solve the system { ((((4(√(1+x^2 )))/x) =((5(√(1+y^2 )))/y)=((6(√(1+z^2 )))/z))),((x+y+z=xyz.)) :} {: (),() }


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Question Number 49809 by Abdo msup. last updated on 10/Dec/18
$solve the system { ((((4(√(1+x^2 )))/x) =((5(√(1+y^2 )))/y)=((6(√(1+z^2 )))/z))),((x+y+z=xyz.)) :} {: (),() }$
$s o l v e t h e s y s t e m {\begin{cases} \frac{4 \sqrt{1 + x^{2}}}{x} = \frac{5 \sqrt{1 + y^{2}}}{y} = \frac{6 \sqrt{1 + z^{2}}}{z} \\ x + y + z = x y z . \end{cases}$ $\begin{matrix} \end{matrix}}$
Commented by MJS last updated on 12/Dec/18

$\frac{a \sqrt{1 + x^{2}}}{x} = \frac{b \sqrt{1 + y^{2}}}{y} = \frac{c \sqrt{1 + z^{2}}}{z}$ $x + y + z = x y z$ $leads to$ $x = \pm \frac{\sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}}{a^{2} - b^{2} - c^{2}}$ $y = \pm \frac{\sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}}{b^{2} - a^{2} - c^{2}}$ $z = \pm \frac{\sqrt{(a + b + c) (a + b - c) (a + c - b) (b + c - a)}}{c^{2} - a^{2} - b^{2}}$ $\Rightarrow a, b, c must form a triangle but not a$ $rectangular one . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

	$x = t a n a y = t a n b z = t a n c$ $\frac{4 \sqrt{1 + {t a n}^{2} a}}{t a n a} = \frac{5 \sqrt{1 + {t a n}^{2} b}}{t a n b} = \frac{6 \sqrt{1 + {t a n}^{2} c}}{t a n c}$ $\frac{4}{c o s a \times \frac{s i n a}{c o s a}} = \frac{5}{s i n b} = \frac{6}{s i n c} = \frac{1}{k}$ $s i n a = 4 k$ $s i n b = 5 k$ $s i n c = 6 k$ $x + y + z = x y z$ $x + y = x y z - z$ $x + y = z (x y - 1)$ $x + y = - z (1 - x y)$ $\frac{x + y}{1 - x y} = - z$ $\frac{t a n a + t a n b}{1 - t a n a t a n b} = - t a n c$ $t a n (a + b) = - t a n c$ $t a n (a + b) = t a n (π - c)$ $a + b + c = π$ $w a i t . . .$ $s i n (a + b) = s i n (π - c)$ $s i n a c o s b + c o s a s i n b = s i n c$ $4 k \times \sqrt{1 - 25 b^{2}} + \sqrt{1 - 16 k^{2}} \times 5 k = 6 k$ $4 \sqrt{1 - 25 k^{2}} + 5 \sqrt{1 - 16 k^{2}} = 6$ $16 (1 - 25 k^{2}) = 36 + 25 (1 - 16 k^{2}) - 60 \sqrt{1 - 16 k^{2}}$ $16 - 400 k^{2} = 36 + 25 - 400 k^{2} - 60 \sqrt{1 - 16 k^{2}}$ $- 45 = - 60 \sqrt{1 - 16 k^{2}}$ $9 = 16 (1 - 16 k^{2})$ $9 - 16 = - 16 \times 16 k^{2}$ $256 k^{2} = 7$ $k = \pm \frac{\sqrt{7}}{16}$ $s i n a = \frac{\pm 4 \sqrt{7}}{16} x = t a n a = \frac{\pm 4 \sqrt{7}}{\sqrt{256 - 112}} = \frac{\pm 4 \sqrt{7}}{12}$ $s i n b = \frac{\pm 5 \sqrt{7}}{16} y = t a n b = \frac{\pm 4 \sqrt{7}}{\sqrt{256 - 175}} = \frac{\pm 5 \sqrt{7}}{9}$ $s i n c = \frac{\pm 6 \sqrt{7}}{16} z = t a n c = \frac{\pm 4 \sqrt{7}}{\sqrt{256 - 252}} = \frac{\pm 4 \sqrt{7}}{2} \leftarrow s i l l y e r r o r$ $i n p l a c e o f 6 i p u t 4$ $s o c o r r e c t e d . .$ $z = \frac{\pm 6 \sqrt{7}}{2}$ $p l s c h e c k . . .$
	Answered by MJS last updated on 12/Dec/18

	$x > 0 \land y > 0 \land z > 0 \lor x < 0 \land y < 0 \land z < 0$ $\frac{4 \sqrt{1 + x^{2}}}{x} = \frac{5 \sqrt{1 + y^{2}}}{y} \Rightarrow y = \pm \frac{5 x}{\sqrt{16 - 9 x^{2}}}$ $\frac{4 \sqrt{1 + x^{2}}}{x} = \frac{6 \sqrt{1 + z^{2}}}{z} \Rightarrow z = \pm \frac{3 x}{\sqrt{4 - 5 x^{2}}}$ $let x, y, x greater than zero, due to squaring$ ${we}^{'} ll get the other option anyway$ $x + y + x - x y z = 0$ $x + \frac{5 x}{\sqrt{16 - 9 x^{2}}} + \frac{3 x}{\sqrt{4 - 5 x^{2}}} - x \times \frac{5 x}{\sqrt{16 - 9 x^{2}}} \times \frac{3 x}{\sqrt{4 - 5 x^{2}}} = 0$ $\sqrt{4 - 5 x^{2}} = p; \sqrt{16 - 9 x^{2}} = q$ $\Rightarrow x \frac{5 p + p q + 3 q - 15 x^{2}}{p q} = 0$ $p = 3 \frac{(5 x^{2} - q)}{q + 5}$ $p^{2} = 3 \frac{25 x^{4} - 10 {q x}^{2} + q^{2}}{q^{2} + 10 q + 25}$ $p^{2} = 4 - 5 x^{2}; q^{2} = 16 - 9 x^{2}$ $\Rightarrow q = \frac{9 x^{2} - 1}{2}$ $q^{2} = \frac{81 x^{4} - 18 x^{2} + 1}{4}$ $q^{2} = 16 - 9 x^{2}$ $\Rightarrow x^{4} + \frac{2}{9} x^{2} - \frac{7}{9} = 0$ $\Rightarrow x^{2} = - 1 \lor x^{2} = \frac{7}{9}$ $\Rightarrow x = \pm \frac{\sqrt{7}}{3} \Rightarrow y = \pm \frac{5 \sqrt{7}}{9}; z = \pm 3 \sqrt{7}$ $solutions (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} - \frac{\sqrt{7}}{3} \\ - \frac{5 \sqrt{7}}{9} \\ - 3 \sqrt{7} \end{matrix}) \lor (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} \frac{\sqrt{7}}{3} \\ \frac{5 \sqrt{7}}{9} \\ 3 \sqrt{7} \end{matrix})$
	Commented by Abdo msup. last updated on 12/Dec/18

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