Question Number 188059 by normans last updated on 25/Feb/23
Answered by nikif99 last updated on 25/Feb/23
$${A}+{B}+{C}=\mathrm{6}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{2}{A}+{B}=\mathrm{8}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{C}+{E}=\mathrm{6}\:\left(\mathrm{3}\right) \\ $$$${A}+{D}+{E}=\mathrm{9}\:\left(\mathrm{4}\right) \\ $$$${B}+\mathrm{2}{D}=\mathrm{6}\:\left(\mathrm{5}\right) \\ $$$${B}+\frac{{D}}{{C}}=\mathrm{6}\:\left(\mathrm{6}\right) \\ $$$$\left(\mathrm{5}\right)\left(\mathrm{6}\right)\:\Rightarrow\mathrm{2}{D}=\frac{{D}}{{C}}\:\Rightarrow \\ $$$${a}:\:{C}=\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:{b}:\:{D}=\mathrm{0} \\ $$$${a}:\:\left(\mathrm{3}\right)\:\Rightarrow{E}=\mathrm{5},\:\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\Rightarrow{A}=\frac{\mathrm{5}}{\mathrm{2}},\:{B}=\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\Rightarrow{D}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow{C}+{E}−{A}=\mathrm{3} \\ $$$${b}:\:\left(\mathrm{5}\right)\:\Rightarrow{B}=\mathrm{6},\:\left(\mathrm{2}\right)\:\Rightarrow{A}=\mathrm{1} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow{C}=−\mathrm{1},\:\left(\mathrm{3}\right)\:\Rightarrow{E}=\mathrm{8} \\ $$$${C}+{E}−{A}=\mathrm{6} \\ $$