Question-59787

Question Number 59787 by Khairun Nisa last updated on 14/May/19

Answered by MJS last updated on 14/May/19

$$\mathrm{both}\:\mathrm{should}\:\mathrm{be}\:\mathrm{square}\:\mathrm{numbers},\:\mathrm{trying}\:\mathrm{I}\:\mathrm{get} \\ $$$${x}=\mathrm{9} \\ $$$${y}=\mathrm{4} \\ $$$$ \\ $$$$\sqrt{{x}}+{y}=\mathrm{7}\:\Rightarrow\:{y}=\mathrm{7}−\sqrt{{x}} \\ $$$$\:\:\:\:\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{half}\:\mathrm{parabola} \\ $$$${x}+\sqrt{{y}}=\mathrm{11}\:\Rightarrow\:{x}=\mathrm{11}−\sqrt{{y}} \\ $$$$\:\:\:\:\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{half}\:\mathrm{parabola}\:\mathrm{too} \\ $$$$\Rightarrow\:\mathrm{there}'\mathrm{s}\:\mathrm{only}\:\mathrm{one}\:\mathrm{intersection} \\ $$

Answered by behi83417@gmail.com last updated on 15/May/19

$$\left(\sqrt{{x}}−\mathrm{3}\right)+\left({y}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{9}\right)+\left(\sqrt{{y}}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\left(\sqrt{{x}}−\mathrm{3}\right)+\left(\sqrt{{y}}−\mathrm{2}\right)\left(\sqrt{{y}}+\mathrm{2}\right)=\mathrm{0}}\\{\left(\sqrt{{x}}−\mathrm{3}\right)\left(\sqrt{{x}}+\mathrm{3}\right)+\left(\sqrt{{y}}−\mathrm{2}\right)=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\left(\sqrt{{x}}−\mathrm{3}\right)−\left(\sqrt{{x}}−\mathrm{3}\right)\left(\sqrt{{x}}+\mathrm{3}\right)\left(\sqrt{{y}}+\mathrm{2}\right)=\mathrm{0}}\\{\Rightarrow\left(\sqrt{{x}}−\mathrm{3}\right)\left[\mathrm{1}−\left(\sqrt{{x}}+\mathrm{3}\right)\left(\sqrt{{y}}+\mathrm{2}\right)\right]=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\sqrt{{x}}−\mathrm{3}=\mathrm{0}\Rightarrow\sqrt{{x}}=\mathrm{3}\Rightarrow\left[\boldsymbol{\mathrm{x}}=\mathrm{9}\right] \\ $$$${x}=\mathrm{9}\Rightarrow\sqrt{{y}}=\mathrm{11}−\mathrm{9}=\mathrm{2}\Rightarrow\left[\boldsymbol{\mathrm{y}}=\mathrm{4}\right].\:\:\:\blacksquare \\ $$

Commented by Kunal12588 last updated on 15/May/19

$${great}! \\ $$

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