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Question-193235

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Question Number 193235 by mathlove last updated on 08/Jun/23
Answered by a.lgnaoui last updated on 08/Jun/23
Totale Area=22+A (A=xy) 22+xy=6x+(8−x)y 2xy=8y+6x−22 xy=4y+3x−11 (1) { (((8−x)y=10)),(((6−y)x=12 ⇒10−8y=12−6x)) :} x=((4y+1)/3) (2) (1)⇒x(y−3)=4y−11 x=((4y−11)/(y−3)) (1)⇔ ((y(4y+1))/3)=8y−10 (1)=(2)⇒ ((4y+1)/3)=((4y−11)/(y−3)) 4y^2 −12y+y−3=12y−33 4y^2 −23y+30=0 y^2 −((23)/4)y+((15)/2)=0 △=((23^2 )/(16))−30=((23^2 −480)/(16))=((7/4))^2 y=((((23)/4)±(7/4))/2)= y=(((15)/4),2) a− y=2 x=3 (x;y)= (3,2) b− y=((15)/4) x=((16)/3) (x;y)= (((16)/3) ; ((15)/4)) ⇒xy= { ((6 pour (x,y)=(3,2))),((20 pour (x,y)=(((16)/3),((15)/4)))) :}
$$\boldsymbol{\mathrm{T}}\mathrm{otale}\:\mathrm{Area}=\mathrm{22}+\mathrm{A}\:\:\:\:\:\:\left(\mathrm{A}=\boldsymbol{\mathrm{xy}}\right) \\ $$$$\mathrm{22}+\boldsymbol{\mathrm{xy}}=\mathrm{6}\boldsymbol{\mathrm{x}}+\left(\mathrm{8}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{y}} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{xy}}=\mathrm{8}\boldsymbol{\mathrm{y}}+\mathrm{6}\boldsymbol{\mathrm{x}}−\mathrm{22} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{xy}}=\mathrm{4}\boldsymbol{\mathrm{y}}+\mathrm{3}\boldsymbol{\mathrm{x}}−\mathrm{11}\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\begin{cases}{\left(\mathrm{8}−\boldsymbol{\mathrm{x}}\right)\boldsymbol{\mathrm{y}}=\mathrm{10}}\\{\left(\mathrm{6}−\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{x}}=\mathrm{12}\:\:\:\:\:\Rightarrow\mathrm{10}−\mathrm{8y}=\mathrm{12}−\mathrm{6x}}\end{cases} \\ $$$$\:\:\:\mathrm{x}=\frac{\mathrm{4y}+\mathrm{1}}{\mathrm{3}}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\Rightarrow\mathrm{x}\left(\mathrm{y}−\mathrm{3}\right)=\mathrm{4y}−\mathrm{11}\:\:\:\:\mathrm{x}=\frac{\mathrm{4y}−\mathrm{11}}{\mathrm{y}−\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\:\:\:\frac{\mathrm{y}\left(\mathrm{4y}+\mathrm{1}\right)}{\mathrm{3}}=\mathrm{8y}−\mathrm{10} \\ $$$$\left(\mathrm{1}\right)=\left(\mathrm{2}\right)\Rightarrow\:\:\frac{\mathrm{4y}+\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{4y}−\mathrm{11}}{\mathrm{y}−\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4y}^{\mathrm{2}} −\mathrm{12y}+\mathrm{y}−\mathrm{3}=\mathrm{12y}−\mathrm{33} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4y}^{\mathrm{2}} −\mathrm{23y}+\mathrm{30}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\frac{\mathrm{23}}{\mathrm{4}}\boldsymbol{\mathrm{y}}+\frac{\mathrm{15}}{\mathrm{2}}=\mathrm{0} \\ $$$$\bigtriangleup=\frac{\mathrm{23}^{\mathrm{2}} }{\mathrm{16}}−\mathrm{30}=\frac{\mathrm{23}^{\mathrm{2}} −\mathrm{480}}{\mathrm{16}}=\left(\frac{\mathrm{7}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{y}}=\frac{\frac{\mathrm{23}}{\mathrm{4}}\pm\frac{\mathrm{7}}{\mathrm{4}}}{\mathrm{2}}=\:\mathrm{y}=\left(\frac{\mathrm{15}}{\mathrm{4}},\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{a}−\:\:\:\mathrm{y}=\mathrm{2}\:\:\:\:\:\:\:\mathrm{x}=\mathrm{3}\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{x}};\mathrm{y}\right)=\:\:\left(\mathrm{3},\mathrm{2}\right) \\ $$$$\mathrm{b}−\:\:\:\mathrm{y}=\frac{\mathrm{15}}{\mathrm{4}}\:\:\:\:\mathrm{x}=\frac{\mathrm{16}}{\mathrm{3}}\:\:\:\:\:\:\left(\mathrm{x};\mathrm{y}\right)=\:\:\left(\frac{\mathrm{16}}{\mathrm{3}}\:;\:\:\frac{\mathrm{15}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\boldsymbol{\mathrm{xy}}=\begin{cases}{\mathrm{6}\:\:\:\:\mathrm{pour}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x},\mathrm{y}\right)=\left(\mathrm{3},\mathrm{2}\right)}\\{\mathrm{20}\:\:\:\mathrm{pour}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{\mathrm{16}}{\mathrm{3}},\frac{\mathrm{15}}{\mathrm{4}}\right)}\end{cases} \\ $$$$ \\ $$$$ \\ $$

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