Question Number 155562 by 0731619 last updated on 02/Oct/21
Commented by tabata last updated on 02/Oct/21
$$\boldsymbol{{Solve}}\:::\:\sqrt{\boldsymbol{{x}}}\:+\:\boldsymbol{{y}}\:=\:\mathrm{5}\:\:,\:\sqrt{\boldsymbol{{y}}}\:+\:\boldsymbol{{x}}\:=\:\mathrm{3}\: \\ $$$$ \\ $$$$\boldsymbol{{Solution}}:: \\ $$$$ \\ $$$$\boldsymbol{{let}}:\:\boldsymbol{{a}}^{\mathrm{2}} \:=\:\boldsymbol{{x}}\:\:\:\:,\:\:\:\boldsymbol{{b}}^{\mathrm{2}} \:=\:\boldsymbol{{y}}\: \\ $$$$ \\ $$$$\boldsymbol{{a}}+\boldsymbol{{b}}^{\mathrm{2}} =\:\mathrm{5}\:\rightarrow\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}\:=\:\mathrm{3}\:\rightarrow\left(\mathrm{2}\right)\:×\left(\boldsymbol{{b}}\:\right) \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} \:\boldsymbol{{b}}\:+\:\boldsymbol{{b}}^{\mathrm{2}} \:=\:\mathrm{3}\:\boldsymbol{{b}}\:\rightarrow\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{3}\right)\:−\:\left(\mathrm{1}\right)\:\Rightarrow\:\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}\:−\:\boldsymbol{{a}}\:=\:\mathrm{3}\:\boldsymbol{{b}}\:−\:\mathrm{5}\:\Rightarrow\boldsymbol{{b}}\:=\:\frac{\boldsymbol{{a}}\:−\:\mathrm{5}}{\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{3}} \\ $$$$\boldsymbol{{a}}^{\mathrm{2}} \:+\:\frac{\boldsymbol{{a}}−\mathrm{5}}{\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{3}}\:=\:\mathrm{3}\:\Rightarrow\:\boldsymbol{{a}}^{\mathrm{2}} \left(\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{3}\right)\:+\:\left(\boldsymbol{{a}}−\mathrm{5}\right)\:=\:\mathrm{3}\:\left(\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{3}\right) \\ $$$$\Rightarrow\boldsymbol{{a}}^{\mathrm{4}} −\mathrm{6}\boldsymbol{{a}}^{\mathrm{2}} +\:\boldsymbol{{a}}+\:\mathrm{4}\:=\:\mathrm{0}\:\Rightarrow\:\left(\boldsymbol{{a}}−\mathrm{1}\right)\:\left(\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{a}}−\mathrm{4}\right)=\:\mathrm{0}\:\Rightarrow\:\boldsymbol{{a}}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\:\boldsymbol{{b}}\:=\:\frac{\mathrm{1}−\mathrm{5}}{\mathrm{1}−\mathrm{3}}\:=\:\frac{−\mathrm{4}}{−\mathrm{2}}\:=\:\mathrm{2}\: \\ $$$$ \\ $$$$\because\:\boldsymbol{{x}}\:=\:\boldsymbol{{a}}^{\mathrm{2}} \:\Rightarrow\:\boldsymbol{{x}}\:=\:\left(\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{1} \\ $$$$\because\:\boldsymbol{{y}}\:=\:\boldsymbol{{b}}^{\mathrm{2}} \:\Rightarrow\:\boldsymbol{{y}}\:=\:\left(\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$ \\ $$$$\therefore\:\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)\:=\:\left(\mathrm{1},\mathrm{4}\right) \\ $$$$ \\ $$$$\langle\:\boldsymbol{{M}}\:.\:\boldsymbol{{T}}\:\:\rangle \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/Oct/21
$$\mathrm{nice} \\ $$
Answered by amin96 last updated on 02/Oct/21
$$\begin{cases}{\sqrt{{x}}+{y}=\mathrm{5}}\\{\sqrt{{y}}+{x}=\mathrm{3}}\end{cases}\:\:\Rightarrow\:\:\begin{cases}{\sqrt{{y}}={b}}\\{\sqrt{{x}}={a}}\end{cases}\Rightarrow\:\begin{cases}{{y}={b}^{\mathrm{2}} }\\{{x}={a}^{\mathrm{2}} }\end{cases}\:\:\Rightarrow \\ $$$$\begin{cases}{{b}^{\mathrm{2}} +{a}=\mathrm{5}}\\{{a}^{\mathrm{2}} +{b}=\mathrm{3}}\end{cases}\:\:\:\Rightarrow\:\left({b}−{a}\right)\left({b}+{a}\right)+\left({a}−{b}\right)=\mathrm{2} \\ $$$$\left({b}−{a}\right)\left({b}+{a}−\mathrm{1}\right)=\mathrm{2}\:\:\Rightarrow\:\begin{cases}{{b}−{a}=\mathrm{1}}\\{{b}+{a}−\mathrm{1}=\mathrm{2}}\end{cases}\:\Rightarrow\begin{cases}{{b}−{a}=\mathrm{1}}\\{{b}+{a}=\mathrm{3}}\end{cases} \\ $$$$\mathrm{2}{b}=\mathrm{4}\:\:{b}=\mathrm{2} \\ $$$${a}=\mathrm{1}\:\:\:\:\:\:\Rightarrow\:\begin{cases}{{y}={b}^{\mathrm{2}} }\\{{x}={a}^{\mathrm{2}} }\end{cases}\:\:\Rightarrow\:\:{y}=\mathrm{4}\:\:\:{x}=\mathrm{1}\:\:\: \\ $$
Commented by 0731619 last updated on 02/Oct/21
$${thanks} \\ $$