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Question Number 205471 by Fridunatjan08 last updated on 21/Mar/24

Solve the equation: (x/(21))+(x/(77))+(x/(165))+(x/(285))=200

$${Solve}\:{the}\:{equation}:\:\frac{{x}}{\mathrm{21}}+\frac{{x}}{\mathrm{77}}+\frac{{x}}{\mathrm{165}}+\frac{{x}}{\mathrm{285}}=\mathrm{200} \\ $$

Answered by Rasheed.Sindhi last updated on 23/Mar/24

(x/(21))+(x/(77))+(x/(165))+(x/(285))=200 x((1/(21))+(1/(77))+(1/(165))+(1/(285)))=200 x( (1/7)((1/3)+(1/(11)))+(1/(15))((1/(11))+(1/(19))))=200 x( (1/( 7_(1) ))(((14^(2) )/(3×11)))+(1/(15_(1) ))(((30^(2) )/(11×19))))=200 x( (1/(33))+(1/(209)))=100 x((1/(11))( (1/3)+(1/(19))))=100 x((1/(11_(1) ))( ((22^(2) )/(3×19))))=100 x((1/(57)))=50 x=50×57=2850

$$\frac{{x}}{\mathrm{21}}+\frac{{x}}{\mathrm{77}}+\frac{{x}}{\mathrm{165}}+\frac{{x}}{\mathrm{285}}=\mathrm{200} \\ $$$${x}\left(\frac{\mathrm{1}}{\mathrm{21}}+\frac{\mathrm{1}}{\mathrm{77}}+\frac{\mathrm{1}}{\mathrm{165}}+\frac{\mathrm{1}}{\mathrm{285}}\right)=\mathrm{200} \\ $$$${x}\left(\:\frac{\mathrm{1}}{\mathrm{7}}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{11}}\right)+\frac{\mathrm{1}}{\mathrm{15}}\left(\frac{\mathrm{1}}{\mathrm{11}}+\frac{\mathrm{1}}{\mathrm{19}}\right)\right)=\mathrm{200} \\ $$$${x}\left(\:\frac{\mathrm{1}}{\cancel{\underset{\mathrm{1}} {\:\mathrm{7}}}}\left(\frac{\cancel{\overset{\mathrm{2}} {\mathrm{14}}}}{\mathrm{3}×\mathrm{11}}\right)+\frac{\mathrm{1}}{\cancel{\underset{\mathrm{1}} {\mathrm{15}}}}\left(\frac{\cancel{\overset{\mathrm{2}} {\mathrm{30}}}}{\mathrm{11}×\mathrm{19}}\right)\right)=\mathrm{200} \\ $$$${x}\left(\:\frac{\mathrm{1}}{\mathrm{33}}+\frac{\mathrm{1}}{\mathrm{209}}\right)=\mathrm{100} \\ $$$${x}\left(\frac{\mathrm{1}}{\mathrm{11}}\left(\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{19}}\right)\right)=\mathrm{100} \\ $$$${x}\left(\frac{\mathrm{1}}{\cancel{\underset{\mathrm{1}} {\mathrm{11}}}}\left(\:\frac{\cancel{\overset{\mathrm{2}} {\mathrm{22}}}}{\mathrm{3}×\mathrm{19}}\right)\right)=\mathrm{100} \\ $$$${x}\left(\frac{\mathrm{1}}{\mathrm{57}}\right)=\mathrm{50} \\ $$$${x}=\mathrm{50}×\mathrm{57}=\mathrm{2850} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Mar/24

Similar method(grouping in another way) (x/(21))+(x/(285))+(x/(77))+(x/(165))=200 (1/3)((x/7)+(x/(95)))+(1/(11))((x/7)+(x/(15)))=200 (1/3)∙((102^(34) x)/(665))+(1/(11))∙((22^(2) x)/(105))=200 ((17x)/(665))+(x/(105))=100 (1/(35))(((17x)/(19))+(x/3))=100 (1/(35))(((70^(2) x)/(57)))=100 (x/(57))=50⇒x=2850

$${Similar}\:{method}\left({grouping}\:{in}\:{another}\:{way}\right) \\ $$$$\frac{{x}}{\mathrm{21}}+\frac{{x}}{\mathrm{285}}+\frac{{x}}{\mathrm{77}}+\frac{{x}}{\mathrm{165}}=\mathrm{200} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{x}}{\mathrm{7}}+\frac{{x}}{\mathrm{95}}\right)+\frac{\mathrm{1}}{\mathrm{11}}\left(\frac{{x}}{\mathrm{7}}+\frac{{x}}{\mathrm{15}}\right)=\mathrm{200} \\ $$$$\frac{\mathrm{1}}{\cancel{\mathrm{3}}}\centerdot\frac{\cancel{\overset{\mathrm{34}} {\mathrm{102}}}{x}}{\mathrm{665}}+\frac{\mathrm{1}}{\cancel{\mathrm{11}}}\centerdot\frac{\cancel{\overset{\mathrm{2}} {\mathrm{22}}}{x}}{\mathrm{105}}=\mathrm{200} \\ $$$$\:\:\:\:\:\frac{\mathrm{17}{x}}{\mathrm{665}}+\frac{{x}}{\mathrm{105}}=\mathrm{100} \\ $$$$\:\:\:\frac{\mathrm{1}}{\mathrm{35}}\left(\frac{\mathrm{17}{x}}{\mathrm{19}}+\frac{{x}}{\mathrm{3}}\right)=\mathrm{100} \\ $$$$\:\:\:\frac{\mathrm{1}}{\cancel{\mathrm{35}}}\left(\frac{\cancel{\overset{\mathrm{2}} {\mathrm{70}}}{x}}{\mathrm{57}}\right)=\mathrm{100} \\ $$$$\:\:\:\:\:\:\:\:\frac{{x}}{\mathrm{57}}=\mathrm{50}\Rightarrow{x}=\mathrm{2850} \\ $$

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