## Solve the following equations and verify your answer:(i) $\frac{7x-2}{5x-1}=\frac{7x+3}{5x+4}$(ii) $(\frac{x+1}{x+2})^2=\frac{x+2}{x+4}$

Updated on 13-Apr-2023 23:45:48
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)$$35x^2-10x+28x-8=35x^2-7x+15x-3On rearranging, we get, 35x^2-35x^2+18x-8x=-3+8$$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}$$=1RHS =\frac{7x+3}{5x+4}$$=\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

## Solve the following equations and verify your answer:(i) $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$(ii) $\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$

Updated on 13-Apr-2023 23:24:36

## Solve the following equations and verify your answer:(i) $\frac{1-9y}{19-3y}=\frac{5}{8}$(ii) $\frac{2x}{3x+1}=1$

Updated on 13-Apr-2023 23:23:17
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)$$8-72y=95-15yOn rearranging, we get, 72y-15y=8-95$$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)$$2x=3x+1On rearranging, we get, 3x-2x=-1$$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

## Solve the following equations and verify your answer:(i) $\frac{2y+5}{y+4}=1$(ii) $\frac{2x+1}{3x-2}=\frac{5}{9}$

Updated on 13-Apr-2023 23:22:39
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$$\frac{2y+5}{y+4}=1On cross multiplication, we get, 2y+5=(1)(y+4)$$2y+5=y+4$On rearranging, we get, $2y-y=4-5$$y=-1Verification:LHS =\frac{2y+5}{y+4}$$=\frac{2(-1)+5}{(-1)+4}$$=\frac{-2+5}{-1+4}$$=\frac{3}{3}$$=1RHS =1LHS = RHSHence verified.(ii) The given equation is \frac{2x+1}{3x-2}=\frac{5}{9}$$\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

## Solve the following equations and verify your answer:(i) $\frac{5x-7}{3x}=2$(ii) $\frac{3x+5}{2x+7}=4$

Updated on 13-Apr-2023 23:21:21

## Solve each of the following equations and also check your results in each case:(i) $6.5x+(\frac{19.5x-32.5}{2})=6.5x+13+\frac{13x-26}{2}$(ii) $(3x-8)(3x+2)-(4x-11)(2x+1)=(x-3)(x+7)$

Updated on 13-Apr-2023 23:18:51
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$$\frac{6.5x-6.5}{2}=13On cross multiplication, we get, 6.5x-6.5=13\times2$$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$$=65RHS =6.5x+13+\frac{13x-26}{2}$$=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}$$x=4Verification:LHS =(3x-8)(3x+2)-(4x-11)(2x+1)$$=[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)$$=11LHS = RHSHence verified.Read More ## Solve each of the following equations and also check your results in each case:(i) \frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}(ii) 0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})=x+6.1 Updated on 13-Apr-2023 23:17:26 Given:The given equations are:(i) \frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}(ii) 0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})=x+6.1To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is \frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}$$\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$$x=-2.8Verification:LHS =0.5\frac{(x-0.4)}{0.35}-0.6(\frac{x-2.71}{0.42})$$=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
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