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

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

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

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

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

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

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

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

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

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