{ ((((15)/(x^2 +xy)) + (3/(y^2 +xy)) = 5,5)),((((12)/(x^2 +xy))


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Question Number 146535 by mathdanisur last updated on 13/Jul/21
${ ((((15)/(x^2 +xy)) + (3/(y^2 +xy)) = 5,5)),((((12)/(x^2 +xy)) - (6/(y^2 +xy)) = 3)) :} ⇒ ∣x+y∣=?$
${\begin{cases} \frac{15}{x^{2} + x y} + \frac{3}{y^{2} + x y} = 5, 5 \\ \frac{12}{x^{2} + x y} - \frac{6}{y^{2} + x y} = 3 \end{cases} \Rightarrow ∣ x + y ∣= ?$
	Answered by liberty last updated on 14/Jul/21
	${ ((((30)/(x^2 +xy))+(6/(y^2 +xy))=11)),((((12)/(x^2 +xy))−(6/(y^2 +xy))=3)) :} (1)+(2)→((42^3 )/(x^2 +xy))=14^1 ⇒x^2 +xy=3 ⇒y^2 +xy=6 ⇒(1)+(2) (x+y)^2 =9 { ((x=3−y or)),((x=−3−y)) :}⇒∣x+y∣=(√((x+y)^2 ))=3$
	${\begin{cases} \frac{30}{x^{2} + x y} + \frac{6}{y^{2} + x y} = 11 \\ \frac{12}{x^{2} + x y} - \frac{6}{y^{2} + x y} = 3 \end{cases}$ $(1) + (2) \to \frac{42^{3}}{x^{2} + x y} = 14^{1}$ $\Rightarrow x^{2} + x y = 3$ $\Rightarrow y^{2} + x y = 6$ $\Rightarrow (1) + (2) {(x + y)}^{2} = 9$ ${\begin{cases} x = 3 - y o r \\ x = - 3 - y \end{cases} \Rightarrow∣ x + y ∣= \sqrt{{(x + y)}^{2}} = 3$
	Commented by mathdanisur last updated on 14/Jul/21

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	Answered by Olaf_Thorendsen last updated on 14/Jul/21
	${ ((((15)/(x^2 +xy))+(3/(y^2 +xy)) = 5,5 (×2))),((((12)/(x^2 +xy))−(6/(y^2 +xy)) = 3)) :} { ((((30)/(x^2 +xy))+(6/(y^2 +xy)) = 11 (1))),((((12)/(x^2 +xy))−(6/(y^2 +xy)) = 3 (2))) :} (1)+(2) : ((42)/(x^2 +xy)) = 14 (1/(x^2 +xy)) = (1/3) x^2 +xy = 3 (3) (1) : (6/(y^2 +xy)) = 11−((30)/(x^2 +xy)) (1) : (6/(y^2 +xy)) = 11−30×(1/3) = 1 y^2 +xy = 6 (4) (3)+(4) : x^2 +xy+y^2 +xy = 3+6 (x+y)^2 = 9 ∣x+y∣ = 3$
	${\begin{cases} \frac{15}{x^{2} + x y} + \frac{3}{y^{2} + x y} = 5, 5 (\times 2) \\ \frac{12}{x^{2} + x y} - \frac{6}{y^{2} + x y} = 3 \end{cases}$ ${\begin{cases} \frac{30}{x^{2} + x y} + \frac{6}{y^{2} + x y} = 11 (1) \\ \frac{12}{x^{2} + x y} - \frac{6}{y^{2} + x y} = 3 (2) \end{cases}$ $(1) + (2) : \frac{42}{x^{2} + x y} = 14$ $\frac{1}{x^{2} + x y} = \frac{1}{3}$ $x^{2} + x y = 3 (3)$ $(1) : \frac{6}{y^{2} + x y} = 11 - \frac{30}{x^{2} + x y}$ $(1) : \frac{6}{y^{2} + x y} = 11 - 30 \times \frac{1}{3} = 1$ $y^{2} + x y = 6 (4)$ $(3) + (4) : x^{2} + x y + y^{2} + x y = 3 + 6$ ${(x + y)}^{2} = 9$ $∣ x + y ∣ = 3$
	Commented by mathdanisur last updated on 14/Jul/21

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