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Question Number 197475 by mnjuly1970 last updated on 19/Sep/23

x , y ∈ R , x^( 2) + xy = 12 y^( 2) + 2xy = 7 −−−−− x , y =?

$$ \\ $$$$\:\:\:\:\:\:{x}\:,\:{y}\:\in\:\mathbb{R}\:\:\:, \\ $$$$\:\:\:\:{x}^{\:\mathrm{2}} \:+\:{xy}\:=\:\mathrm{12} \\ $$$$\:\:\:\:\:{y}^{\:\mathrm{2}} \:+\:\mathrm{2}{xy}\:=\:\mathrm{7}\: \\ $$$$\:\:\:\:−−−−−\:\:\:{x}\:,\:{y}\:=? \\ $$

Answered by Sutrisno last updated on 19/Sep/23

x^( 2) + xy = 12→y=((12)/x)−x (((12)/x)−x)^2 + 2x(((12)/x)−x)= 7 ((144)/x^2 )−24+x^2 +24−2x^2 =7 x^4 −7x^2 −144=0 (x+3)(x−3)(x^2 +16)=0 x=−3 and y=−1 x=3 and y =1

$$\:\:\:\:{x}^{\:\mathrm{2}} \:+\:{xy}\:=\:\mathrm{12}\rightarrow{y}=\frac{\mathrm{12}}{{x}}−{x} \\ $$$$\:\:\:\:\left(\frac{\mathrm{12}}{{x}}−{x}\right)^{\mathrm{2}} \:+\:\mathrm{2}{x}\left(\frac{\mathrm{12}}{{x}}−{x}\right)=\:\mathrm{7}\: \\ $$$$\frac{\mathrm{144}}{{x}^{\mathrm{2}} }−\mathrm{24}+{x}^{\mathrm{2}} +\mathrm{24}−\mathrm{2}{x}^{\mathrm{2}} =\mathrm{7} \\ $$$${x}^{\mathrm{4}} −\mathrm{7}{x}^{\mathrm{2}} −\mathrm{144}=\mathrm{0} \\ $$$$\left({x}+\mathrm{3}\right)\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{16}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{3}\:{and}\:{y}=−\mathrm{1} \\ $$$${x}=\mathrm{3}\:{and}\:{y}\:=\mathrm{1} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 19/Sep/23

thx alot sir

$${thx}\:{alot}\:{sir} \\ $$

Answered by Frix last updated on 19/Sep/23

y=px x^2 (1+p)=12 x^2 (p^2 +2p)=7 7(p+1)=12(p^2 +2p) p^2 +((17)/(12))p−(7/(12))=0 p=−(7/4)∨p=(1/3) x^2 =((12)/(p+1)) x^2 =−16∨x^2 =9 x∈R ⇒ x=±3 ⇒ x=−3∧y=−1∨x=3∧y=1

$${y}={px} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}+{p}\right)=\mathrm{12} \\ $$$${x}^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{2}{p}\right)=\mathrm{7} \\ $$$$\mathrm{7}\left({p}+\mathrm{1}\right)=\mathrm{12}\left({p}^{\mathrm{2}} +\mathrm{2}{p}\right) \\ $$$${p}^{\mathrm{2}} +\frac{\mathrm{17}}{\mathrm{12}}{p}−\frac{\mathrm{7}}{\mathrm{12}}=\mathrm{0} \\ $$$${p}=−\frac{\mathrm{7}}{\mathrm{4}}\vee{p}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{12}}{{p}+\mathrm{1}} \\ $$$${x}^{\mathrm{2}} =−\mathrm{16}\vee{x}^{\mathrm{2}} =\mathrm{9} \\ $$$${x}\in\mathbb{R}\:\Rightarrow\:{x}=\pm\mathrm{3}\:\Rightarrow \\ $$$${x}=−\mathrm{3}\wedge{y}=−\mathrm{1}\vee{x}=\mathrm{3}\wedge{y}=\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 21/Sep/23

7x^2 +7xy=84 ∧ 12y^2 +24xy=84 7x^2 −17xy−12y^2 =0 (x−3y)(7x+4y)=0 x=3y ∨ x=−(4/7)y y^( 2) + 2xy = 7 ⇒ (a) y^2 +2(3y)y=7⇒7y^2 =7⇒y=±1⇒x=±3 { ((x=3,y=1)),((x=−3,y=−1)) :} (b) y^2 +2(((−4y)/7))y=7⇒7y^2 −8y^2 =49⇒y=±7i∉R

$$\mathrm{7}{x}^{\mathrm{2}} +\mathrm{7}{xy}=\mathrm{84}\:\wedge\:\mathrm{12}{y}^{\mathrm{2}} +\mathrm{24}{xy}=\mathrm{84} \\ $$$$\mathrm{7}{x}^{\mathrm{2}} −\mathrm{17}{xy}−\mathrm{12}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}−\mathrm{3}{y}\right)\left(\mathrm{7}{x}+\mathrm{4}{y}\right)=\mathrm{0} \\ $$$${x}=\mathrm{3}{y}\:\vee\:{x}=−\frac{\mathrm{4}}{\mathrm{7}}{y} \\ $$$$\:{y}^{\:\mathrm{2}} \:+\:\mathrm{2}{xy}\:=\:\mathrm{7} \\ $$$$\Rightarrow \\ $$$$\left({a}\right)\:\:\:{y}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}{y}\right){y}=\mathrm{7}\Rightarrow\mathrm{7}{y}^{\mathrm{2}} =\mathrm{7}\Rightarrow{y}=\pm\mathrm{1}\Rightarrow{x}=\pm\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\begin{cases}{{x}=\mathrm{3},{y}=\mathrm{1}}\\{{x}=−\mathrm{3},{y}=−\mathrm{1}}\end{cases} \\ $$$$\left({b}\right)\:\:{y}^{\mathrm{2}} +\mathrm{2}\left(\frac{−\mathrm{4}{y}}{\mathrm{7}}\right){y}=\mathrm{7}\Rightarrow\mathrm{7}{y}^{\mathrm{2}} −\mathrm{8}{y}^{\mathrm{2}} =\mathrm{49}\Rightarrow{y}=\pm\mathrm{7}{i}\notin\mathbb{R} \\ $$

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