Question Number 226144 by Linton last updated on 20/Nov/25
$$\mathrm{6}^{{x}} \:+\:\mathrm{6}^{{y}} \:=\:\mathrm{42} \\ $$$${x}\:+\:{y}\:=\mathrm{3} \\ $$$${Find}\:{x}\:{and}\:{y}\:? \\ $$
Commented by mr W last updated on 20/Nov/25
$${not}\:{clear}? \\ $$$$\left(\mathrm{1},\:\mathrm{2}\right)\:{or}\:\left(\mathrm{2},\:\mathrm{1}\right) \\ $$
Answered by ibrahimmatematic last updated on 20/Nov/25
$${SOLUTION}:{Abdullayev}.{I} \\ $$$${x}=\mathrm{3}−{y} \\ $$$$\mathrm{6}^{\mathrm{3}−{y}} +\mathrm{6}^{{y}} =\mathrm{42} \\ $$$$\mathrm{6}^{\mathrm{3}} +\mathrm{6}^{\mathrm{2}{y}} =\mathrm{42}\centerdot\mathrm{6}^{{y}} \\ $$$$\mathrm{6}^{{y}} ={t} \\ $$$${t}^{\mathrm{2}} −\mathrm{42}{t}+\mathrm{216}=\mathrm{0} \\ $$$${t}=\mathrm{6};\mathrm{36} \\ $$$$\mathrm{6}^{{y}} =\mathrm{6}\:\Rightarrow{y}=\mathrm{1};{x}=\mathrm{2} \\ $$$$\mathrm{6}^{{y}} =\mathrm{36}\:\Rightarrow{y}=\mathrm{2};{x}=\mathrm{1} \\ $$
Answered by Kademi last updated on 20/Nov/25
$$\: \\ $$$$\:{y}\:=\:\mathrm{3}−{x} \\ $$$$\:\mathrm{6}^{{x}} +\mathrm{6}^{{y}} \:=\:\mathrm{42}\: \\ $$$$\:\mathrm{6}^{{x}} +\mathrm{6}^{\mathrm{3}−{x}} \:=\:\mathrm{42} \\ $$$$\:\mathrm{6}^{{x}} −\mathrm{42}+\frac{\mathrm{6}^{\mathrm{3}} }{\mathrm{6}^{{x}} }\:=\:\mathrm{0} \\ $$$$\:\left(\mathrm{6}^{{x}} \right)^{\mathrm{2}} −\mathrm{42}\left(\mathrm{6}^{{x}} \right)+\mathrm{6}^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\:\mathrm{6}^{{x}} \:=\:\frac{\mathrm{42}\pm\sqrt{\mathrm{42}^{\mathrm{2}} −\mathrm{4}×\mathrm{6}^{\mathrm{3}} }}{\mathrm{2}}\:=\:\mathrm{21}\pm\mathrm{15}\:\:\rightarrow\:\:\begin{cases}{\left.\mathrm{1}\right)\:\mathrm{36}}\\{\left.\mathrm{2}\right)\:\mathrm{6}}\end{cases} \\ $$$$\left.\:\mathrm{1}\right)\:\mathrm{6}^{{x}} \:=\:\mathrm{36}\:\:\Rightarrow\:\:{x}_{\mathrm{1}} \:=\:\mathrm{2}\:\:\Rightarrow\:\:{y}_{\mathrm{1}} \:=\:\mathrm{1} \\ $$$$\left.\:\mathrm{2}\right)\:\mathrm{6}^{{x}} \:=\:\mathrm{6}\:\:\Rightarrow\:\:{x}_{\mathrm{2}} \:=\:\mathrm{1}\:\:\Rightarrow\:\:{y}_{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:\begin{array}{|c|}{{x}_{\mathrm{1}} \wedge{y}_{\mathrm{2}} \:=\:\mathrm{2}}&\hline{{x}_{\mathrm{2}} \wedge{y}_{\mathrm{1}} \:=\:\mathrm{1}}\\\hline\end{array} \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$