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Question Number 107999 by Don08q last updated on 13/Aug/20

Commented by udaythool last updated on 13/Aug/20

$$\Rightarrow 4x^2+3xy-y^2=0$$$$\Rightarrow (x+y)(4x-y)=0$$$$\Rightarrow 4x=y;(\because x+y\ne 0)$$$$\Rightarrow 4x^2=xy$$$$\Rightarrow 5xy=5$$$$\Rightarrow xy=1$$

Commented by Don08q last updated on 13/Aug/20

$$ThanksSir$$

Commented by Rasheed.Sindhi last updated on 13/Aug/20

$$Whyx+y\ne 0?Suchconditionis$$$$notincludedinquestion.$$

Commented by Sarah85 last updated on 13/Aug/20

$$\mathrm{if}x+y=0\iff y=-x\iff x=-y$$$$4x^2-4x^2=5$$$$y^2-y^2=5$$$$\mathrm{both}\mathrm{are}\mathrm{impossible}$$

Commented by Rasheed.Sindhi last updated on 14/Aug/20

$$\mathcal{THANKS}madam!$$

Answered by Sarah85 last updated on 13/Aug/20

$$x=\alpha -\beta \wedge y=\alpha +\beta$$$$8{\alpha }^2-8\alpha \beta =5\Rightarrow \beta =\frac{8{\alpha }^2-5}{8\alpha }$$$$2{\alpha }^2+2\alpha \beta =5\Rightarrow \beta =\frac{5-2{\alpha }^2}{2\alpha }$$$$\frac{8{\alpha }^2-5}{8\alpha }=\frac{5-2{\alpha }^2}{2\alpha }$$$$8{\alpha }^2-5=20-8{\alpha }^2$$$$16{\alpha }^2=25$$$$\alpha =\pm \frac{5}{4}\Rightarrow \beta =\pm \frac{3}{4}$$$$\Rightarrow x=\pm \frac{1}{2}\wedge y=\pm 2$$$$$$$$\mathrm{or}y=\alpha x$$$$4x^2+4\alpha x^2=5$$$${\alpha }^2{x}^2+\alpha x^2=5$$$$x^2=\frac{5}{4(1+\alpha )}$$$$x^2=\frac{5}{\alpha (1+\alpha )}$$$$\Rightarrow \alpha =4\Rightarrow y=4x$$$$4x^2+16x^2=5\Rightarrow x^2=\frac{1}{4}\Rightarrow y^2=4$$$$$$$$\mathrm{or}\mathrm{just}\mathrm{test}\mathrm{if}y=\frac{1}{x}\mathrm{solves}\mathrm{both}\mathrm{equations}$$$$4x^2+4=5\Rightarrow x^2=\frac{1}{4}$$$$\frac{1}{x^2}+1=5\Rightarrow x^2=\frac{1}{4}$$$$\Rightarrow y^2=4$$

Answered by 1549442205PVT last updated on 14/Aug/20

$${4\mathrm{x}}^2+4\mathrm{xy}={\mathrm{y}}^2+\mathrm{xy}\Rightarrow {4\mathrm{x}}^2+3\mathrm{xy}-{\mathrm{y}}^2=0$$$$\mathrm{\Delta }={\left(3\mathrm{y}\right)}^2-4.4.(-{\mathrm{y}}^2)={25\mathrm{y}}^2$$$$\Rightarrow \mathrm{x}=\frac{-3\mathrm{y}\pm 5\mathrm{y}}{8}\in \{-2\mathrm{y};\frac{\mathrm{y}}{4}\}$$$$\mathrm{i})\mathrm{If}\mathrm{x}=-2\mathrm{y}\mathrm{then}\mathrm{replace}\mathrm{into}\mathrm{the}$$$$\mathrm{second}\mathrm{equation}\mathrm{we}\mathrm{get}$$$${\mathrm{y}}^2+(-2\mathrm{y})\mathrm{y}=5\iff -{\mathrm{y}}^2=5\left(\mathrm{rejected}\right)$$$$\mathrm{ii})\mathrm{If}\mathrm{x}=\frac{\mathrm{y}}{4}\iff \mathrm{y}=4\mathrm{x}\mathrm{then}\mathrm{we}\mathrm{get}$$$${\left(4\mathrm{x}\right)}^2+\mathrm{x}\left(4\mathrm{x}\right)=5\iff {20\mathrm{x}}^2=5\iff {\mathrm{x}}^2=\frac{1}{4}$$$$\iff \mathrm{x}=\pm \frac{1}{2}\Rightarrow \mathrm{y}=4\mathrm{x}=\pm 2$$$$\mathrm{Clearly}\mathrm{we}\mathrm{have}\mathrm{xy}=(\pm \frac{1}{2})\times (\pm 2)=1$$$$\mathrm{Hence}\mathrm{y}\mathrm{is}\mathrm{the}\mathrm{multiplicative}\mathrm{inverse}$$$$\mathrm{of}\mathrm{x}(\mathbf{q}.\mathbf{e}.\mathbf{d})$$

Commented by Don08q last updated on 14/Aug/20

$$ThanksSir$$

Answered by Rasheed.Sindhi last updated on 14/Aug/20

$$yisinverseofx\Rightarrow xy=1.\mathrm{So}\mathrm{now}$$$$\mathrm{the}\mathrm{question}\mathrm{in}\mathrm{other}\mathrm{words}\mathrm{is}:$$$$‘‘Toshowthefollowingsystem$$$$\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{i}\mathit{s}\mathit{t}\mathit{e}\mathit{n}\mathit{t}{}^{″}:$$$$\{\begin{array}{l}4x^2+4xy=5\\ y^2+xy=5\\ xy=1\end{array}$$$$I-Ethesystemhascommon$$$$solution.$$$$\Rightarrow \{\begin{array}{l}4x^2+4\left(1\right)=5\Rightarrow x=\pm \frac{1}{2}\\ y^2+\left(1\right)=5\Rightarrow y=\pm 2\end{array}$$$$Since(\frac{1}{2},2)\mathrm{\&}(-\frac{1}{2},-2)are$$$$solutionssatisfyingthesystem$$$$sothesystemis\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{i}\mathit{s}\mathit{t}\mathit{e}\mathit{n}\mathit{t}.$$$$(Evidentfromsolutionsthat$$$$yismultiplicativeinverseofx.)$$

Commented by Don08q last updated on 14/Aug/20

$$ThanksSir$$

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