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Question-199732

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Question Number 199732 by Rupesh123 last updated on 08/Nov/23
Answered by Rasheed.Sindhi last updated on 08/Nov/23
y=((x^2 +2)/(x+3))=((x^2 −9+9+2)/(x+3)) =(((x−3)(x+3)+11)/(x+3)) =(x−3)+((11)/(x+3))∈Z⇒(x+3)∣11 x+3=±1,±11 x=−2,−4,8,−14 y=((x^2 +2)/(x+3)) y=(((−2)^2 +2)/((−2)+3)),(((−4)^2 +2)/((−4)+3)),(((8)^2 +2)/((8)+3)),(((−14)^2 +2)/((−14)+3)) =6,−18,6,−18 (x,y)=(−2,6),(−4,−18),(8,6),(−14,−18)
$${y}=\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}+\mathrm{3}}=\frac{{x}^{\mathrm{2}} −\mathrm{9}+\mathrm{9}+\mathrm{2}}{{x}+\mathrm{3}} \\ $$$$\:\:\:=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)+\mathrm{11}}{{x}+\mathrm{3}} \\ $$$$=\left({x}−\mathrm{3}\right)+\frac{\mathrm{11}}{{x}+\mathrm{3}}\in\mathbb{Z}\Rightarrow\left({x}+\mathrm{3}\right)\mid\mathrm{11} \\ $$$${x}+\mathrm{3}=\pm\mathrm{1},\pm\mathrm{11} \\ $$$${x}=−\mathrm{2},−\mathrm{4},\mathrm{8},−\mathrm{14} \\ $$$${y}=\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}+\mathrm{3}} \\ $$$${y}=\frac{\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}}{\left(−\mathrm{2}\right)+\mathrm{3}},\frac{\left(−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}}{\left(−\mathrm{4}\right)+\mathrm{3}},\frac{\left(\mathrm{8}\right)^{\mathrm{2}} +\mathrm{2}}{\left(\mathrm{8}\right)+\mathrm{3}},\frac{\left(−\mathrm{14}\right)^{\mathrm{2}} +\mathrm{2}}{\left(−\mathrm{14}\right)+\mathrm{3}} \\ $$$$\:=\mathrm{6},−\mathrm{18},\mathrm{6},−\mathrm{18} \\ $$$$\left({x},{y}\right)=\left(−\mathrm{2},\mathrm{6}\right),\left(−\mathrm{4},−\mathrm{18}\right),\left(\mathrm{8},\mathrm{6}\right),\left(−\mathrm{14},−\mathrm{18}\right) \\ $$
Commented by Rupesh123 last updated on 08/Nov/23
Perfect, sir!
Answered by Rasheed.Sindhi last updated on 08/Nov/23
AnOther Way y(x+3)−x^2 =2 y(x+3)−x^2 +9=2+9 y(x+3)−(x^2 −9)=11 y(x+3)−(x−3)(x+3)=11 (x+3)(y−x+3)=11=1×11=−1×−11 { ((x+3=1 ∧ y−x+3=11)),((x+3=11 ∧ y−x+3=1)),((x+3=−1 ∧ y−x+3=−11)),((x+3=−11 ∧ y−x+3=−1)) :} { ((x=−2 ∧ y=x+8=−2+8=6)),((x=8 ∧ y=x−2=8−2=6)),((x=−4 ∧ y=x−14=−4−14=−18)),((x=−14 ∧ y=x−4=−14−4=−18)) :} (x,y)=(−2,6),(8,6),(−4,−18),(−14,−18)
$$\boldsymbol{\mathrm{AnOther}}\:\boldsymbol{\mathrm{Way}} \\ $$$${y}\left({x}+\mathrm{3}\right)−{x}^{\mathrm{2}} =\mathrm{2} \\ $$$${y}\left({x}+\mathrm{3}\right)−{x}^{\mathrm{2}} +\mathrm{9}=\mathrm{2}+\mathrm{9} \\ $$$${y}\left({x}+\mathrm{3}\right)−\left({x}^{\mathrm{2}} −\mathrm{9}\right)=\mathrm{11} \\ $$$${y}\left({x}+\mathrm{3}\right)−\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)=\mathrm{11} \\ $$$$\left({x}+\mathrm{3}\right)\left({y}−{x}+\mathrm{3}\right)=\mathrm{11}=\mathrm{1}×\mathrm{11}=−\mathrm{1}×−\mathrm{11} \\ $$$$\begin{cases}{{x}+\mathrm{3}=\mathrm{1}\:\wedge\:{y}−{x}+\mathrm{3}=\mathrm{11}}\\{{x}+\mathrm{3}=\mathrm{11}\:\wedge\:{y}−{x}+\mathrm{3}=\mathrm{1}}\\{{x}+\mathrm{3}=−\mathrm{1}\:\wedge\:{y}−{x}+\mathrm{3}=−\mathrm{11}}\\{{x}+\mathrm{3}=−\mathrm{11}\:\wedge\:{y}−{x}+\mathrm{3}=−\mathrm{1}}\end{cases}\: \\ $$$$\begin{cases}{{x}=−\mathrm{2}\:\wedge\:{y}={x}+\mathrm{8}=−\mathrm{2}+\mathrm{8}=\mathrm{6}}\\{{x}=\mathrm{8}\:\wedge\:{y}={x}−\mathrm{2}=\mathrm{8}−\mathrm{2}=\mathrm{6}}\\{{x}=−\mathrm{4}\:\wedge\:{y}={x}−\mathrm{14}=−\mathrm{4}−\mathrm{14}=−\mathrm{18}}\\{{x}=−\mathrm{14}\:\wedge\:{y}={x}−\mathrm{4}=−\mathrm{14}−\mathrm{4}=−\mathrm{18}}\end{cases}\: \\ $$$$\left({x},{y}\right)=\left(−\mathrm{2},\mathrm{6}\right),\left(\mathrm{8},\mathrm{6}\right),\left(−\mathrm{4},−\mathrm{18}\right),\left(−\mathrm{14},−\mathrm{18}\right) \\ $$

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