Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 85104 by jagoll last updated on 19/Mar/20

$$\mathrm{Given}$$$$\{\begin{array}{l}{\mathrm{x}}^2-2\mathrm{xy}-3\mathrm{x}=-1\\ {4\mathrm{y}}^2-2\mathrm{xy}+6\mathrm{y}=-1\end{array}$$$$\mathrm{find}2\mathrm{y}-\mathrm{x}$$

Commented by john santu last updated on 19/Mar/20

$$\mathrm{let}2\mathrm{y}-\mathrm{x}=\mathrm{k}$$$$\Rightarrow {4\mathrm{y}}^2-4\mathrm{xy}+{\mathrm{x}}^2={\mathrm{k}}^2$$$$\left(1\right)+\left(2\right)\Rightarrow {4\mathrm{y}}^2-4\mathrm{xy}+{\mathrm{x}}^2-3\mathrm{x}+6\mathrm{y}=-2$$$$\Rightarrow {\mathrm{k}}^2-3(\mathrm{x}-2\mathrm{y})=-2$$$$\Rightarrow {\mathrm{k}}^2+3\mathrm{k}+2=0$$$$\Rightarrow (\mathrm{k}+2)(\mathrm{k}+1)=0\to \{\begin{array}{l}\mathrm{k}=-2\\ \mathrm{k}=-1\end{array}$$

Commented by jagoll last updated on 19/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{mr}\mathrm{john}\mathrm{\&}\mathrm{mjs}$$

Answered by MJS last updated on 19/Mar/20

$$\left(1\right)\Rightarrow y=\frac{1}{2x}-\frac{3}{2}+\frac{x}{2}$$$$\left(2\right)$$$$2-\frac{3}{x}+\frac{1}{x^2}=0$$$$x^2-\frac{3}{2}x+\frac{1}{2}=0$$$$\Rightarrow x=\frac{1}{2}\vee x=1\Rightarrow y=-\frac{1}{4}\vee y=-\frac{1}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com