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Question Number 14396 by tawa tawa last updated on 31/May/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17

Commented by tawa tawa last updated on 31/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by nume1114 last updated on 31/May/17

$$y^2+x^2=2xy+1$$$$\Rightarrow y^2-2xy+x^2-1=0$$$$\Rightarrow {(y-x)}^2-1^2=0$$$$\Rightarrow (y-x+1)(y-x-1)=0$$$$\Rightarrow y-x+1=0\vee y-x-1=0$$$$\Rightarrow x=y+1\vee x=y-1$$$$$$$$xy-4=x$$$$\Rightarrow x(y-1)-4=0...\left(\mathrm{a}\right)$$$$\left(\mathrm{i}\right)\mathrm{if}x=y+1\mathrm{then}$$$$\left(\mathrm{a}\right)\Rightarrow (y+1)(y-1)-4=0$$$$\Rightarrow y^2-1-4=0$$$$\Rightarrow y^2=5$$$$\Rightarrow y=\pm \sqrt{5},x=y+1=1\pm \sqrt{5}$$$$\left(\mathrm{ii}\right)\mathrm{if}x=y-1\mathrm{then}$$$$\left(\mathrm{a}\right)\Rightarrow {(y-1)}^2-4=0$$$$\Rightarrow y-1=\pm 2$$$$\Rightarrow y=1\pm 2=-1,3$$$$x=y-1=-2,2$$$$\left(\mathrm{i}\right),\left(\mathrm{ii}\right)$$$$\Rightarrow (x,y)=(-2,-1),(2,3),(1\pm \sqrt{5},\pm \sqrt{5})$$

Commented by tawa tawa last updated on 31/May/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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