Given:The given equations are:(i) $\frac{(3a-2)}{3}+\frac{(2a+3)}{2}=a+\frac{7}{6}$(ii) $x-\frac{(x-1)}{2}=1-\frac{(x-2)}{3}$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{(3a-2)}{3}+\frac{(2a+3)}{2}=a+\frac{7}{6}$$\frac{(3a-2)}{3}+\frac{(2a+3)}{2}=a+\frac{7}{6}$On rearranging, we get, $\frac{(3a-2)}{3}+\frac{(2a+3)}{2}-a=\frac{7}{6}$LCM of denominators $3$ and $2$ is $6$$\frac{(3a-2)\times2+(2a+3)\times3-a \times6}{6}=\frac{7}{6}$$\frac{2(3a)-2(2)+(2a(3)+3(3)-6a}{6}=\frac{7}{6}$$\frac{6a-4+6a+9-6a}{6}=\frac{7}{6}$$\frac{6a-4+9}{6}=\frac{7}{6}$$\frac{6a+5}{6}=\frac{7}{6}$On cross multiplication, we get, $6a+5=\frac{7\times6}{6}$$6a+5=7$$6a+5=7$$6a=7-5$$6a=2$$a=\frac{2}{6}$$a=\frac{1}{3}$Verification:LHS $=\frac{(3a-2)}{3}+\frac{(2a+3)}{2}$$=\frac{(3(\frac{1}{3})-2)}{3}+\frac{(2(\frac{1}{3})+3)}{2}$$=\frac{1-2}{3}+\frac{\frac{2}{3}+3}{2}$$=\frac{-1}{3}+\frac{\frac{2+3\times3}{3}}{2}$$=\frac{-1}{3}+\frac{\frac{2+9}{3}}{2}$$=\frac{-1}{3}+\frac{11}{3\times2}$$=\frac{-1}{3}+\frac{11}{6}$$=\frac{-1\times2+11}{6}$           (LCM of $3$ and $6$ is $6$)$=\frac{-2+11}{6}$$=\frac{9}{6}$$=\frac{3}{2}$RHS $=a+\frac{7}{6}$$=\frac{1}{3}+\frac{7}{6}$$=\frac{1\times2+7}{6}$                    (LCM of $3$ and $6$ is $6$)$=\frac{2+7}{6}$$=\frac{9}{6}$$=\frac{3}{2}$LHS $=$ ... Read More

Given:The given equations are:(i) $\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$(ii) $\frac{6x+1}{2}+1=\frac{7x-3}{3}$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 $\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$.$\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$On rearranging, we get, $\frac{7x}{2}-\frac{5x}{2}-\frac{20x}{3}=10$LCM of $2$ and $3$ is $6$$\frac{7x \times3-5x \times 3-20x \times2}{6}=10$$\frac{21x-15x-40x}{6}=10$$\frac{21x-55x}{6}=10$$\frac{-34x}{6}=10$$\frac{-17x}{3}=10$On cross multiplication, we get, $-17x=3(10)$$-17x=30$$x=\frac{30}{-17}$$x=\frac{-30}{17}$Verification:LHS $=\frac{7x}{2}-\frac{5x}{2}$$=\frac{7(\frac{-30}{17})}{2}-\frac{5(\frac{-30}{17})}{2}$$=\frac{-210}{34}-\frac{-150}{34}$$=\frac{-210+150}{34}$$=\frac{-60}{34}$$=\frac{-30}{17}$RHS $=\frac{20x}{3}+10$$=\frac{20(\frac{-30}{17})}{3}+10$$=\frac{20\times(-30)}{17\times3}+10$$=\frac{-600}{51}+10$$=\frac{-600+51\times10}{51}$                        (LCM of $51$ and $1$ is $51$)$=\frac{-600+510}{51}$$=\frac{-90}{51}$$=\frac{-30}{17}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{6x+1}{2}+1=\frac{7x-3}{3}$$\frac{6x+1}{2}+1=\frac{7x-3}{3}$On ... Read More

Given:The given equations are:(i) $\frac{1}{2}x+7x-6=7x+\frac{1}{4}$(ii) $\frac{3}{4}x+4x=\frac{7}{8}+6x-6$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 $\frac{1}{2}x+7x-6=7x+\frac{1}{4}$.$\frac{1}{2}x+7x-6=7x+\frac{1}{4}$On rearranging, we get, $\frac{1}{2}x+7x-7x=\frac{1}{4}+6$$\frac{1}{2}x=\frac{1+6\times4}{4}$               (LCM of $4$ and $1$ is $4$)$\frac{1}{2}x=\frac{1+24}{4}$$\frac{1}{2}x=\frac{25}{4}$On cross multiplication, we get, $x=\frac{25\times2}{4}$$x=\frac{25}{2}$Verification:LHS $=\frac{1}{2}x+7x-6$$=\frac{1}{2}(\frac{25}{2})+7(\frac{25}{2})-6$$=\frac{25}{2\times2}+\frac{25\times7}{2}-6$$=\frac{25}{4}+\frac{175}{2}-6$$=\frac{25+175\times2-6\times4}{4}$              (LCM of $2$ and $4$ is $4$)$=\frac{25+350-24}{4}$$=\frac{351}{4}$RHS $=7x+\frac{1}{4}$$=7(\frac{25}{2})+\frac{1}{4}$$=\frac{25\times7}{2}+\frac{1}{4}$$=\frac{175}{2}+\frac{1}{4}$              $=\frac{175\times2+1}{4}$        ... Read More

Given:The given equations are:(i) $\frac{7y+2}{5}=\frac{6y-5}{11}$(ii) $x-2x+2-\frac{16}{3}x+5=3-\frac{7}{2}x$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 $\frac{7y+2}{5}=\frac{6y-5}{11}$.$\frac{7y+2}{5}=\frac{6y-5}{11}$On cross multiplication, we get, $(7y+2)\times11=5(6y-5)$$11(7y)+11(2)=5(6y)-5(5)$$77y+22=30y-25$$77y-30y=-25-22$$47y=-47$$y=\frac{-47}{47}$$y=-1$Verification:LHS $=\frac{7y+2}{5}$$=\frac{7(-1)+2}{5}$$=\frac{-7+2}{5}$$=\frac{-5}{5}$$=-1$RHS $=\frac{6y-5}{11}$$=\frac{6(-1)-5}{11}$$=\frac{-6-5}{11}$$=\frac{-11}{11}$$=-1$LHS $=$ RHSHence verified.(ii) The given equation is $x-2x+2-\frac{16}{3}x+5=3-\frac{7}{2}x$$x-2x+2-\frac{16}{3}x+5=3-\frac{7}{2x}$On rearranging, we get, $x-2x-\frac{16}{3}x+\frac{7}{2}x=3-2-5$$-x-\frac{16}{3}x+\frac{7}{2}x=3-7$$x(-1-\frac{16}{3}+\frac{7}{2})=-4$LCM of denominators $3$ and $2$ is $6$$x(\frac{-1\times6-16\times2+7\times3}{6})=-4$$x(\frac{-6-32+21}{6})=-4$$x(\frac{-38+21}{6})=-4$$x(\frac{-17}{6})=-4$On cross multiplication, we get, $-17x=(-4)\times6$$-17x=-24$$x=\frac{-24}{-17}$$x=\frac{24}{17}$Verification:LHS $=x-2x+2-\frac{16}{3}x+5$$=\frac{24}{17}-2(\frac{24}{17})+2-\frac{16}{3}(\frac{24}{17})+5$$=\frac{24}{17}-\frac{48}{17}+2-\frac{16\times24}{3\times17}+5$$=\frac{24-48}{17}+7-\frac{16\times8}{17}$$=\frac{-24}{17}-\frac{128}{17}+7$$=\frac{-24-128+7\times17}{17}$$=\frac{-152+119}{17}$$=\frac{-33}{17}$RHS $=3-\frac{7}{2}x$$=3-\frac{7}{2}(\frac{24}{17})$$=3-\frac{7\times24}{2\times17}$$=3-\frac{7\times12}{17}$$=\frac{3\times17-84}{17}$$=\frac{51-84}{17}$$=\frac{-33}{17}$LHS $=$ RHSHence verified.Read More

Given:The given equations are:(i) $\frac{2x+5}{3}=3x-10$(ii) $\frac{a-8}{3}=\frac{a-3}{2}$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 $\frac{2x+5}{3}=3x-10$.$\frac{2x+5}{3}=3x-10$On cross multiplication, we get, $2x+5=3(3x-10)$$2x+5=3(3x)-3(10)$$2x+5=9x-30$$9x-2x=5+30$$7x=35$$x=\frac{35}{7}$$x=5$Verification:LHS $=\frac{2x+5}{3}$$=\frac{2\times5+5}{3}$$=\frac{10+5}{3}$$=\frac{15}{3}$$=5$RHS $=3x-10$$=3(5)-10$$=15-10$$=5$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{a-8}{3}=\frac{a-3}{2}$$\frac{a-8}{3}=\frac{a-3}{2}$On cross multiplication, we get, $(a-8)\times2=(a-3)\times3$$a(2)-8(2)=a(3)-3(3)$$2a-16=3a-9$$3a-2a=9-16$$a=-7$Verification:LHS $=\frac{a-8}{3}$$=\frac{-7-8}{3}$$=\frac{-15}{3}$$=-5$RHS $=\frac{a-3}{2}$$=\frac{-7-3}{2}$$=\frac{-10}{2}$$=-5$LHS $=$ RHSHence verified.Read More

Given:The given equation is $\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$.To do:We have to solve the given equation and verify the solution.Solution:To verify the solution we have to find the value of the variable and substitute it in the equation. Find the value of LHS and the value of RHS and check whether both are equal.The given equation is $\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$.$\frac{2(x-5)}{3}-\frac{1(x-2)}{4}=\frac{9}{2}$$\frac{2x-10}{3}-\frac{x-2}{4}=\frac{9}{2}$LCM of denominators $3$ and $4$ is $12$.$\frac{(2x-10)\times4-(x-2)\times3}{12}=\frac{9}{2}$$\frac{8x-40-3x+6}{12}=\frac{9}{2}$$\frac{5x-34}{12}=\frac{9}{2}$On cross multiplication, we get, $(5x-34=\frac{9\times12}{2}$$5x-34=9\times6$$5x-34=54$$5x=54+34$$5x=88$$x=\frac{88}{5}$Verification:LHS $=\frac{2}{3}(x-5)-\frac{1}{4}(x-2)$$=\frac{2}{3}(\frac{88}{5}-5)-\frac{1}{4}(\frac{88}{5}-2)$$=\frac{2}{3}(\frac{88-5\times5}{5})-\frac{1}{4}(\frac{88-2\times5}{5})$$=\frac{2}{3}(\frac{88-25}{5})-\frac{1}{4}(\frac{88-10}{5})$$=\frac{2}{3}(\frac{63}{5})-\frac{1}{4}(\frac{78}{5})$$=\frac{2}{1}(\frac{21}{5})-\frac{1}{2}(\frac{39}{5})$$=\frac{42}{5}-\frac{39}{10}$$=\frac{42\times2-39}{10}$$=\frac{84-39}{10}$$=\frac{45}{10}$$=\frac{9}{2}$RHS $=\frac{9}{2}$LHS $=$ RHSHence verified.Read More

Given:The given equations are:(i) $\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$(ii) $13(y-4)-3(y-9)-5(y+4)=0$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions 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 $\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$.$\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$LCM of denominators $3$ and $5$ is $15$$\frac{(2x-1)\times5-(6x-2) \times3}{15}=\frac{1}{3}$$\frac{10x-5-18x+6}{15}=\frac{1}{3}$$\frac{-8x+1}{15}=\frac{1}{3}$On cross multiplication, we get, $-8x+1=\frac{1\times15}{3}$$-8x+1=5$$8x=1-5$$8x=-4$$x=\frac{-4}{8}$$x=\frac{-1}{2}$Verification:LHS $=\frac{(2x-1)}{3}-\frac{(6x-2)}{5}$$=\frac{(2\times\frac{-1}{2}-1)}{3}-\frac{(6\times\frac{-1}{2}-2)}{5}$$=\frac{-1-1}{3}-\frac{-3-2}{5}$$=\frac{-2}{3}-\frac{-5}{5}$$=\frac{-2}{3}+1$$=\frac{-2+1\times3}{3}$$=\frac{-2+3}{3}$$=\frac{1}{3}$RHS $=\frac{1}{3}$LHS $=$ RHSHence verified.(ii) The given equation is $13(y-4)-3(y-9)-5(y+4)=0$.$13(y-4)-3(y-9)-5(y+4)=0$$13y-52-3y+27-5y-20=0$$13y-8y-72+27=0$$5y-45=0$$5y=45$$y=\frac{45}{5}$$y=9$Verification:LHS $=13(y-4)-3(y-9)-5(y+4)$$=13(9-4)-3(9-9)-5(9+4)$$=13(5)-3(0)-5(13)$$=65-0-65$$=0$RHS $=0$LHS $=$ RHSHence verified.Read More

Given:The given equations are:(i) $\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}$(ii) $\frac{7}{x}+35=\frac{1}{10}$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions 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 $\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}$.$\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}$$\frac{x}{2}+\frac{x}{5}+\frac{3x}{10}=\frac{1}{5}+\frac{4}{5}$     (Transposing $\frac{4}{5}$ to RHS)LCM of denominators $2, 5$ and $10$ is $10$$\frac{x \times5+x \times2+3x \times1}{10}=\frac{1+4}{5}$$\frac{5x+2x+3x}{10}=\frac{5}{5}$$\frac{10x}{10}=1$$x=1$Verification:LHS $=\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3x}{10}$$=\frac{1}{2}-\frac{4}{5}+\frac{1}{5}+\frac{3(1)}{10}$$=\frac{1\times5-4\times2+1\times2+3}{10}$    (LCM of $2, 5$ and $10$ is $10$)$=\frac{5-8+2+3}{10}$$=\frac{10-8}{10}$$=\frac{2}{10}$$=\frac{1}{5}$RHS $=\frac{1}{5}$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{7}{x}+35=\frac{1}{10}$.$\frac{7}{x}+35=\frac{1}{10}$$\frac{7}{x}=\frac{1}{10}-35$                    (Transposing $35$ to ... Read More

Given:The given equations are:(i) $\frac{2x}{3}-\frac{3x}{8}=\frac{7}{12}$(ii) $(x+2)(x+3)+(x-3)(x-2)-2x(x+1)=0$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions 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 $\frac{2x}{3}-\frac{3x}{8}=\frac{7}{12}$.$\frac{2x}{3}-\frac{3x}{8}=\frac{7}{12}$LCM of $3$ and $8$ is $24$$\frac{2x \times 8-3x \times3}{24}=\frac{7}{12}$$\frac{16x-9x}{24}=\frac{7}{12}$$\frac{7x}{24}=\frac{7}{12}$On cross multiplication, we get, $7x =\frac{7\times24}{12}$$7x=\frac{7\times2}{1}$$7x=14$$x=\frac{14}{7}$$x=2$Verification:LHS $=\frac{2x}{3}-\frac{3x}{8}$$=\frac{2\times2}{3}-\frac{3\times2}{8}$$=\frac{4}{3}-\frac{3}{4}$$=\frac{4\times4-3\times3}{12}$               (LCM of $3$ and $4$ is $12$)$=\frac{16-9}{12}$$=\frac{7}{12}$RHS $=\frac{7}{12}$LHS $=$ RHSHence verified.(ii) The given equation is $(x+2)(x+3)+(x-3)(x-2)-2x(x+1)=0$.$(x+2)(x+3)+(x-3)(x-2)-2x(x+1)=0$$x(x+3)+2(x+3)+x(x-2)-3(x-2)-2x(x)-2x(1)=0$$x^2+3x+2x+6+x^2-2x-3x+6-2x^2-2x=0$$2x^2-2x^2+5x-7x+12=0$$-2x+12=0$$2x=12$$x=\frac{12}{2}$$x=6$Verification:LHS $=(x+2)(x+3)+(x-3)(x-2)-2x(x+1)$$=(6+2)(6+3)+(6-3)(6-2)-2(6)(6+1)$$=(8)(9)+(3)(4)-12(7)$$=72+12-84$$=84-84$$=0$RHS $=0$LHS $=$ RHSHence verified.Read More

Given:The given equations are:(i) $\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$(ii) $\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$To do:We have to solve the given equations and verify the solutions.Solution:To verify the solutions 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 $\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$.$\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13$LCM of denominators $2, 3$  and $4$ is $12$Therefore, $\frac{x \times6+x \times4+x \times3}{12}=13$$\frac{6x+4x+3x}{12}=13$$\frac{13x}{12}=13$On cross multiplication, we get, $13x=12\times13$$x=\frac{12\times13}{13}$$x=12$Verification:LHS $=\frac{x}{2}+\frac{x}{3}+\frac{x}{4}$$=\frac{12}{2}+\frac{12}{3}+\frac{12}{4}$$=6+4+3$$=13$RHS $=13$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$.$\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$LCM of denominators $2$ and $8$ is $8$$\frac{x \times 4+x}{8}=\frac{1}{8}$$\frac{4x+x}{8}=\frac{1}{8}$$\frac{5x}{8}=\frac{1}{8}$On cross multiplication, we get, $5x=\frac{1\times8}{8}$$5x=1$$x=\frac{1}{5}$Verification:LHS $=\frac{x}{2}+\frac{x}{8}$$=\frac{\frac{1}{5}}{2}+\frac{\frac{1}{5}}{8}$$=\frac{1}{5\times2}+\frac{1}{5\times8}$$=\frac{1}{10}+\frac{1}{40}$$=\frac{1\times4+1}{40}$            ... Read More