{ ((x^2 + xy = 11)),((y^2 + xy = 24)) :} ⇒ ∣x+y∣=?


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Question Number 164496 by HongKing last updated on 18/Jan/22
${ ((x^2 + xy = 11)),((y^2 + xy = 24)) :} ⇒ ∣x+y∣=?$
${\begin{cases} x^{2} + xy = 11 \\ y^{2} + xy = 24 \end{cases} \Rightarrow ∣ x + y ∣= ?$
	Answered by Rasheed.Sindhi last updated on 18/Jan/22
	${ ((x^2 + xy = 11)),((y^2 + xy = 14)) :} ⇒ ∣x+y∣=? { ((x(x+y)=11)),((y(x + y = 14)) :}⇒(x/y)=((11)/(14))⇒x=((11y)/(14)) ((11y)/(14))(((11y)/(14))+y)=11 11y(25y)=11×14^2 y^2 =((14^2 )/(25)) y=±((14)/5) x=((11)/(14))(±((14)/5) )=±((11)/( 5)) ∣x+y∣=∣±((11)/( 5))±((14)/5) ∣=((25)/5)=5$
	${\begin{cases} x^{2} + xy = 11 \\ y^{2} + xy = 14 \end{cases} \Rightarrow ∣ x + y ∣= ?$ ${\begin{cases} x (x + y) = 11 \\ y (x + y = 14 \end{cases} \Rightarrow \frac{x}{y} = \frac{11}{14} \Rightarrow x = \frac{11 y}{14}$ $\frac{11 y}{14} (\frac{11 y}{14} + y) = 11$ $11 y (25 y) = 11 \times 14^{2}$ $y^{2} = \frac{14^{2}}{25}$ $y = \pm \frac{14}{5}$ $x = \frac{11}{14} (\pm \frac{14}{5}) = \pm \frac{11}{5}$ $∣ x + y ∣=∣ \pm \frac{11}{5} \pm \frac{14}{5} ∣= \frac{25}{5} = 5$
	Commented by HongKing last updated on 18/Jan/22

	$sorry dear Sir, y^{2} + xy = 14$
	Answered by nurtani last updated on 18/Jan/22

	$x^{2} + y^{2} + 2 x y = 25$ ${(x + y)}^{2} = 25 \Rightarrow x + y = \pm 5 \Rightarrow∣ x + y ∣= 5$
	Commented by Rasheed.Sindhi last updated on 18/Jan/22

	$C o m p a c t! . . . \cap i \subset\in!$
	Commented by nurtani last updated on 18/Jan/22

	$t h a n k s s i r$
	Commented by HongKing last updated on 18/Jan/22

	$thank you dear Sir cool$
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