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x-2-y-2-25-x-2y-3-0-




Question Number 211574 by MrGaster last updated on 13/Sep/24
                          { ((x^2 +y^2 =25)),((x+2y−3=0)) :}
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{cases}{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{25}}\\{\boldsymbol{{x}}+\mathrm{2}\boldsymbol{\mathrm{y}}−\mathrm{3}=\mathrm{0}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 13/Sep/24
   { ((x^2 +y^2 =25...(i))),((x+2y−3=0...(ii))) :}  (ii)⇒x=3−2y  (i)⇒(3−2y)^2 +y^2 =25           5y^2 −12y−16=0      y=((12±(√(144+320)))/(10))      y=((6±2(√(29)))/5)  x=3−2(((6±2(√(29)))/5))=((15−12±4(√(29)))/5)       =((3±4(√(29)))/5)  (x,y)=(((3±4(√(29)))/5),((6±2(√(29)))/5))
$$\:\:\begin{cases}{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{25}…\left({i}\right)}\\{\boldsymbol{{x}}+\mathrm{2}\boldsymbol{\mathrm{y}}−\mathrm{3}=\mathrm{0}…\left({ii}\right)}\end{cases} \\ $$$$\left({ii}\right)\Rightarrow{x}=\mathrm{3}−\mathrm{2}{y} \\ $$$$\left({i}\right)\Rightarrow\left(\mathrm{3}−\mathrm{2}{y}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{5}{y}^{\mathrm{2}} −\mathrm{12}{y}−\mathrm{16}=\mathrm{0} \\ $$$$\:\:\:\:{y}=\frac{\mathrm{12}\pm\sqrt{\mathrm{144}+\mathrm{320}}}{\mathrm{10}} \\ $$$$\:\:\:\:{y}=\frac{\mathrm{6}\pm\mathrm{2}\sqrt{\mathrm{29}}}{\mathrm{5}} \\ $$$${x}=\mathrm{3}−\mathrm{2}\left(\frac{\mathrm{6}\pm\mathrm{2}\sqrt{\mathrm{29}}}{\mathrm{5}}\right)=\frac{\mathrm{15}−\mathrm{12}\pm\mathrm{4}\sqrt{\mathrm{29}}}{\mathrm{5}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{29}}}{\mathrm{5}} \\ $$$$\left({x},{y}\right)=\left(\frac{\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{29}}}{\mathrm{5}},\frac{\mathrm{6}\pm\mathrm{2}\sqrt{\mathrm{29}}}{\mathrm{5}}\right) \\ $$

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