Question-76169

Question Number 76169 by Master last updated on 24/Dec/19

Answered by mr W last updated on 24/Dec/19

yx+27y=154 xy+27x=30 2xy+27(x+y)=184 ...(i) y−x=((124)/(27)) y^2 +x^2 −2xy=(((124)/(27)))^2 (x+y)^2 −4xy=(((124)/(27)))^2 ...(ii) 2(i)+(ii): (x+y)^2 +54(x+y)−2×184−(((124)/(27)))^2 =0 ⇒x+y=−27±(√(27^2 +2×184+(((124)/(27)))^2 )) ⇒x+y=−27±((√(815089))/(27))=6.437 or −60.437

$${yx}+\mathrm{27}{y}=\mathrm{154} \\ $$$${xy}+\mathrm{27}{x}=\mathrm{30} \\ $$$$\mathrm{2}{xy}+\mathrm{27}\left({x}+{y}\right)=\mathrm{184}\:\:\:…\left({i}\right) \\ $$$${y}−{x}=\frac{\mathrm{124}}{\mathrm{27}} \\ $$$${y}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{xy}=\left(\frac{\mathrm{124}}{\mathrm{27}}\right)^{\mathrm{2}} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}=\left(\frac{\mathrm{124}}{\mathrm{27}}\right)^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\mathrm{2}\left({i}\right)+\left({ii}\right): \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} +\mathrm{54}\left({x}+{y}\right)−\mathrm{2}×\mathrm{184}−\left(\frac{\mathrm{124}}{\mathrm{27}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{x}+{y}=−\mathrm{27}\pm\sqrt{\mathrm{27}^{\mathrm{2}} +\mathrm{2}×\mathrm{184}+\left(\frac{\mathrm{124}}{\mathrm{27}}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}+{y}=−\mathrm{27}\pm\frac{\sqrt{\mathrm{815089}}}{\mathrm{27}}=\mathrm{6}.\mathrm{437}\:{or}\:−\mathrm{60}.\mathrm{437} \\ $$

Answered by behi83417@gmail.com last updated on 24/Dec/19

y=((154)/(x+27))⇒x.(((154)/(x+27))+27)=30 ⇒x(154+27x+27^2 )=30x+30×27 ⇒27x^2 +853x−810=0 ⇒x=((−853±(√(853^2 +4×27×810)))/(2×27)) ⇒ { ((x=0.923⇒y=5.52⇒(x+y=6.12))),((x=−32.5⇒y=−27.90⇒(x+y=−60.4))) :}

$$\mathrm{y}=\frac{\mathrm{154}}{\mathrm{x}+\mathrm{27}}\Rightarrow\mathrm{x}.\left(\frac{\mathrm{154}}{\mathrm{x}+\mathrm{27}}+\mathrm{27}\right)=\mathrm{30} \\ $$$$\Rightarrow\mathrm{x}\left(\mathrm{154}+\mathrm{27x}+\mathrm{27}^{\mathrm{2}} \right)=\mathrm{30x}+\mathrm{30}×\mathrm{27} \\ $$$$\Rightarrow\mathrm{27x}^{\mathrm{2}} +\mathrm{853x}−\mathrm{810}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\frac{−\mathrm{853}\pm\sqrt{\mathrm{853}^{\mathrm{2}} +\mathrm{4}×\mathrm{27}×\mathrm{810}}}{\mathrm{2}×\mathrm{27}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}=\mathrm{0}.\mathrm{923}\Rightarrow\mathrm{y}=\mathrm{5}.\mathrm{52}\Rightarrow\left(\mathrm{x}+\mathrm{y}=\mathrm{6}.\mathrm{12}\right)}\\{\mathrm{x}=−\mathrm{32}.\mathrm{5}\Rightarrow\mathrm{y}=−\mathrm{27}.\mathrm{90}\Rightarrow\left(\mathrm{x}+\mathrm{y}=−\mathrm{60}.\mathrm{4}\right)}\end{cases} \\ $$

Commented by Master last updated on 24/Dec/19

$$\mathrm{thanks} \\ $$

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