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Found 225 Articles for Class 8
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
88 Views
Given:The given equations are:(i) $\frac{45-2x}{15}-\frac{4x+10}{5}=\frac{15-14x}{9}$(ii) $\frac{5(7x+5)}{3}-\frac{23}{3}=13-\frac{4x-2}{3}$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{45-2x}{15}-\frac{4x+10}{5}=\frac{15-14x}{9}$$\frac{45-2x}{15}-\frac{4x+10}{5}=\frac{15-14x}{9}$On rearranging, we get, $\frac{45-2x}{15}-\frac{4x+10}{5}-\frac{15-14x}{9}=0$LCM of denominators $15, 5$ and $9$ is $45$$\frac{(45-2x)\times3-(4x+10)\times9-(15-14x) \times5}{45}=0$$\frac{3(45)-3(2x)-9(4x)-9(10)-5(15)+5(14x)}{45}=0$$\frac{135-6x-36x-90-75+70x}{45}=0$$\frac{135-165-42x+70x}{45}=0$$\frac{-30+28x}{45}=0$On cross multiplication, we get, $28x-30=45(0)$$28x-30=0$$28x=30$$x=\frac{30}{28}$$x=\frac{15}{14}$Verification:LHS $=\frac{45-2x}{15}-\frac{4x+10}{5}$$=\frac{45-2(\frac{15}{14})}{15}-\frac{4(\frac{15}{14})+10}{5}$$=\frac{45-\frac{15}{7}}{15}-\frac{\frac{30}{7}+10}{5}$$=\frac{45\times7-15}{7\times15}-\frac{30+10\times7}{7\times5}$$=\frac{315-15}{105}-\frac{30+70}{35}$$=\frac{300}{105}-\frac{100}{35}$$=\frac{60}{21}-\frac{20}{7}$$=\frac{60-20\times3}{21}$$=\frac{60-60}{21}$$=0$RHS $=\frac{15-14x}{9}$$=\frac{15-14(\frac{15}{14})}{9}$$=\frac{15-15}{9}$$=0$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{5(7x+5)}{3}-\frac{23}{3}=13-\frac{4x-2}{3}$$\frac{5(7x+5)}{3}-\frac{23}{3}=13-\frac{4x-2}{3}$On rearranging, we get, $\frac{5(7x+5)}{3}+\frac{4x-2}{3}=\frac{23}{3}+13$LCM of $3$ and $1$ is $3$$\frac{5(7x)+5(5)+4x-2}{3}=\frac{23+13\times3}{3}$$\frac{35x+25+4x-2}{3}=\frac{23+39}{3}$$\frac{39x+23}{3}=\frac{62}{3}$On cross multiplication, we get, $39x+23=62$$39x=62-23$$39x=39$$x=\frac{39}{39}$$x=1$Verification:LHS $=\frac{5(7x+5)}{3}-\frac{23}{3}$$=\frac{5(7(1)+5)}{3}-\frac{23}{3}$$=\frac{5(7+5)}{3}-\frac{23}{3}$$=\frac{5(12)}{3}-\frac{23}{3}$$=\frac{60}{3}-\frac{23}{3}$$=\frac{60-23}{3}$$=\frac{37}{3}$RHS $=13-\frac{4x-2}{3}$$=13-\frac{4(1)-2}{3}$$=13-\frac{4-2}{3}$$=13-\frac{2}{3}$$=\frac{13\times3-2}{3})$$=\frac{39-2}{3}$$=\frac{37}{3}$LHS $=$ RHSHence ... Read More
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
80 Views
Given:The given equations are:(i) $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$(ii) $\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$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{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\times2-3\times3}{6x}=\frac{1}{12}$$\frac{4-9}{6x}=\frac{1}{12}$$\frac{-5}{6x}=\frac{1}{12}$On cross multiplication, we get, $-5\times12=1\times6x$$6x=-60$$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}$$=\frac{-1}{15}+\frac{3}{20}$$=\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{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$$\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$On rearranging, we get, $\frac{4x}{9}+\frac{13x}{108}-\frac{8x+19}{18}=-\frac{1}{3}$LCM of $9, 108$ and $18$ is $108$$\frac{4x \times 12+13x \times1- (8x+19)\times6}{108}=-\frac{1}{3}$$\frac{48x+13x-48x-114}{108}=-\frac{1}{3}$$\frac{13x-114}{108}=-\frac{1}{3}$On ... Read More
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
95 Views
Given:The given equations are:(i) $\frac{9x+7}{2}-(x-\frac{(x-2)}{7})=36$(ii) $0.18(5x-4)=0.5x+0.8$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{9x+7}{2}-(x-\frac{(x-2)}{7})=36$$\frac{9x+7}{2}-(x-\frac{(x-2)}{7})=36$$\frac{9x+7}{2}-(\frac{7x-(x-2)}{7})=36$$\frac{9x+7}{2}-(\frac{7x-x+2}{7})=36$$\frac{9x+7}{2}-(\frac{6x+2)}{7})=36$LCM of denominators $2$ and $7$ is $14$$\frac{(9x+7)\times7-(6x+2)\times2}{14}=36$$\frac{7(9x)+7(7)-2(6x)-2(2)}{14}=36$$\frac{63x+49-12x-4}{14}=36$$\frac{51x+45}{14}=36$On cross multiplication, we get, $51x+45=36\times14$$51x+45=504$$51x=504-45$$51x=459$$x=\frac{459}{51}$$x=9$Verification:LHS $=\frac{9x+7}{2}-(x-\frac{(x-2)}{7})$$=\frac{9(9)+7}{2}-(9-\frac{(9-2)}{7})$$=\frac{81+7}{2}-(9-\frac{7}{7})$$=\frac{88}{2}-(9-1)$$=44-8$$=36$RHS $=36$LHS $=$ RHSHence verified.(ii) The given equation is $0.18(5x-4)=0.5x+0.8$$0.18(5x-4)=0.5x+0.8$$0.18(5x)-0.18(4)=0.5x+0.8$$0.9x-0.72=0.5x+0.8$On rearranging, we get, $0.9x-0.5x=0.8+0.72$$0.4x=1.52$$x=\frac{1.52}{0.4}$$x=3.8$Verification:LHS $=0.18(5x-4)$$=0.18(5(3.8)-4)$$=0.18(19-4)$$=0.18(15)$$=2.7$RHS $=0.5x+0.8$$=0.5(3.8)+0.8$$=1.9+0.8$$=2.7$LHS $=$ RHSHence verified.Read More
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
102 Views
Given:The given equations are:(i) $\frac{3x+1}{16}+\frac{2x-3}{7}=\frac{x+3}{8}+\frac{3x-1}{14}$(ii) $\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{3x+1}{16}+\frac{2x-3}{7}=\frac{x+3}{8}+\frac{3x-1}{14}$$\frac{3x+1}{16}+\frac{2x-3}{7}=\frac{x+3}{8}+\frac{3x-1}{14}$On rearranging, we get, $\frac{3x+1}{16}+\frac{2x-3}{7}-\frac{x+3}{8}-\frac{3x-1}{14}=0$LCM of denominators $16, 7, 8$ and $14$ is $112$$\frac{(3x+1)\times7+(2x-3)\times16-(x+3) \times14-(3x-1)\times8}{112}=0$$\frac{7(3x)+7(1)+16(2x)-16(3)-14(x)-14(3)-8(3x)+8(1)}{112}=0$$\frac{21x+7+32x-48-14x-42-24x+8}{112}=0$$\frac{53x-38x+15-90}{112}=0$$\frac{15x-75}{112}=0$On cross multiplication, we get, $15x-75=112(0)$$15x-75=0$$15x=75$$x=\frac{75}{15}$$x=5$Verification:LHS $=\frac{3x+1}{16}+\frac{2x-3}{7}$$=\frac{3(5)+1}{16}+\frac{2(5)-3}{7}$$=\frac{15+1}{16}+\frac{10-3}{7}$$=\frac{16}{16}+\frac{7}{7}$$=1+1$$=2$RHS $=\frac{x+3}{8}+\frac{3x-1}{14}$$=\frac{5+3}{8}+\frac{3(5)-1}{14}$$=\frac{8}{8}+\frac{15-1}{14}$$=1+\frac{14}{14}$$=1+1$$=2$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$$\frac{1-2x}{7}-\frac{2-3x}{8}=\frac{3}{2}+\frac{x}{4}$On rearranging, we get, $\frac{1-2x}{7}-\frac{2-3x}{8}-\frac{x}{4}=\frac{3}{2}$LCM of $7, 8$ and $4$ is $56$$\frac{8\times (1-2x)-(2-3x)\times7-(x)\times14}{56}=\frac{3}{2}$$\frac{8-16x-14+21x-14x}{56}=\frac{3}{2}$$\frac{-9x-6}{56}=\frac{3}{2}$On cross multiplication, we get, $(-9x-6)\times2=3\times56$$-18x-12=168$$-18x=168+12$$-18x=180$$x=\frac{180}{-18}$$x=-10$Verification:LHS $=\frac{1-2x}{7}-\frac{2-3x}{8}$$=\frac{1-2(-10)}{7}-\frac{2-3(-10)}{8}$$=\frac{1+20}{7}-\frac{2+30}{8}$$=\frac{21}{7}-\frac{32}{8}$$=3-4$$=-1$RHS ... Read More
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
104 Views
Given:The given equations are:(i) $\frac{3x}{4}-\frac{x-1}{2}=\frac{x-2}{3}$(ii) $\frac{5x}{3}-\frac{(x-1)}{4}=\frac{(x-3)}{5}$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{3x}{4}-\frac{x-1}{2}=\frac{x-2}{3}$$\frac{3x}{4}-\frac{x-1}{2}=\frac{x-2}{3}$On rearranging, we get, $\frac{3x}{4}-\frac{x-1}{2}-\frac{x-2}{3}=0$LCM of denominators $4, 2$ and $3$ is $12$$\frac{(3x)\times3-(x-1)\times6-(x-2) \times4}{4}=0$$\frac{9x-6(x)+6(1)-4(x)+4(2)}{12}=0$$\frac{9x-6x+6-4x+8}{12}=0$$\frac{-x+14}{12}=0$On cross multiplication, we get, $-x+14=0(12)$$-x+14=0$$x=14$Verification:LHS $=\frac{3x}{4}-\frac{x-1}{2}$$=\frac{3(14)}{4}-\frac{14-1}{2}$$=\frac{42}{4}-\frac{13}{2}$$=\frac{21}{2}-\frac{13}{2}$$=\frac{21-13}{2}$$=\frac{8}{2}$$=4$RHS $=\frac{x-2}{3}$$=\frac{14-2}{3}$$=\frac{12}{3}$$=4$LHS $=$ RHSHence verified.(ii) The given equation is $\frac{5x}{3}-\frac{(x-1)}{4}=\frac{(x-3)}{5}$.$\frac{5x}{3}-\frac{(x-1)}{4}=\frac{(x-3)}{5}$On rearranging, we get, $\frac{5x}{3}-\frac{(x-1)}{4}-\frac{(x-3)}{5}=0$LCM of $3, 4$ and $5$ is $60$$\frac{5x \times 20-(x-1)\times15-(x-3)\times12}{60}=0$$\frac{100x-15x+15-12x+36}{60}=0$$\frac{73x+51}{60}=0$On cross multiplication, we get, $73x+51=60(0)$$73x+51=0$$73x=-51$$x=\frac{-51}{73}$Verification:LHS $=\frac{5x}{3}-\frac{(x-1)}{4}$$=\frac{5(\frac{-51}{73})}{3}-\frac{(\frac{-51}{73}-1)}{4}$$=\frac{\frac{5\times(-51)}{73}}{3}-\frac{\frac{-51-1\times73}{73}}{4}$$=\frac{-255}{219}-\frac{-51-73}{73\times4}$$=\frac{-255}{219}-\frac{-124}{292}$$=\frac{-255\times4+124\times3}{876}$ ... Read More
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
95 Views
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}$$=\frac{3}{2}$LHS $=$ ... Read More
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
76 Views
Given:The given equations are:(i) $\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$(ii) $\frac{6x+1}{2}+1=\frac{7x-3}{3}$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$.$\frac{7x}{2}-\frac{5x}{2}=\frac{20x}{3}+10$On rearranging, we get, $\frac{7x}{2}-\frac{5x}{2}-\frac{20x}{3}=10$LCM of $2$ and $3$ is $6$$\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}$$\frac{6x+1}{2}+1=\frac{7x-3}{3}$On ... Read More
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
86 Views
Given:The given equations are:(i) $\frac{1}{2}x+7x-6=7x+\frac{1}{4}$(ii) $\frac{3}{4}x+4x=\frac{7}{8}+6x-6$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{1}{2}x+7x-6=7x+\frac{1}{4}$.$\frac{1}{2}x+7x-6=7x+\frac{1}{4}$On rearranging, we get, $\frac{1}{2}x+7x-7x=\frac{1}{4}+6$$\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
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
75 Views
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}$$=\frac{-33}{17}$LHS $=$ RHSHence verified.Read More
![Akhileshwar Nani](https://www.tutorialspoint.com/assets/profiles/629140/profile/60_2164282-1680251555.png)
66 Views
Given:The given equations are:(i) $\frac{2x+5}{3}=3x-10$(ii) $\frac{a-8}{3}=\frac{a-3}{2}$To do:We have to solve the given equations and check the results.Solution:To check the results we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.(i) The given equation is $\frac{2x+5}{3}=3x-10$.$\frac{2x+5}{3}=3x-10$On cross multiplication, we get, $2x+5=3(3x-10)$$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