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Question-204389




Question Number 204389 by Ngarmadji last updated on 15/Feb/24
Answered by Frix last updated on 15/Feb/24
((2x)/3)−((2x)/5)=1  (x/3)−(x/5)=(1/2)  5x−3x=((15)/2)  x=((15)/4)  ((3x)/5)−((2x)/(15))+(x/3)=−(7/(15))  ((9x−2x+5x)/(15))=−(7/(15))  ... x=−(7/(12))
$$\frac{\mathrm{2}{x}}{\mathrm{3}}−\frac{\mathrm{2}{x}}{\mathrm{5}}=\mathrm{1}\:\:\frac{{x}}{\mathrm{3}}−\frac{{x}}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{5}{x}−\mathrm{3}{x}=\frac{\mathrm{15}}{\mathrm{2}}\:\:{x}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}{x}}{\mathrm{5}}−\frac{\mathrm{2}{x}}{\mathrm{15}}+\frac{{x}}{\mathrm{3}}=−\frac{\mathrm{7}}{\mathrm{15}}\:\:\frac{\mathrm{9}{x}−\mathrm{2}{x}+\mathrm{5}{x}}{\mathrm{15}}=−\frac{\mathrm{7}}{\mathrm{15}}\:\:…\:{x}=−\frac{\mathrm{7}}{\mathrm{12}} \\ $$

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