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Question Number 11459 by @ANTARES_VY last updated on 26/Mar/17

$$\mathbf{If}$$$$\frac{4{\mathbf{x}}^2-4\mathbf{xy}+3{\mathbf{y}}^2}{2{\mathbf{y}}^2+2\mathbf{xy}-5{\mathbf{x}}^2}=1\mathbf{if}\mathbf{the}$$$$\frac{\mathbf{x}+\mathbf{y}}{\mathbf{x}-\mathbf{y}}\mathbf{what}\mathbf{is}\mathbf{th}\mathbf{value}?$$

Answered by ajfour last updated on 26/Mar/17

$$-2$$

Commented by ajfour last updated on 26/Mar/17

$$\frac{{4\mathrm{x}}^2-4\mathrm{xy}+{3\mathrm{y}}^2}{{2\mathrm{y}}^2+2\mathrm{xy}-{5\mathrm{x}}^2};\mathrm{dividing}{\mathrm{N}}^{\mathrm{r}}\mathrm{and}{\mathrm{D}}^{\mathrm{r}}\mathrm{by}{\mathrm{y}}^2\mathrm{yields}$$$$=\frac{4{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^2-4\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+3}{2+2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-5{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^2}=1$$$$\mathrm{let}\mathrm{m}=\frac{\mathrm{x}}{\mathrm{y}}$$$$\frac{{4\mathrm{m}}^2-4\mathrm{m}+3}{2+2\mathrm{m}-{5\mathrm{m}}^2}=1$$$$\Rightarrow$$$${4\mathrm{m}}^2-4\mathrm{m}+3=2+2\mathrm{m}-{5\mathrm{m}}^2$$$${9\mathrm{m}}^2-6\mathrm{m}+1=0$$$${(3\mathrm{m}-1)}^2=1\Rightarrow \mathrm{m}=\frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{3}$$$$\mathrm{Now},\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}=\frac{1+3}{1-3}=-2$$

Answered by ridwan balatif last updated on 26/Mar/17

$${4\mathrm{x}}^2-4\mathrm{xy}+{3\mathrm{y}}^2={2\mathrm{y}}^2+2\mathrm{xy}-{5\mathrm{x}}^2$$$${9\mathrm{x}}^2-6\mathrm{xy}=-{\mathrm{y}}^2$$$${9\mathrm{x}}^2-6\mathrm{xy}+{\mathrm{y}}^2=0$$$$(3\mathrm{x}-\mathrm{y})=0$$$$3\mathrm{x}=\mathrm{y}$$$$\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}-\mathrm{y}}=\frac{\mathrm{x}+3\mathrm{x}}{\mathrm{x}-3\mathrm{x}}=-2$$

Commented by @ANTARES_VY last updated on 26/Mar/17

$$\mathbb{THENKS}$$

Commented by @ANTARES_VY last updated on 26/Mar/17

$$\mathbb{THENKS}$$

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