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Found 225 Articles for Class 8
339 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
188 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}$$=\frac{-1}{5}$LHS ... Read More
173 Views
Given:The given equations are:(i) $\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$(ii) $\frac{6}{2x-(3-4x)}=\frac{2}{3}$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{y-(7-8y)}{9y-(3+4y)}=\frac{2}{3}$$\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
143 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
107 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
201 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}$$=4$RHS $=4$LHS $=$ RHSHence verified.Read More
124 Views
Given:The given equations are:(i) $\frac{2x-3}{3x+2}=\frac{-2}{3}$(ii) $\frac{2-y}{y+7}=\frac{3}{5}$To do:We have to solve the given equations and verify the answers.Solution:To verify the answer we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{2x-3}{3x+2}=\frac{-2}{3}$$\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}$$=\frac{3}{5}$RHS $=\frac{3}{5}$LHS $=$ RHSHence verified.Read More
190 Views
Given:The given equation is $[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$To do:We have to solve the given equation and check the result.Solution:To check the result we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.The given equation is $[(2x+3)+(x+5)]^2+[(2x+3)-(x+5)]^2=10x^2 + 92$$[(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
116 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)$$=11$LHS $=$ RHSHence verified.Read More
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Given:The given equations are:(i) $\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}$(ii) $0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})=x+6.1$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}$$\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