting }x\text{= 11 in the given equation, we get:}\\ \\ \text{LHS=}\frac{x+2}{6}-(\frac{11-x}{3}-\frac{1}{4})\\ \text{ =}\frac{11+2}{6}-(\frac{11-11}{3}-\frac{1}{4})\\ \text{=}\frac{13}{6}-(-\frac{1}{4})\\ \text{=}\frac{13}{6}+\frac{1}{4}\\ \text{=}\frac{13\times 2+3}{12}\\ =\frac{29}{12}\\ \text{RHS=}\frac{3x-4}{12}\\ =\frac{3\times 11-4}{12}\\ =\frac{33-4}{12}\\ =\frac{29}{2}\\ \text{∴ LHS=RHS }\\ \text{Hence, }x\text{ = 11 is a solution of the given equation}\text{. }\\ \text{Verified.}\end{array}$

Question 26:

$\frac{9x+7}{2}-\left(x-\frac{x-2}{7}\right)=36$

Answer 26:

$\begin{array}{l}\\ \text{We have:}\\ \frac{9x+7}{2}-(x-\frac{x-2}{7})=36\\ \Rightarrow \frac{9x+7}{2}-x+\frac{x-2}{7}=36\\ \Rightarrow \frac{7(9x+7)-14\times x+2\times (x-2)}{14}=36\\ \Rightarrow \frac{63x+49-14x+2x-4}{14}=36\\ \Rightarrow 51x+45=36\times 14\\ \Rightarrow 51x=504-45\\ \Rightarrow x=\frac{459}{51}\\ \Rightarrow x=9\\ \Rightarrow x=9\\ \text{CHECK: Substituting }x\text{= 9 in the given equation, we get:}\\ \\ \text{LHS=}\frac{9x+7}{2}-\left(x-\frac{x-2}{7}\right)\\ \text{ =}\frac{9\times 9+7}{2}-\left(9-\frac{9-2}{7}\right)\\ \text{=}\frac{88}{2}-9+\frac{7}{7}\\ \text{=44}-9+1\\ \text{=36}\\ \text{RHS =36}\\ \\ \text{∴ LHS=RHS }\\ \text{Hence, }x\text{ = 11 is a solution of the given equation}\text{. }\\ \text{Verified.}\end{array}$

Question 27:

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

Answer 27:

$\begin{array}{l}\\ \text{We have:}\\ 0.5x+\frac{x}{3}=0.25x+7\\ \Rightarrow \frac{1}{2}x+\frac{x}{3}=\frac{x}{4}+7\\ \Rightarrow \frac{x}{2}+\frac{x}{3}-\frac{x}{4}=7\\ \Rightarrow \frac{6x+4x-3x}{12}=7\\ \Rightarrow \frac{7x}{12}=7\\ \Rightarrow x=12\\ \\ \text{CHECK: Substituting }x\text{= 9 in the given equation, we get:}\\ \\ \text{LHS=}0.5x+\frac{x}{3}\\ \text{ =}0.5\times 12+\frac{12}{3}\\ \text{=}\frac{1}{2}\times 12+4\\ \text{=6+4}\\ \text{=10}\\ \text{RHS=}0.25x+7\\ =0.25\times 12+7\\ =3+7\\ =10\\ \\ \text{∴ LHS=RHS }\\ \text{Hence, }x\text{ = 12 is a solution of the given equation}\text{. }\\ \text{Verified.}\end{array}$

Question 28:

0.18(5x − 4) = 0.5x + 0.8

Answer 28:

$\begin{array}{l}\\ \text{We have:}\\ 0.18(5x-4)=0.5x+0.8\\ \Rightarrow 100\times 0.18(5x-4)=100(0.5x+0.8)\left(\mathrm{M}\text{ultipling both sides by 100}\right)\\ \Rightarrow 18(5x-4)=100\times 0.5x+100\times 0.8\\ \Rightarrow 90x-72=50x+80\\ \Rightarrow 90x-50x=80+72\\ \Rightarrow 40x=152\\ \Rightarrow x=\frac{152}{40}\\ \Rightarrow \text{x=}\frac{19}{5}\text{=3.8 }\\ \\ \text{CHECK: Substituting }x\text{= }3.8\text{ in the given equation, we get:}\\ \\ \text{LHS=}0.18(5x-4)\\ \text{ =}0.18(5\times 3.8-4)\\ \text{=0}\text{.18}\times \text{15}\\ \text{=2}\text{.7}\\ \\ \text{RHS=}0.5x+0.8\\ =0.5\times 3.8+0.8\\ =1.9+0.8\\ =2.7\\ \text{∴ LHS=RHS }\\ \text{Hence, }x\text{ = }3.8\text{ is a solution of the given equation}\text{. }\\ \text{Verified.}\end{array}$

Question 29:

2.4(3 − x) − 0.6(2x − 3) = 0

Answer 29:

$\begin{array}{l}\\ \text{We have:}\\ \Rightarrow 2.4(3-x)-0.6(2x-3)=0\\ \Rightarrow 10\times 2.4(3-x)-10\times 0.6(2x-3)\text{=0 }\left(\mathrm{M}\text{ultiplying both sides by 10 to remove decimals}\right)\\ \Rightarrow 24(3-x)-6(2x-3)=0\\ \Rightarrow 6\left[4\right(3-x)-(2x-3\left)\right]=0\\ \Rightarrow 4(3-x)-(2x-3)=0\\ \Rightarrow 12-4x-2x+3=0\\ \Rightarrow 15-6x=0\\ \Rightarrow -\text{6x=}-15\\ \Rightarrow x=\frac{15}{6}\\ \Rightarrow \text{x=}\frac{5}{2}\text{=2.5 }\\ \\ \text{CHECK: Substituting }x\text{= }2.5\text{ in the given equation, we get:}\\ \\ \text{LHS=}2.4(3-x)-0.6(2x-3)\\ \text{ =}2.4(3-2.5)-0.6(2\times 2.5-3)\\ \text{=2.4}\times 0.5-0.6\times 2\\ \text{=1}\text{.2-1}\text{.2}\\ =0\\ \text{RHS=}0\\ \text{∴ LHS = RHS }\\ \text{Hence, }x\text{ = }\frac{19}{5}\text{ is a solution of the given equation}\text{. }\\ \text{Verified.}\end{array}$

Question 30:

0.5x −(0.8 − 0.2x) = 0.2 − 0.3x

Answer 30:

$\begin{array}{l}\\ \text{We have:}\\ 0.5x-(0.8-0.2x)=0.2-0.3x\\ \Rightarrow 0.5x+0.3x-0.8+0.2x\text{=0}\text{.2 (By transposition)}\\ \Rightarrow (0.5+0.3+0.2)x=0.2+0.8\\ \Rightarrow 1x=1\\ \Rightarrow x=1\\ \\ \text{CHECK: Substituting }x\text{= 1 in the given equation, we get:}\\ \\ \text{ LHS=}0.5x-(0.8-0.2x)\\ \text{ =}0.5\times 1-(0.8-0.2\times 1)\\ \text{=0}\text{.5}-0.8+0.2\\ \text{=}-\text{0}\text{.1}\\ \text{ RHS=}0.2-0.3x\\ =0.2-0.3\times 1\\ =-0.1\\ \text{∴ LHS=RHS }\\ \text{Hence, }x\text{ = 1 is a solution of the given equation}\text{. }\\ \text{Verified.}\end{array}$

Question 31:

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

Answer 31:

$\begin{array}{l}\\ \text{We have:}\\ \frac{x+2}{x-2}=\frac{7}{3}\\ \Rightarrow (x+2)\times \text{3=7}\times (x-2)\left(\mathrm{C}\text{ross multiplication}\right)\\ \Rightarrow 3x+6=7x-14\\ \Rightarrow 4x=20\\ \Rightarrow x=\frac{20}{4}\\ \Rightarrow x=5\\ \\ \text{CHECK: Substituting }x\text{= 5 in the given equation, we get}\text{.}\\ \\ \text{ LHS=}\frac{x+2}{x-2}\\ \text{ =}\frac{5+2}{5-2}\\ \text{=}\frac{7}{3}\\ \text{ RHS=}\frac{7}{3}\\ \\ \text{∴ LHS=RHS }\\ \text{Hence, }x\text{ = 5 is a solution of the given equation}\text{. }\\ \text{Verified.}\end{array}$

Question 32:

$\frac{2x+5}{3x+4}=3$

Answer 32:

$\begin{array}{l}\\ \text{We have:}\\ \frac{2x+5}{3x+4}=3\\ \Rightarrow \frac{2x+5}{3x+4}\text{=}\frac{3}{1}\\ \Rightarrow 1\times (2x+5)=3\times (3x+4)\\ \Rightarrow 2x+5=9x+12\\ \Rightarrow 7x=-7\\ \Rightarrow x=-1\\ \\ \text{CHECK: Substituting }x\text{=}-1\text{ in the given equation, we get:}\\ \\ \mathrm{LHS}: \frac{2x+5}{3x+4}\\ \text{ =}\frac{2\times (-1)+5}{3\times (-1)+4}\\ \text{=}\frac{-2+5}{-3+4}\\ =\frac{3}{1}\\ \text{RHS =3}\\ \\ \text{∴ LHS = RHS }\\ \text{Hence, }x\text{ = 5 is a solution of the given equation}\text{. }\\ \text{Verified.}\end{array}$