Question Number 182093 by Fridunatjan08 last updated on 04/Dec/22
$${Solve}\:{the}\:{equation}: \\ $$$$\frac{{x}−\mathrm{6}}{\mathrm{2020}}+\frac{{x}−\mathrm{5}}{\mathrm{2021}}+\frac{{x}−\mathrm{4}}{\mathrm{2022}}=\mathrm{3} \\ $$
Commented by Fridunatjan08 last updated on 04/Dec/22
$${Please}\:{show}\:{me}\:{the}\:{easy}\:{way}\:{to}\:{solve}\:{this}\:{equation} \\ $$
Commented by CElcedricjunior last updated on 04/Dec/22
$$\boldsymbol{{x}}=\mathrm{2026} \\ $$$$\boldsymbol{{car}} \\ $$$$\frac{\mathrm{2026}−\mathrm{6}}{\mathrm{2020}}+\frac{\mathrm{2026}−\mathrm{5}}{\mathrm{2021}}+\frac{\mathrm{2026}−\mathrm{4}}{\mathrm{2022}}=\frac{\mathrm{2020}}{\mathrm{2020}}+\frac{\mathrm{2021}}{\mathrm{2021}}+\frac{\mathrm{2022}}{\mathrm{2022}}=\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{3} \\ $$
Answered by Rasheed.Sindhi last updated on 04/Dec/22
$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{most}}\:\boldsymbol{\mathrm{tricky}}\:\boldsymbol{\mathrm{way}} \\ $$$$\frac{{x}−\mathrm{6}}{\mathrm{2020}}+\frac{{x}−\mathrm{5}}{\mathrm{2021}}+\frac{{x}−\mathrm{4}}{\mathrm{2022}}=\mathrm{3} \\ $$$$\frac{{x}−\mathrm{6}}{\mathrm{2020}}−\mathrm{1}+\frac{{x}−\mathrm{5}}{\mathrm{2021}}−\mathrm{1}+\frac{{x}−\mathrm{4}}{\mathrm{2022}}−\mathrm{1}=\mathrm{0} \\ $$$$\frac{{x}−\mathrm{2026}}{\mathrm{2020}}+\frac{{x}−\mathrm{2026}}{\mathrm{2021}}+\frac{{x}−\mathrm{2026}}{\mathrm{2022}}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2026}\right)\left(\frac{\mathrm{1}}{\mathrm{2020}}+\frac{\mathrm{1}}{\mathrm{2021}}+\frac{\mathrm{1}}{\mathrm{2022}}\right)=\mathrm{0} \\ $$$${x}−\mathrm{2026}=\mathrm{0} \\ $$$${x}=\mathrm{2026} \\ $$
Commented by Fridunatjan08 last updated on 04/Dec/22
$${thanks}\:{you}! \\ $$