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Question Number 188059 by normans last updated on 25/Feb/23

Answered by nikif99 last updated on 25/Feb/23

A+B+C=6 (1)  2A+B=8 (2)  2C+E=6 (3)  A+D+E=9 (4)  B+2D=6 (5)  B+(D/C)=6 (6)  (5)(6) ⇒2D=(D/C) ⇒  a: C=(1/2) or b: D=0  a: (3) ⇒E=5, (1)(2) ⇒A=(5/2), B=3  (4) ⇒D=(3/2) ⇒C+E−A=3  b: (5) ⇒B=6, (2) ⇒A=1  (1) ⇒C=−1, (3) ⇒E=8  C+E−A=6

$${A}+{B}+{C}=\mathrm{6}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{2}{A}+{B}=\mathrm{8}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{C}+{E}=\mathrm{6}\:\left(\mathrm{3}\right) \\ $$$${A}+{D}+{E}=\mathrm{9}\:\left(\mathrm{4}\right) \\ $$$${B}+\mathrm{2}{D}=\mathrm{6}\:\left(\mathrm{5}\right) \\ $$$${B}+\frac{{D}}{{C}}=\mathrm{6}\:\left(\mathrm{6}\right) \\ $$$$\left(\mathrm{5}\right)\left(\mathrm{6}\right)\:\Rightarrow\mathrm{2}{D}=\frac{{D}}{{C}}\:\Rightarrow \\ $$$${a}:\:{C}=\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:{b}:\:{D}=\mathrm{0} \\ $$$${a}:\:\left(\mathrm{3}\right)\:\Rightarrow{E}=\mathrm{5},\:\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\Rightarrow{A}=\frac{\mathrm{5}}{\mathrm{2}},\:{B}=\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\Rightarrow{D}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow{C}+{E}−{A}=\mathrm{3} \\ $$$${b}:\:\left(\mathrm{5}\right)\:\Rightarrow{B}=\mathrm{6},\:\left(\mathrm{2}\right)\:\Rightarrow{A}=\mathrm{1} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow{C}=−\mathrm{1},\:\left(\mathrm{3}\right)\:\Rightarrow{E}=\mathrm{8} \\ $$$${C}+{E}−{A}=\mathrm{6} \\ $$

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