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Question Number 110365 by mathdave last updated on 28/Aug/20

	Answered by Rasheed.Sindhi last updated on 29/Aug/20

	$w x y z = 120, w x z - y = 26,$ $x y z - w = 58, w z + x y = 22$ $\frac{w x y z}{y} - y = 26 \Rightarrow \frac{120}{y} - y = 26 . . . . (i i)$ $\frac{w x y z}{w} - w = 58 \Rightarrow \frac{120}{w} - w = 58 . . . (i i i)$ $\frac{w x y z}{x y} + x y = 22 \Rightarrow \frac{120}{x y} + x y = 22 . . . (i v)$ $(i i) \Rightarrow y^{2} + 26 y - 120 = 0$ $y = 4, - 30$ $(i i i) \Rightarrow w^{2} + 58 w - 120 = 0$ $w = 2, - 60$ $(i v) \Rightarrow {(x y)}^{2} - 22 (x y) + 120 = 0$ $y = 4 \to 16 x^{2} - 88 x + 120 = 0$ $2 x^{2} - 11 x + 15 = 0$ $(x - 3) (2 x - 5) = 0$ $x = 3, 5 / 2$ $y = - 30 \to 900 x^{2} + 660 x + 120 = 0$ $15 x^{2} + 11 x + 2 = 0$ $(3 x + 1) (5 x + 2) = 0$ $x = - 1 / 3, - 2 / 5$
	Answered by kkc last updated on 28/Aug/20

	$W = 2 X = 3 Y = 4 z = 5$
	Answered by 1549442205PVT last updated on 29/Aug/20
	$Set xyz=a,w=b⇒ { ((ab=120)),((a−b=58)) :}⇔ { ((a.(−b)=−120)),((a+(−b)=58)) :} a,(−b) are roots of the equation t^2 −58t−120=0 ,Δ^′ =29^2 +120=31^2 ⇒(a,−b)∈{(60,−2),(−2,60)} ⇒(xyz,w)∈{(60,2),(−2,−60)} Set xy=c,wz=d⇒ { ((c+d=22)),((cd=120)) :} c,d are roots of the equation t^2 −22t+120=0 Δ′=11^2 −120=1 ⇒(c,d)∈{(12,10),(10,12)} •a=xyz=60,b=w=2,c=xy=12,d=wz=10 ⇒z=5,wxz−y=26⇒10x−y=26 ⇒x(10x−26)=12⇔10x^2 −26x−12=0 5x^2 −13x−6=0⇒x=(13±17)/10 x(1⇒x∈{3,−2/5}⇒y∈{4,−30} we get (x,y,z,w)∈{(3,4,5,2),(((−2)/5),−30,5,2)} •a=xyz=−2,b=w=−60,c=xy=10,d=wz=12 ⇒z=−(1/5),wxz−y=26⇒12x−y=26 x(12x−26)=10⇔12x^2 −26x=10 ⇔6x^2 −13x−5=0, Δ=169+120=17^2 x∈{(5/2),((−2)/6)}⇒y∈{4;−30} (x,y,z,w)∈{((5/2),4,((−1)/5),−60),(((−1)/3),−30,((−1)/5),−60)} •a=xyz=60,b=w=2,c=xy=10,d=wz=12 ⇒z=6,wxz−y=26⇒12x−y=26 ⇒x(12x−26)=10⇔12x^2 −26x−10=0 ⇔6x^2 −13x−5=0.Like above we get x∈{(5/2),((−2)/6)}⇒y∈{4;−30} ⇒(x,y,z,w)∈{((5/2),4,6,2),(((−1)/3),−30,6,2)} •a=xyz=−2,b=w=−60,c=xy=12,d=wz=10 ⇒z=((−1)/6),wxz−y=26⇔10x−y=12 x(10x−26)=12⇔5x^2 −13x−6=0 ⇒x∈{3,−2/5}⇒y∈{4,−30} ⇒(x,y,z,w)∈{(3,4,((−1)/6),−60),(((−2)/5),−30,((−1)/6);−60)}$
	$Set xyz = a, w = b \Rightarrow {\begin{cases} ab = 120 \\ a - b = 58 \end{cases} \Leftrightarrow {\begin{cases} a . (- b) = - 120 \\ a + (- b) = 58 \end{cases}$ $a, (- b) are roots of the equation$ $t^{2} - 58 t - 120 = 0, Δ^{^{'}} = 29^{2} + 120 = 31^{2}$ $\Rightarrow (a, - b) \in {(60, - 2), (- 2, 60)}$ $\Rightarrow (xyz, w) \in {(60, 2), (- 2, - 60)}$ $Set xy = c, wz = d \Rightarrow {\begin{cases} c + d = 22 \\ cd = 120 \end{cases}$ $c, d are roots of the equation$ $t^{2} - 22 t + 120 = 0 Δ^{'} = 11^{2} - 120 = 1$ $\Rightarrow (c, d) \in {(12, 10), (10, 12)}$ $∙ a = xyz = 60, b = w = 2, c = xy = 12, d = wz = 10$ $\Rightarrow z = 5, wxz - y = 26 \Rightarrow 10 x - y = 26$ $\Rightarrow x (10 x - 26) = 12 \Leftrightarrow {10 x}^{2} - 26 x - 12 = 0$ ${5 x}^{2} - 13 x - 6 = 0 \Rightarrow x = (13 \pm 17) / 10$ $x (1 \Rightarrow x \in {3, - 2 / 5} \Rightarrow y \in {4, - 30}$ $we get (x, y, z, w) \in {(3, 4, 5, 2), (\frac{- 2}{5}, - 30, 5, 2)}$ $∙ a = xyz = - 2, b = w = - 60, c = xy = 10, d = wz = 12$ $\Rightarrow z = - \frac{1}{5}, wxz - y = 26 \Rightarrow 12 x - y = 26$ $x (12 x - 26) = 10 \Leftrightarrow {12 x}^{2} - 26 x = 10$ $\Leftrightarrow {6 x}^{2} - 13 x - 5 = 0, Δ = 169 + 120 = 17^{2}$ $x \in {\frac{5}{2}, \frac{- 2}{6}} \Rightarrow y \in {4; - 30}$ $(x, y, z, w) \in {(\frac{5}{2}, 4, \frac{- 1}{5}, - 60), (\frac{- 1}{3}, - 30, \frac{- 1}{5}, - 60)}$ $∙ a = xyz = 60, b = w = 2, c = xy = 10, d = wz = 12$ $\Rightarrow z = 6, wxz - y = 26 \Rightarrow 12 x - y = 26$ $\Rightarrow x (12 x - 26) = 10 \Leftrightarrow {12 x}^{2} - 26 x - 10 = 0$ $\Leftrightarrow {6 x}^{2} - 13 x - 5 = 0 . Like above we get$ $x \in {\frac{5}{2}, \frac{- 2}{6}} \Rightarrow y \in {4; - 30}$ $\Rightarrow (x, y, z, w) \in {(\frac{5}{2}, 4, 6, 2), (\frac{- 1}{3}, - 30, 6, 2)}$ $∙ a = xyz = - 2, b = w = - 60, c = xy = 12, d = wz = 10$ $\Rightarrow z = \frac{- 1}{6}, wxz - y = 26 \Leftrightarrow 10 x - y = 12$ $x (10 x - 26) = 12 \Leftrightarrow {5 x}^{2} - 13 x - 6 = 0$ $\Rightarrow x \in {3, - 2 / 5} \Rightarrow y \in {4, - 30}$ $\Rightarrow (x, y, z, w) \in {(3, 4, \frac{- 1}{6}, - 60), (\frac{- 2}{5}, - 30, \frac{- 1}{6}; - 60)}$
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