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Question Number 10755 by Joel576 last updated on 24/Feb/17

x^2 + y^2 + xy + 2(x − y) = 9 How many solution that fulfilled the equation above ? x, y ∈ N

$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{xy}\:+\:\mathrm{2}\left({x}\:−\:{y}\right)\:=\:\mathrm{9} \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{fulfilled}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{above}\:? \\ $$$${x},\:{y}\:\in\:\mathbb{N} \\ $$

Answered by mrW1 last updated on 24/Feb/17

x^2 + y^2 − 2xy + 2(x − y)+3xy = 9 (x−y)^2 +2(x−y)+3xy−9=0 u=x−y v=xy ⇒u^2 +2u+3v−9=0 u=((−2±(√(4−4(3v−9))))/2)=−1±(√(10−3v)) 10−3v=k^2 v=((10−k^2 )/3)>0 k=1: v=((10−1^2 )/3)=3 ok k=2: v=((10−2^2 )/3)=2 ok k=3: v=((10−3^2 )/3)=(1/3) not ok with v=3 u=−1±1= { (0),((−2)) :} { ((x−y=0),( )),((xy=3),( ⇒no integer solution!)) :} { ((x−y=−2),(⇒x=1, y=3)),((xy=3),(⇒ right solution!)) :} with v=2 u=−1±2= { (1),((−3)) :} { ((x−y=1),(⇒x=2, y=1)),((xy=2),(⇒ right solution!)) :} { ((x−y=−3),),((xy=2),(⇒ no integer solution!)) :} all solutions: ((x),(y) )= ((1),(3) ) or ((2),(1) )

$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:+\:\mathrm{2}\left({x}\:−\:{y}\right)+\mathrm{3}{xy}\:=\:\mathrm{9} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +\mathrm{2}\left({x}−{y}\right)+\mathrm{3}{xy}−\mathrm{9}=\mathrm{0} \\ $$$${u}={x}−{y} \\ $$$${v}={xy} \\ $$$$\Rightarrow{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{3}{v}−\mathrm{9}=\mathrm{0} \\ $$$${u}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{4}\left(\mathrm{3}{v}−\mathrm{9}\right)}}{\mathrm{2}}=−\mathrm{1}\pm\sqrt{\mathrm{10}−\mathrm{3}{v}} \\ $$$$\mathrm{10}−\mathrm{3}{v}={k}^{\mathrm{2}} \\ $$$${v}=\frac{\mathrm{10}−{k}^{\mathrm{2}} }{\mathrm{3}}>\mathrm{0} \\ $$$${k}=\mathrm{1}:\:{v}=\frac{\mathrm{10}−\mathrm{1}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{3}\:\:\:{ok} \\ $$$${k}=\mathrm{2}:\:{v}=\frac{\mathrm{10}−\mathrm{2}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{2}\:\:\:{ok} \\ $$$${k}=\mathrm{3}:\:{v}=\frac{\mathrm{10}−\mathrm{3}^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}\:\:{not}\:{ok} \\ $$$$ \\ $$$${with}\:{v}=\mathrm{3} \\ $$$${u}=−\mathrm{1}\pm\mathrm{1}=\begin{cases}{\mathrm{0}}\\{−\mathrm{2}}\end{cases} \\ $$$$\begin{cases}{{x}−{y}=\mathrm{0}}&{\:\:\:}\\{{xy}=\mathrm{3}}&{\:\:\:\Rightarrow{no}\:{integer}\:{solution}!}\end{cases} \\ $$$$ \\ $$$$\begin{cases}{{x}−{y}=−\mathrm{2}}&{\Rightarrow{x}=\mathrm{1},\:{y}=\mathrm{3}}\\{{xy}=\mathrm{3}}&{\Rightarrow\:{right}\:{solution}!}\end{cases} \\ $$$$ \\ $$$${with}\:{v}=\mathrm{2} \\ $$$${u}=−\mathrm{1}\pm\mathrm{2}=\begin{cases}{\mathrm{1}}\\{−\mathrm{3}}\end{cases} \\ $$$$\begin{cases}{{x}−{y}=\mathrm{1}}&{\Rightarrow{x}=\mathrm{2},\:{y}=\mathrm{1}}\\{{xy}=\mathrm{2}}&{\Rightarrow\:{right}\:{solution}!}\end{cases} \\ $$$$ \\ $$$$\begin{cases}{{x}−{y}=−\mathrm{3}}&{}\\{{xy}=\mathrm{2}}&{\Rightarrow\:{no}\:{integer}\:{solution}!}\end{cases} \\ $$$$ \\ $$$${all}\:{solutions}: \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:{or}\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix} \\ $$

Commented by Joel576 last updated on 24/Feb/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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