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Question Number 171837 by Mikenice last updated on 21/Jun/22

solve:  2x+3y=2  x+y=5xy

$${solve}: \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{2} \\ $$$${x}+{y}=\mathrm{5}{xy} \\ $$

Answered by Rasheed.Sindhi last updated on 21/Jun/22

2x+3y=2⇒y=((2−2x)/3)  x+y=5xy⇒x+((2−2x)/3)=5x(((2−2x)/3))  ⇒3x+2−2x=10x−10x^2   ⇒10x^2 −9x+2=0       10x^2 −5x−4x+2=0       5x(2x−1)−2(2x−1)=0       (2x−1)(5x−2)=0       x=(1/2) ∣ x=(2/5)  •y=((2−2x)/3)=((2−2((1/2)))/3)=(1/3)  •y=((2−2((2/5)))/3)=(6/(15))=(2/5)  (x,y)={((1/2),(1/3)),((2/5),(2/5))}

$$\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{2}\Rightarrow{y}=\frac{\mathrm{2}−\mathrm{2}{x}}{\mathrm{3}} \\ $$$${x}+{y}=\mathrm{5}{xy}\Rightarrow{x}+\frac{\mathrm{2}−\mathrm{2}{x}}{\mathrm{3}}=\mathrm{5}{x}\left(\frac{\mathrm{2}−\mathrm{2}{x}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{3}{x}+\mathrm{2}−\mathrm{2}{x}=\mathrm{10}{x}−\mathrm{10}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{10}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{10}{x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{4}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{5}{x}\left(\mathrm{2}{x}−\mathrm{1}\right)−\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\left(\mathrm{2}{x}−\mathrm{1}\right)\left(\mathrm{5}{x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\mid\:{x}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\bullet{y}=\frac{\mathrm{2}−\mathrm{2}{x}}{\mathrm{3}}=\frac{\mathrm{2}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\bullet{y}=\frac{\mathrm{2}−\mathrm{2}\left(\frac{\mathrm{2}}{\mathrm{5}}\right)}{\mathrm{3}}=\frac{\mathrm{6}}{\mathrm{15}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\left({x},{y}\right)=\left\{\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{3}}\right),\left(\frac{\mathrm{2}}{\mathrm{5}},\frac{\mathrm{2}}{\mathrm{5}}\right)\right\} \\ $$

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