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Question Number 122002 by liberty last updated on 13/Nov/20

$$\mathrm{Solve}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}$$$$\{\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+\frac{2\mathrm{xy}}{\mathrm{x}+\mathrm{y}}=1\\ \sqrt{\mathrm{x}+\mathrm{y}}={\mathrm{x}}^2-\mathrm{y}\end{array}\mathrm{in}\mathrm{real}\mathrm{numbers}\mathrm{x},\mathrm{y}.$$

Answered by MJS_new last updated on 13/Nov/20

$$\sqrt{x+y}=x^2-y$$$$\mathrm{squaring}\mathrm{\&}\mathrm{transforming}$$$$y^2-(2x^2+1)y+(x^4-x)=0$$$$\Rightarrow \left(\mathrm{a}\right)y=x^2-x\vee \left(\mathrm{b}\right)y=x^2+x+1$$$$[\mathrm{we}\mathrm{have}\mathrm{some}\mathrm{limitations}\mathrm{but}{\mathrm{let}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{go}\mathrm{on}$$$$\mathrm{and}\mathrm{see}\mathrm{what}\mathrm{will}\mathrm{happen}]$$$$\mathrm{inserting}\mathrm{in}\left(1\right)\mathrm{we}\mathrm{get}$$$$\left(\mathrm{a}\right)x^6-2x^5+2x^4+2x^3-3x^2=0$$$$\left(\mathrm{b}\right)x^6+4x^5+9x^4+14x^3+10x^2+4x=0$$$$\Rightarrow$$$$\left(\mathrm{a}\right)x^2(x-1)(x+1)(x^2-2x+3)=0$$$$\left(\mathrm{b}\right)x(x+2)(x^4+2x^3+5x^2+4x+2)=0$$$$\Rightarrow$$$$\left(\mathrm{a}\right)x=-1\vee x=0\vee x=1$$$$\Rightarrow y=2{}^{}{}^{}\vee y=0\vee y=0$$$$\left(\mathrm{b}\right)x=-2\vee x=0$$$$\Rightarrow y=3\vee y=1$$$$\mathrm{testing}\mathrm{all}\mathrm{pairs}\mathrm{we}\mathrm{get}$$$$(x\mid y)=(1\mid 0)\vee (-2\mid 3)$$

Commented by liberty last updated on 13/Nov/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by liberty last updated on 13/Nov/20

$$\mathrm{my}\mathrm{way}$$$$\mathrm{multiply}\mathrm{the}\mathrm{first}\mathrm{eq}\mathrm{by}\mathrm{x}+\mathrm{y}\mathrm{give}$$$$({\mathrm{x}}^2+{\mathrm{y}}^2)(\mathrm{x}+\mathrm{y})+2\mathrm{xy}=\mathrm{x}+\mathrm{y}$$$$\mathrm{adding}{\mathrm{x}}^2+{\mathrm{y}}^2\mathrm{to}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{this}\mathrm{eq},\mathrm{we}\mathrm{are}$$$$\mathrm{driven}\mathrm{by}\mathrm{standard}\mathrm{manipulation}\mathrm{to}$$$$\mathrm{nice}\mathrm{factorization}:$$$$({\mathrm{x}}^2+{\mathrm{y}}^2)(\mathrm{x}+\mathrm{y})+{(\mathrm{x}+\mathrm{y})}^2=({\mathrm{x}}^2+{\mathrm{y}}^2)+(\mathrm{x}+\mathrm{y})$$$$({\mathrm{x}}^2+{\mathrm{y}}^2)(\mathrm{x}+\mathrm{y}-1)+(\mathrm{x}+\mathrm{y})(\mathrm{x}+\mathrm{y}-1)=0$$$$({\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x}+\mathrm{y})(\mathrm{x}+\mathrm{y}-1)=0$$$$\mathrm{The}\mathrm{first}\mathrm{factor}\mathrm{cannot}\mathrm{be}\mathrm{zero}\mathrm{because}\mathrm{x}+\mathrm{y}>0$$$$\mathrm{so}\mathrm{x}+\mathrm{y}-1\mathrm{must}\mathrm{be}\mathrm{zero}.\mathrm{Inserting}$$$$\mathrm{y}=1-\mathrm{x}\mathrm{into}\mathrm{second}\mathrm{eq}\mathrm{of}\mathrm{the}\mathrm{system}\mathrm{we}$$$$\mathrm{find}\mathrm{the}\mathrm{two}\mathrm{solutions}(\mathrm{x},\mathrm{y}):(1,0)\mathrm{and}(-2,3).▴$$

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