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2-x-y-3-2-x-y-6-6-x-2-5y-2-6xy-

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Question Number 87516 by mary_ last updated on 04/Apr/20
{ ((2^((x+y)/3) +2^((x+y)/6) =6)),((x^2 +5y^2 =6xy)) :}
$$\begin{cases}{\mathrm{2}^{\frac{{x}+{y}}{\mathrm{3}}} +\mathrm{2}^{\frac{{x}+{y}}{\mathrm{6}}} =\mathrm{6}}\\{{x}^{\mathrm{2}} +\mathrm{5}{y}^{\mathrm{2}} =\mathrm{6}{xy}}\end{cases} \\ $$
Answered by TANMAY PANACEA. last updated on 04/Apr/20
x^2 −6xy+5y^2 =0 x^2 −xy−5xy+5y^2 =0 x(x−y)−5y(x−y)=0 (x−y)(x−5y)=0 either x=y or x=5y 2^((x+y)/3) +2^((x+y)/6) =6 2^((2x)/3) +2^((2x)/6) =6 2^((2x)/6) =a a^2 +a−6=0 (a+3)(a−2)=0 a=2=2^((2x)/6) →((2x)/6)=1 so x=y=(6/2)=3 if a≠−3 now when x=5y 2^((x+y)/3) +2^((x+y)/6) =6 2^(2y) +2^y =6 2^y =b b^2 +b−6=0 (b+3)(b−2)=0 when b=2 2^y =2 y=1 so x=5×1=5 so when x=3 y=3 and x=5 y=1
$${x}^{\mathrm{2}} −\mathrm{6}{xy}+\mathrm{5}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{xy}−\mathrm{5}{xy}+\mathrm{5}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}\left({x}−{y}\right)−\mathrm{5}{y}\left({x}−{y}\right)=\mathrm{0} \\ $$$$\left({x}−{y}\right)\left({x}−\mathrm{5}{y}\right)=\mathrm{0} \\ $$$${either}\:{x}={y}\:\:\:{or}\:{x}=\mathrm{5}{y} \\ $$$$\mathrm{2}^{\frac{{x}+{y}}{\mathrm{3}}} +\mathrm{2}^{\frac{{x}+{y}}{\mathrm{6}}} =\mathrm{6} \\ $$$$\mathrm{2}^{\frac{\mathrm{2}{x}}{\mathrm{3}}} +\mathrm{2}^{\frac{\mathrm{2}{x}}{\mathrm{6}}} =\mathrm{6} \\ $$$$\mathrm{2}^{\frac{\mathrm{2}{x}}{\mathrm{6}}} ={a} \\ $$$${a}^{\mathrm{2}} +{a}−\mathrm{6}=\mathrm{0} \\ $$$$\left({a}+\mathrm{3}\right)\left({a}−\mathrm{2}\right)=\mathrm{0} \\ $$$${a}=\mathrm{2}=\mathrm{2}^{\frac{\mathrm{2}{x}}{\mathrm{6}}} \rightarrow\frac{\mathrm{2}{x}}{\mathrm{6}}=\mathrm{1}\:\:{so}\:\:{x}={y}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3} \\ $$$${if}\:{a}\neq−\mathrm{3} \\ $$$$\boldsymbol{{now}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{5}\boldsymbol{{y}} \\ $$$$\mathrm{2}^{\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{3}}} \:+\mathrm{2}^{\frac{\boldsymbol{{x}}+\boldsymbol{{y}}}{\mathrm{6}}} =\mathrm{6} \\ $$$$\mathrm{2}^{\mathrm{2}\boldsymbol{{y}}} +\mathrm{2}^{\boldsymbol{{y}}} =\mathrm{6} \\ $$$$\mathrm{2}^{\boldsymbol{{y}}} =\boldsymbol{{b}} \\ $$$$\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{b}}−\mathrm{6}=\mathrm{0} \\ $$$$\left(\boldsymbol{{b}}+\mathrm{3}\right)\left(\boldsymbol{{b}}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{b}}=\mathrm{2} \\ $$$$\mathrm{2}^{\boldsymbol{{y}}} =\mathrm{2}\:\:\boldsymbol{{y}}=\mathrm{1}\:\:\boldsymbol{{so}}\:\boldsymbol{{x}}=\mathrm{5}×\mathrm{1}=\mathrm{5} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{3}\:\:\boldsymbol{{y}}=\mathrm{3} \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{x}}=\mathrm{5}\:\:\:\boldsymbol{{y}}=\mathrm{1} \\ $$$$ \\ $$

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