- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- MS Excel
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP
- Physics
- Chemistry
- Biology
- Mathematics
- English
- Economics
- Psychology
- Social Studies
- Fashion Studies
- Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Found 225 Articles for Class 8
370 Views
Given:The given equations are:(i) $\frac{7x-2}{5x-1}=\frac{7x+3}{5x+4}$(ii) $(\frac{x+1}{x+2})^2=\frac{x+2}{x+4}$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{7x-2}{5x-1}=\frac{7x+3}{5x+4}$$\frac{7x-2}{5x-1}=\frac{7x+3}{5x+4}$On cross multiplication, we get, $(5x+4)(7x-2)=(7x+3)(5x-1)$$5x(7x-2)+4(7x-2)=7x(5x-1)+3(5x-1)$$35x^2-10x+28x-8=35x^2-7x+15x-3$On rearranging, we get, $35x^2-35x^2+18x-8x=-3+8$$10x=5$$x=\frac{5}{10}$$x=\frac{1}{2}$Verification:LHS $=\frac{7x-2}{5x-1}$$=\frac{7(\frac{1}{2})-2}{5(\frac{1}{2})-1}$$=\frac{\frac{7}{2}-2}{\frac{5}{2}-1}$$=\frac{\frac{7-2\times2}{2}}{\frac{5-2\times1}{2}}$$=\frac{\frac{7-4}{2}}{\frac{5-2}{2}}$$=\frac{\frac{3}{2}}{\frac{3}{2}}$$=\frac{3}{2}\times\frac{2}{3}$$=1$RHS $=\frac{7x+3}{5x+4}$$=\frac{7(\frac{1}{2})+3}{5(\frac{1}{2})+4}$$=\frac{\frac{7}{2}+3}{\frac{5}{2}+4}$$=\frac{\frac{7+2\times3}{2}}{\frac{5+2\times4}{2}}$$=\frac{\frac{7+6}{2}}{\frac{5+8}{2}}$$=\frac{\frac{13}{2}}{\frac{13}{2}}$$=\frac{13}{2}\times\frac{2}{13}$$=1$LHS $=$ RHSHence verified.(ii) The given equation is $(\frac{x+1}{x+2})^2=\frac{x+2}{x+4}$$(\frac{x+1}{x+2})^2=\frac{x+2}{x+4}$On cross multiplication, we get, $(x+1)^2(x+4)=(x+2)^2(x+2)$$(x^2+2(x)(1)+1^2)(x+4)=(x^2+2(x)(2)+2^2)(x+2)$$x(x^2+2x+1)+4(x^2+2x+1)=x(x^2+4x+4)+2(x^2+4x+4)$$x^3+2x^2+x+4x^2+8x+4=x^3+4x^2+4x+2x^2+8x+8$On rearranging, we get, $x^3-x^3+6x^2-6x^2+9x-12x=8-4$$-3x=4$$x=\frac{-4}{3}$Verification:LHS $=(\frac{x+1}{x+2})^2$$=(\frac{\frac{-4}{3}+1}{\frac{-4}{3}+2})^2$$=(\frac{\frac{-4+3\times1}{3}}{\frac{-4+2\times3}{3}})^2$$=(\frac{\frac{-4+3}{3}}{\frac{-4+6}{3}})^2$$=(\frac{\frac{-1}{3}}{\frac{2}{3}})^2$$=(\frac{-1}{3})^2\times(\frac{3}{2})^2$$=\frac{1}{9}\times\frac{9}{4}$$=\frac{1}{4}$RHS $=\frac{x+2}{x+4}$$=\frac{\frac{-4}{3}+2}{\frac{-4}{3}+4}$$=\frac{\frac{-4+2\times3}{3}}{\frac{-4+4\times3}{3}}$$=\frac{\frac{-4+6}{3}}{\frac{-4+12}{3}}$$=\frac{\frac{2}{3}}{\frac{8}{3}}$$=\frac{2}{3}\times\frac{3}{8}$$=\frac{1}{1}\times\frac{1}{4}$$=\frac{1}{4}$LHS $=$ RHSHence verified.Read More
206 Views
Given:The given equations are:(i) $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$(ii) $\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$$\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$LCM of denominators $3x$ and $2x$ is $6x$$\frac{2(2)-3(3)}{6x}=\frac{1}{12}$$\frac{4-9}{6x}=\frac{1}{12}$$\frac{-5}{6x}=\frac{1}{12}$On cross multiplication, we get, $12(-5)=(1)(6x)$$-60=6x$$6x=-60$$x=\frac{-60}{6}$$x=-10$Verification:LHS $=\frac{2}{3x}-\frac{3}{2x}$$=\frac{2}{3(-10)}-\frac{3}{2(-10)}$$=\frac{2}{-30}-\frac{3}{-20}$$=\frac{-1}{15}+\frac{3}{20}$$=\frac{-1\times4+3\times3}{60}$ (LCM of $15$ and $20$ is $60$)$=\frac{-4+9}{60}$$=\frac{5}{60}$$=\frac{1}{12}$RHS $=\frac{1}{12}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$$\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$On cross multiplication, we get, $(3x+5)(4x+7)=(3x+4)(4x+2)$$3x(4x+7)+5(4x+7)=3x(4x+2)+4(4x+2)$$12x^2+21x+20x+35=12x^2+6x+16x+8$On rearranging, we get, $12x^2-12x^2+41x-22x=8-35$$19x=-27$$x=\frac{-27}{19}$Verification:LHS $=\frac{3x+5}{4x+2}$$=\frac{3(\frac{-27}{19})+5}{4(\frac{-27}{19})+2}$$=\frac{\frac{3\times(-27)}{19}+5}{\frac{4\times(-27)}{19}+2}$$=\frac{\frac{-81+5\times19}{19}}{\frac{-108+2\times19}{19}}$$=\frac{\frac{-81+95}{19}}{\frac{-108+38}{19}}$$=\frac{\frac{14}{19}}{\frac{-70}{19}}$$=\frac{14}{19}\times\frac{19}{-70}$$=\frac{1}{1}\times\frac{1}{-5}$$=\frac{-1}{5}$RHS $=\frac{3x+4}{4x+7}$$=\frac{3(\frac{-27}{19})+4}{4(\frac{-27}{19})+7}$$=\frac{\frac{3\times(-27)}{19})+4}{\frac{4\times(-27)}{19})+7}$$=\frac{\frac{-81+19\times4}{19}}{\frac{-108+19\times7}{19}}$$=\frac{\frac{-81+76}{19}}{\frac{-108+133}{19}}$$=\frac{\frac{-5}{19}}{\frac{25}{19}}$$=\frac{-5}{19}\times\frac{19}{25}$$=\frac{-1}{1}\times{1}{5}$$=\frac{-1}{5}$LHS ... Read More
191 Views
Given:The given equations are:(i) $\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$(ii) $\frac{6}{2x-(3-4x)}=\frac{2}{3}$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$$\frac{y-7+8y}{9y-3-4y}=\frac{2}{3}$$\frac{9y-7}{5y-3}=\frac{2}{3}$On cross multiplication, we get, $3(9y-7)=(2)(5y-3)$$3(9y)-3(7)=2(5y)-2(3)$$27y-21=10y-6$On rearranging, we get, $27y-10y=-6+21$$17y=15$$y=\frac{15}{17}$Verification:LHS $=\frac{y-(7-8y)}{9y-(3+4y)}$$=\frac{\frac{15}{17}-(7-8(\frac{15}{17}))}{9(\frac{15}{17})-(3+4(\frac{15}{17}))}$$=\frac{\frac{15}{17}-(7-(\frac{-8\times15}{17}))}{\frac{9\times15}{17}-(3+\frac{4\times15}{17})}$$=\frac{\frac{15}{17}-(7-(\frac{120}{17}))}{\frac{135}{17}-(3+\frac{60}{17})}$$=\frac{\frac{15}{17}-(\frac{7\times17-120}{17})}{\frac{135}{17}-(\frac{3\times17+60}{17})}$$=\frac{\frac{15}{17}-(\frac{119-120}{17})}{\frac{135}{17}-(\frac{51+60}{17})}$$=\frac{\frac{15}{17}-(\frac{-1}{17})}{\frac{135}{17}-(\frac{111}{17})}$$=\frac{\frac{15+1}{17}}{\frac{135-111}{17}}$$=\frac{\frac{16}{17}}{\frac{24}{17}}$$=\frac{16}{17}\times\frac{17}{24}$$=\frac{2}{3}$RHS $=\frac{2}{3}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{6}{2x-(3-4x)}=\frac{2}{3}$$\frac{6}{2x-(3-4x)}=\frac{2}{3}$$\frac{6}{2x-3+4x}=\frac{2}{3}$$\frac{6}{6x-3}=\frac{2}{3}$On cross multiplication, we get, $3(6)=2(6x-3)$$18=2(6x)-2(3)$$18=12x-6$On rearranging, we get, $12x=18+6$$12x=24$$x=\frac{24}{12}$$x=2$Verification:LHS $=\frac{6}{2x-(3-4x)}$$=\frac{6}{2(2)-(3-4(2))}$$=\frac{6}{4-(3-8)}$$=\frac{6}{4+5}$$=\frac{6}{9}$$=\frac{2}{3}$RHS $=\frac{2}{3}$LHS $=$ RHSHence verified.Read More
153 Views
Given:The given equations are:(i) $\frac{1-9y}{19-3y}=\frac{5}{8}$(ii) $\frac{2x}{3x+1}=1$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{1-9y}{19-3y}=\frac{5}{8}$$\frac{1-9y}{19-3y}=\frac{5}{8}$On cross multiplication, we get, $8(1-9y)=(5)(19-3y)$$8(1)-8(9y)=5(19)-5(3y)$$8-72y=95-15y$On rearranging, we get, $72y-15y=8-95$$57y=-87$$y=\frac{-87}{57}$$y=\frac{-29}{19}$Verification:LHS $=\frac{1-9y}{19-3y}$$=\frac{1-9(\frac{-29}{19})}{19-3(\frac{-29}{19})}$$=\frac{1+\frac{29\times9}{19}}{19+\frac{3\times29}{19}}$$=\frac{1+\frac{261}{19}}{19+\frac{87}{19}}$$=\frac{\frac{19\times1+261}{19}}{\frac{19\times19+87}{19}}$$=\frac{\frac{19+261}{19}}{\frac{361+87}{19}}$$=\frac{\frac{280}{19}}{\frac{448}{19}}$$=\frac{280}{19}\times\frac{19}{448}$$=\frac{280}{448}$$=\frac{5}{8}$RHS $=\frac{5}{8}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{2x}{3x+1}=1$$\frac{2x}{3x+1}=1$On cross multiplication, we get, $2x=1(3x+1)$$2x=3x+1$On rearranging, we get, $3x-2x=-1$$x=-1$Verification:LHS $=\frac{2x}{3x+1}$$=\frac{2(-1)}{3(-1)+1}$$=\frac{-2}{-3+1}$$=\frac{-2}{-2}$$=1$RHS $=1$LHS $=$ RHSHence verified.Read More
111 Views
Given:The given equations are:(i) $\frac{2y+5}{y+4}=1$(ii) $\frac{2x+1}{3x-2}=\frac{5}{9}$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{2y+5}{y+4}=1$$\frac{2y+5}{y+4}=1$On cross multiplication, we get, $2y+5=(1)(y+4)$$2y+5=y+4$On rearranging, we get, $2y-y=4-5$$y=-1$Verification:LHS $=\frac{2y+5}{y+4}$$=\frac{2(-1)+5}{(-1)+4}$$=\frac{-2+5}{-1+4}$$=\frac{3}{3}$$=1$RHS $=1$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{2x+1}{3x-2}=\frac{5}{9}$$\frac{2x+1}{3x-2}=\frac{5}{9}$On cross multiplication, we get, $9(2x+1)=5(3x-2)$$9(2x)+9(1)=5(3x)-5(2)$$18x+9=15x-10$On rearranging, we get, $18x-15x=-10-9$$3x=-19$$x=\frac{-19}{3}$Verification:LHS $=\frac{2x+1}{3x-2}$$=\frac{2(\frac{-19}{3})+1}{3(\frac{-19}{3})-2}$$=\frac{\frac{2\times(-19)}{3}+1}{-19-2}$$=\frac{\frac{-38+1\times3}{3}}{-21}$$=\frac{\frac{-38+3}{3}}{-21}$$=\frac{\frac{-35}{3}}{-21}$$=\frac{-35}{3\times-21}$$=\frac{5}{3\times3}$$=\frac{5}{9}$RHS $=\frac{5}{9}$LHS $=$ RHSHence verified.Read More
209 Views
Given:The given equations are:(i) $\frac{5x-7}{3x}=2$(ii) $\frac{3x+5}{2x+7}=4$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{5x-7}{3x}=2$$\frac{5x-7}{3x}=2$On cross multiplication, we get, $5x-7=3x(2)$$5x-7=6x$On rearranging, we get, $6x-5x=-7$$x=-7$Verification:LHS $=\frac{5x-7}{3x}$$=\frac{5(-7)-7}{3(-7)}$$=\frac{-35-7}{-21}$$=\frac{-42}{-21}$$=2$RHS $=2$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{3x+5}{2x+7}=4$.$\frac{3x+5}{2x+7}=4$On cross multiplication, we get, $3x+5=4(2x+7)$$3x+5=4(2x)+4(7)$$3x+5=8x+28$On rearranging, we get, $8x-3x=5-28$$5x=-23$$x=\frac{-23}{5}$Verification:LHS $=\frac{3x+5}{2x+7}$$=\frac{3(\frac{-23}{5})+5}{2(\frac{-23}{5})+7}$$=\frac{\frac{-69}{5}+5}{\frac{-46}{5}+7}$$=\frac{\frac{-69+5\times5}{5}}{\frac{-46+5\times7}{5}}$$=\frac{\frac{-69+25}{5}}{\frac{-46+35}{5}}$$=\frac{\frac{-44}{5}}{\frac{-11}{5}}$$=\frac{-44}{5}\times\frac{5}{-11}$$=\frac{4}{1}\times\frac{1}{1}$$=4$RHS $=4$LHS $=$ RHSHence verified.Read More
138 Views
Given:The given equations are:(i) $\frac{2x-3}{3x+2}=\frac{-2}{3}$(ii) $\frac{2-y}{y+7}=\frac{3}{5}$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{2x-3}{3x+2}=\frac{-2}{3}$$\frac{2x-3}{3x+2}=\frac{-2}{3}$On cross multiplication, we get, $3(2x-3)=(-2)(3x+2)$$3(2x)-3(3)=-2(3x)-2(2)$$6x-9=-6x-4$On rearranging, we get, $6x+6x=9-4$$12x=5$$x=\frac{5}{12}$Verification:LHS $=\frac{2x-3}{3x+2}$$=\frac{2(\frac{5}{12})-3}{3(\frac{5}{12}+2}$$=\frac{\frac{5}{6}-3}{\frac{5}{4}+2}$$=\frac{\frac{5-3\times6}{6}}{\frac{5+2\times4}{4}}$$=\frac{5-18}{6}\times\frac{4}{5+8}$$=\frac{-13}{6}\times\frac{4}{13}$$=\frac{-1}{3}\times\frac{2}{1}$$=\frac{-2}{3}$RHS $=\frac{-2}{3}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{2-y}{y+7}=\frac{3}{5}$.$\frac{2-y}{y+7}=\frac{3}{5}$On cross multiplication, we get, $5(2-y)=3(y+7)$$5(2)-5(y)=3(y)+3(7)$$10-5y=3y+21$On rearranging, we get, $5y+3y=10-21$$8y=-11$$y=\frac{-11}{8}$Verification:LHS $=\frac{2-y}{y+7}$$=\frac{2-(\frac{-11}{8})}{\frac{-11}{8}+7}$$=\frac{2+\frac{11}{8}}{\frac{-11}{8}+7}$$=\frac{\frac{2\times8+11}{8}}{\frac{-11+7\times8}{8}}$$=\frac{\frac{16+11}{8}}{\frac{-11+56}{8}}$$=\frac{\frac{27}{8}}{\frac{45}{8}}$$=\frac{27}{8}\times\frac{8}{45}$$=\frac{3}{1}\times\frac{1}{5}$$=\frac{3}{5}$RHS $=\frac{3}{5}$LHS $=$ RHSHence verified.Read More
225 Views
Given:The given equation is $[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$To do:We have to solve the given equation and check the result.Solution:To check the result we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.The given equation is $[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$$[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$$[3x+8]^2+[x-2]^2=10x^2 + 92$$(3x)^2+2(3x)(8)+8^2+x^2-2(x)(2)+2^2=10x^2+92$ [Since $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$]$9x^2+48x+64+x^2-4x+4=10x^2+92$$10x^2+44x+68=10x^2+92$On rearranging, we get, $10x^2-10x^2+44x=92-68$$44x=24$$x=\frac{24}{44}$$x=\frac{6}{11}$Verification:LHS $=[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2$$=[(2(\frac{6}{11})+3)+(\frac{6}{11}+5)]^2+[(2(\frac{6}{11})+3)-(\frac{6}{11}+5)]^2$$=[\frac{12}{11}+3)+(\frac{6}{11}+5)]^2+[(\frac{12}{11})+3)-(\frac{6}{11}+5)]^2$$=[\frac{12+6}{11}+8]^2+[\frac{12-6}{11}-2]^2$$=[\frac{18}{11}+8]^2+[\frac{6}{11}-2]^2$$=[\frac{18+11\times8}{11}]^2+[\frac{6-2\times11}{11}]^2$$=[\frac{18+88}{11}]^2+[\frac{6-22}{11}]^2$$=[\frac{106}{11}]^2+[\frac{-16}{11}]^2$$=\frac{11236}{121}+\frac{256}{121}$$=\frac{11236+256}{121}$$=\frac{11492}{121}$RHS $=10x^2 + 92$$=10(\frac{6}{11})^2 + 92$$=10(\frac{36}{121})+92$$=\frac{360}{121}+92$$=\frac{360+121\times92}{121}$$=\frac{360+11132}{121}$$=\frac{11492}{121}$LHS $=$ RHSHence verified.Read More
143 Views
Given:The given equations are:(i) $6.5x+(\frac{19.5x-32.5}{2})=6.5x+13+\frac{13x-26}{2}$(ii) $(3x-8)(3x+2)-(4x-11)(2x+1)=(x-3)(x+7)$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $6.5x+(\frac{19.5x-32.5}{2})=6.5x+13+\frac{13x-26}{2}$$6.5x+(\frac{19.5x-32.5}{2})=6.5x+13+\frac{13x-26}{2}$On rearranging, we get, $6.5x+(\frac{19.5x-32.5}{2})-6.5x-\frac{13x-26}{2}=13$$\frac{19.5x-32.5}{2}-\frac{13x-26}{2}=13$$\frac{19.5x-32.5-(13x-26)}{2}=13$$\frac{19.5x-32.5-13x+26}{2}=13$$\frac{6.5x-6.5}{2}=13$On cross multiplication, we get, $6.5x-6.5=13\times2$$6.5x-6.5=26$$6.5x=26+6.5$$6.5x=32.5$$x=\frac{32.5}{6.5}$$x=5$Verification:LHS $=6.5x+(\frac{19.5x-32.5}{2})$$=6.5(5)+(\frac{19.5(5)-32.5}{2})$$=32.5+\frac{97.5-32.5}{2}$$=32.5+\frac{65}{2}$$=32.5+32.5$$=65$RHS $=6.5x+13+\frac{13x-26}{2}$$=6.5(5)+13+\frac{13(5)-26}{2}$$=32.5+13+\frac{65-26}{2}$$=32.5+13+\frac{39}{2}$$=45.5+19.5$$=65$LHS $=$ RHSHence verified.(ii) The given equation is $(3x-8)(3x+2)-(4x-11)(2x+1)=(x-3)(x+7)$.$(3x-8)(3x+2)-(4x-11)(2x+1)=(x-3)(x+7)$$3x(3x+2)-8(3x+2)-4x(2x+1)+11(2x+1)=x(x+7)-3(x+7)$$9x^2+6x-24x-16-8x^2-4x+22x+11=x^2+7x-3x-21$$x^2-5=x^2+4x-21$$x^2-x^2+4x=21-5$$4x=16$$x=\frac{16}{4}$$x=4$Verification:LHS $=(3x-8)(3x+2)-(4x-11)(2x+1)$$=[3(4)-8][3(4)+2]-[4(4)-11][2(4)+1]$$=(12-8)(12+2)-(16-11)(8+1)$$=4(14)-5(9)$$=56-45$$=11$RHS $=(x-3)(x+7)$$=(4-3)(4+7)$$=1(11)$$=11$LHS $=$ RHSHence verified.Read More
146 Views
Given:The given equations are:(i) $\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}$(ii) $0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})=x+6.1$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}$$\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}$$\frac{7x-1}{4}-\frac{1}{3}(\frac{2(2x)-(1-x)}{2})=\frac{10}{3}$$\frac{7x-1}{4}-\frac{1}{3}(\frac{4x-1+x}{2})=\frac{10}{3}$$\frac{7x-1}{4}-(\frac{5x-1}{2\times3})=\frac{10}{3}$$\frac{7x-1}{4}-\frac{5x-1}{6}=\frac{10}{3}$LCM of denominators $4$ and $6$ is $12$$\frac{(7x-1)\times3-(5x-1)\times2}{12}=\frac{10}{3}$$\frac{3(7x)-3(1)-2(5x)+2(1)}{12}=\frac{10}{3}$$\frac{21x-3-10x+2}{12}=\frac{10}{3}$$\frac{11x-1}{12}=\frac{10}{3}$On cross multiplication, we get, $11x-1=\frac{10\times12}{3}$$11x-1=10\times4$$11x-1=40$$11x=40+1$$11x=41$$x=\frac{41}{11}$Verification:LHS $=\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})$$=\frac{7(\frac{41}{11})-1}{4}-\frac{1}{3}(2(\frac{41}{11})-\frac{1-(\frac{41}{11})}{2})$$=\frac{\frac{41\times7}{11}-1}{4}-\frac{1}{3}(\frac{41\times2}{11}-\frac{\frac{11\times1-41}{11}}{2})$$=\frac{\frac{287}{11}-1}{4}-\frac{1}{3}(\frac{82}{11}-\frac{\frac{11-41}{11}}{2})$$=\frac{287-11}{11\times4}-\frac{1}{3}(\frac{82}{11}-\frac{-30}{11\times2})$$=\frac{276}{44}-\frac{1}{3}(\frac{82}{11}+\frac{30}{22})$$=\frac{69}{11}-\frac{1}{3}(\frac{82\times2+30}{22})$$=\frac{69}{11}-\frac{1}{3}(\frac{164+30}{22})$$=\frac{69}{11}-\frac{1}{3}(\frac{194}{22})$$=\frac{69}{11}-(\frac{194}{3\times22})$$=\frac{69}{11}-\frac{194}{66}$$=\frac{69\times6-194}{66}$$=\frac{414-194}{66}$$=\frac{220}{66}$$=\frac{10}{3}$RHS $=\frac{10}{3}$LHS $=$ RHSHence verified.(ii) The given equation is $0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})=x+6.1$$0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})=x+6.1$On rearranging, we get, $0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})-x=6.1$$\frac{(x-0.4)}{0.7}-(\frac{x-2.71}{0.7})-x=6.1$$\frac{x-0.4-(x-2.71)}{0.7}-x=6.1$$\frac{x-0.4-x+2.71}{0.7}-x=6.1$$\frac{2.31}{0.7}-x=6.1$$\frac{23.1}{7}-6.1=x$$x=3.3-6.1$$x=-2.8$Verification:LHS $=0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})$$=0.5\frac{(-2.8-0.4)}{0.35}-0.6(\frac{-2.8-2.71}{0.42})$$=\frac{-3.2}{0.7}-\frac{-5.51}{0.7}$$=\frac{-3.2+5.51}{0.7}$$=\frac{2.31}{0.7}$$=3.3$RHS $=x+6.1$$=-2.8+6.1$$=3.3$LHS $=$ RHSHence verified.Read More