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

If ((4x^2 −4xy+3y^2 )/(2y^2 +2xy−5x^2 ))=1 if the ((x+y)/(x−y)) what is th value?

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

Answered by ajfour last updated on 26/Mar/17

−2

$$−\mathrm{2} \\ $$

Commented by ajfour last updated on 26/Mar/17

((4x^2 −4xy+3y^2 )/(2y^2 +2xy−5x^2 )) ; dividing N^r and D^r by y^2 yields =((4((x/y))^2 −4((x/y))+3)/(2+2((x/y))−5((x/y))^2 )) =1 let m=(x/y) ((4m^2 −4m+3)/(2+2m−5m^2 )) =1 ⇒ 4m^2 −4m+3=2+2m−5m^2 9m^2 −6m+1=0 (3m−1)^2 =1 ⇒ m=(x/y)=(1/3) Now, ((x+y)/(x−y)) = ((1+3)/(1−3)) = −2

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

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

4x^2 −4xy+3y^2 =2y^2 +2xy−5x^2 9x^2 −6xy=−y^2 9x^2 −6xy+y^2 =0 (3x−y)=0 3x=y ((x+y)/(x−y))=((x+3x)/(x−3x))=−2

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

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

THENKS

$$\mathbb{THENKS} \\ $$

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

THENKS

$$\mathbb{THENKS} \\ $$

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