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Question Number 76167 by Master last updated on 24/Dec/19

	Answered by behi83417@gmail.com last updated on 24/Dec/19
	$⇒ { ((3x−x^2 −3y+xy=1−3x−xy+3x^2 y)),((2x−x^2 +2y−xy=1−2x+xy−2x^2 y)) :} ⇒ { ((x−5y+2xy=−x−2xy+5x^2 y)),((5x−2x^2 −y=2−5x+x^2 y)) :} ⇒ { ((2x−5y+4xy−5x^2 y=0)),((10x−y−2x^2 −x^2 y−2=0⇒y=((10x−2−2x^2 )/(1+x^2 )))) :} 2x−5((−2x^2 +10x−2)/(1+x^2 ))+4x((−2x^2 +10x−2)/(1+x^2 ))−5x^2 ((−2x^2 +10x−2)/(1+x^2 ))=0 2x(1+x^2 )−5(−2x^2 +10x−2)+4x(−2x^2 +10x−2)−5x^2 (−2x^2 +10x−2)=0 ⇒2x+2x^3 +10x^2 −50x+10−8x^3 +40x^2 −8x+20x^4 −50x^3 +10x^2 =0 ⇒10x^4 −28x^3 +30x^2 −28x+5=0 ⇒ { ((x=0.22⇒y=0.098)),((x=1.93⇒y=2.1)) :}$
	$\Rightarrow {\begin{cases} 3 x - x^{2} - 3 y + xy = 1 - 3 x - xy + {3 x}^{2} y \\ 2 x - x^{2} + 2 y - xy = 1 - 2 x + xy - {2 x}^{2} y \end{cases}$ $\Rightarrow {\begin{cases} x - 5 y + 2 xy = - x - 2 xy + {5 x}^{2} y \\ 5 x - {2 x}^{2} - y = 2 - 5 x + x^{2} y \end{cases}$ $\Rightarrow {\begin{cases} 2 x - 5 y + 4 xy - {5 x}^{2} y = 0 \\ 10 x - y - {2 x}^{2} - x^{2} y - 2 = 0 \Rightarrow y = \frac{10 x - 2 - {2 x}^{2}}{1 + x^{2}} \end{cases}$ $2 x - 5 \frac{- {2 x}^{2} + 10 x - 2}{1 + x^{2}} + 4 x \frac{- {2 x}^{2} + 10 x - 2}{1 + x^{2}} - {5 x}^{2} \frac{- {2 x}^{2} + 10 x - 2}{1 + x^{2}} = 0$ $2 x (1 + x^{2}) - 5 (- {2 x}^{2} + 10 x - 2) + 4 x (- {2 x}^{2} + 10 x - 2) - {5 x}^{2} (- {2 x}^{2} + 10 x - 2) = 0$ $\Rightarrow 2 x + {2 x}^{3} + {10 x}^{2} - 50 x + 10 - {8 x}^{3} + {40 x}^{2} - 8 x + {20 x}^{4} - {50 x}^{3} + {10 x}^{2} = 0$ $\Rightarrow {10 x}^{4} - {28 x}^{3} + {30 x}^{2} - 28 x + 5 = 0$ $\Rightarrow {\begin{cases} x = 0.22 \Rightarrow y = 0.098 \\ x = 1.93 \Rightarrow y = 2.1 \end{cases}$
	Answered by vishalbhardwaj last updated on 24/Dec/19

	$(x - y) (3 - x) = (1 - 3 x) (1 - xy)$ $3 x - x^{2} - 3 y + xy = 1 - xy - 3 x + {3 x}^{2} y$ $- x^{2} + 6 x = - 2 xy + 3 y + {3 x}^{2} y + 1$ $- x^{2} + 6 x - 1 = y (- 2 x + 3 + {3 x}^{2})$ $\Rightarrow y = \frac{6 x - x^{2} - 1}{{3 x}^{2} - 2 x + 3} (i)$ $and in the same way$ $for second equation$ $(x + y) (2 - x) = (1 + xy) (1 - 2 x)$ $2 x - x^{2} + 2 y - xy = 1 - 2 x + xy - {2 x}^{2} y$ $- x^{2} + 4 x - 1 = 2 xy - 2 y - {2 x}^{2} y$ $- x^{2} + 4 x - 1 = y (2 x - 2 - {2 x}^{2})$ $\Rightarrow y = \frac{4 x - x^{2} - 1}{2 x - 2 - {2 x}^{2}} (ii)$ $by eq (i) and (ii)$ $\Rightarrow \frac{4 x - x^{2} - 1}{2 x - 2 - {2 x}^{2}}$ $= \frac{6 x - x^{2} - 1}{{3 x}^{2} - 2 x + 3}$ $\Rightarrow (4 x - x^{2} - 1) ({3 x}^{2} - 2 x + 3)$ $= (6 x - x^{2} - 1) (2 x - 2 - {2 x}^{2})$ $\Rightarrow {12 x}^{3} - {8 x}^{2} + 12 x - {3 x}^{4} + {2 x}^{3} - {3 x}^{2} - {3 x}^{2} + 2 x - 3$ $= {12 x}^{2} - 12 x - {12 x}^{3} - {2 x}^{3} + {2 x}^{2} + {2 x}^{4} - 2 x + 2 + {2 x}^{2}$ $\Rightarrow - {5 x}^{4} + {28 x}^{3} - {30 x}^{2} + 28 x - 5 = 0$ $please help for$ $- {5 x}^{4} + {28 x}^{3} - {30 x}^{2} + 28 x - 5 = 0$
	Answered by mind is power last updated on 24/Dec/19
	$x=coth(a) y=coth(b) coth hase range of ]−∞,+∞[ ⇒ we can use this sbstitution ⇒((x−y)/(1−xy))=coth(a−b) ((x+y)/(1+xy))=coth(a+b) let vothc=((1/3)) ((1−3x)/(3−x))=(((1/3)−x)/(1−(1/3).x))=((coth(c)−coth(a))/(1−coth(c)coth(a)))=th(c−a) coth(d)=(1/2) ⇒((1−2x)/(2−x))=(((1/2)−x)/(1−(1/2).x))=((coth(d)−coth(a))/(1−coth(a).coth(d)))=coth(d−a) ⇔ { ((coth(a−b)=coth(c−a))),((coth(a+b)=coth(d−a))) :} coth(x)=coth(y)⇔x=y since coth′(x)=((−1)/(sh^2 (x)))<0 ⇒ { ((a−b=c−a)),((a+b=d−a)) :} ⇒ { ((2a−b=c)),((2a+b=d)) :} b=((d−c)/2) a=((d−b)/2)$
	$x = \coth (a)$ $y = \coth (b)$ $\coth hase range of] - \infty, + \infty [$ $\Rightarrow we can use this sbstitution$ $\Rightarrow \frac{x - y}{1 - xy} = \coth (a - b)$ $\frac{x + y}{1 + xy} = \coth (a + b)$ $let vothc = (\frac{1}{3})$ $\frac{1 - 3 x}{3 - x} = \frac{\frac{1}{3} - x}{1 - \frac{1}{3} . x} = \frac{\coth (c) - \coth (a)}{1 - \coth (c) \coth (a)} = th (c - a)$ $\coth (d) = \frac{1}{2}$ $\Rightarrow \frac{1 - 2 x}{2 - x} = \frac{\frac{1}{2} - x}{1 - \frac{1}{2} . x} = \frac{\coth (d) - \coth (a)}{1 - \coth (a) . \coth (d)} = \coth (d - a)$ $\Leftrightarrow$ ${\begin{cases} \coth (a - b) = \coth (c - a) \\ \coth (a + b) = \coth (d - a) \end{cases}$ $\coth (x) = \coth (y) \Leftrightarrow x = y since \coth^{'} (x) = \frac{- 1}{{sh}^{2} (x)} < 0$ $\Rightarrow {\begin{cases} a - b = c - a \\ a + b = d - a \end{cases}$ $\Rightarrow {\begin{cases} 2 a - b = c \\ 2 a + b = d \end{cases}$ $b = \frac{d - c}{2}$ $a = \frac{d - b}{2}$
	Commented by Master last updated on 25/Dec/19

	$thanks$
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