15-x-2-xy-3-y-2-xy-5-5-12-x-2-xy-6-y-2-xy-3-x-y-

Question Number 146535 by mathdanisur last updated on 13/Jul/21

{\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

{\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

{\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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