Solve the following system of equations: { ((x^3 y+x^3 y^2 +2x^2 y^2 +x^2 y^3 +xy^3 =30)),((x^2 y+xy+x+y+xy^2 =11)) :}


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Question Number 105704 by 1549442205PVT last updated on 31/Jul/20
$Solve the following system of equations: { ((x^3 y+x^3 y^2 +2x^2 y^2 +x^2 y^3 +xy^3 =30)),((x^2 y+xy+x+y+xy^2 =11)) :}$
$Solve the following system of equations :$ ${\begin{cases} x^{3} y + x^{3} y^{2} + {2 x}^{2} y^{2} + x^{2} y^{3} + {xy}^{3} = 30 \\ x^{2} y + xy + x + y + {xy}^{2} = 11 \end{cases}$
Commented by PRITHWISH SEN 2 last updated on 31/Jul/20
$from eqn. (i) xy(x+y)(x+y+xy)=30.........(iii) from eqn. (ii) (xy+1)(x+y+1)=12.....(iv) now put x+y=a and xy=b then (iii)⇒ ab(a+b)=30 (iv)⇒(a+1)(b+1)=12⇒ ab+a+b=11 solving (a+b,ab)=(5,6) or (6,5) ∴ (a,b)=(5,1),(1,5),(3,2) or (2,3) for (5,1) (x,y)= {((((√7)+(√3))/2))^2 ,((((√7)−(√3))/2))^2 } and {((((√7)−(√3))/2))^2 ,((((√7)+(√3))/2))^2 } there is no real value for (a,b)=(1,5) for (3,2) (x,y)= (2,1) and (1,2) and no real value for (a,b)=(2,3)$
$from eqn . (i)$ $xy (x + y) (x + y + xy) = 30 . . . . . . . . . (iii)$ $from eqn . (ii)$ $(xy + 1) (x + y + 1) = 12 . . . . . (iv)$ $now put x + y = a and xy = b$ $then (iii) \Rightarrow ab (a + b) = 30$ $(iv) \Rightarrow (a + 1) (b + 1) = 12 \Rightarrow ab + a + b = 11$ $solving (a + b, ab) = (5, 6) or (6, 5)$ $∴ (a, b) = (5, 1), (1, 5), (3, 2) or (2, 3)$ $for (5, 1)$ $(x, y) = {{(\frac{\sqrt{7} + \sqrt{3}}{2})}^{2}, {(\frac{\sqrt{7} - \sqrt{3}}{2})}^{2}} and {{(\frac{\sqrt{7} - \sqrt{3}}{2})}^{2}, {(\frac{\sqrt{7} + \sqrt{3}}{2})}^{2}}$ $there is no real value for (a, b) = (1, 5)$ $for (3, 2)$ $(x, y) = (2, 1) and (1, 2)$ $and no real value for (a, b) = (2, 3)$
Commented by PRITHWISH SEN 2 last updated on 31/Jul/20

$welcome$
	Answered by 1549442205PVT last updated on 31/Jul/20
	$Set x+y=u,xy=v we have (1)⇔xy(x^2 +y^2 )+(xy)^2 (x+y)+2(xy)^2 =30 ⇔xy[(x+y)^2 −2xy]+(xy)^2 (x+y)+2(xy)^2 =30 (x+y)^2 (xy)+(xy)^2 (x+y)=30 (2)⇔xy(x+y)+xy+x+y=11. { ((u^2 v+v^2 u=30)),((uv+i+v=11)) :}⇔ { ((uv(u+v)=30)),((uv+u+v=11)) :} Set u+v=m,uv=n⇒ { ((m+n=11)),((mn=30)) :} ⇔(m,n)∈{(5,6),(6,5)} i)(m,n)=(5,6)⇔(u,v)∈{(3,2),(2,3)} +(u,v)=(3,2)⇔ { ((x+y=3)),((xy=2)) :}⇔(x,y)∈{(2,1),(1,2)} +(u,v)=(2,3)⇔ { ((x+y=2)),((xy=3)) :}⇒(x−y)^2 =(x+y)^2 −4xy=−8<0 This case has no roots ii)(m,n)=(6,5)⇔(u,v)∈{(5,1),(1,5)} +(u,v)=(5,1)⇔ { ((x+y=5)),((xy=1)) :}⇔ { ((x+y=5)),(((x−y)^2 =(x+y)^2 −4xy=21)) :} ⇔ { ((x+y=5)),((x−y=±(√(21)))) :} ⇔(x,y)∈{(((5+(√(21)))/2),((5−(√(21)))/2)),(((5−(√(21)))/2),((5+(√(21)))/2))} +(u,v)=(1,5)⇔ { ((x+y=1)),((xy=5)) :}⇒(x−y)^2 =(x+y)^2 −4xy=−19<0 so this case has no roots Thus,the given system has four roots: (x,y)∈{(1,2),(2,1),(((5+(√(21)))/2),((5−(√(21)))/2)),(((5−(√(21)))/2),((5+(√(21)))/2))$
	$Set x + y = u, xy = v we have$ $(1) \Leftrightarrow xy (x^{2} + y^{2}) + {(xy)}^{2} (x + y) + 2 {(xy)}^{2} = 30$ $\Leftrightarrow xy [{(x + y)}^{2} - 2 xy] + {(xy)}^{2} (x + y) + 2 {(xy)}^{2} = 30$ ${(x + y)}^{2} (xy) + {(xy)}^{2} (x + y) = 30$ $(2) \Leftrightarrow xy (x + y) + xy + x + y = 11 .$ ${\begin{cases} u^{2} v + v^{2} u = 30 \\ uv + i + v = 11 \end{cases} \Leftrightarrow {\begin{cases} uv (u + v) = 30 \\ uv + u + v = 11 \end{cases}$ $Set u + v = m, uv = n \Rightarrow {\begin{cases} m + n = 11 \\ mn = 30 \end{cases}$ $\Leftrightarrow (m, n) \in {(5, 6), (6, 5)}$ $i) (m, n) = (5, 6) \Leftrightarrow (u, v) \in {(3, 2), (2, 3)}$ $+ (u, v) = (3, 2) \Leftrightarrow {\begin{cases} x + y = 3 \\ xy = 2 \end{cases} \Leftrightarrow (x, y) \in {(2, 1), (1, 2)}$ $+ (u, v) = (2, 3) \Leftrightarrow {\begin{cases} x + y = 2 \\ xy = 3 \end{cases} \Rightarrow {(x - y)}^{2} = {(x + y)}^{2} - 4 xy = - 8 < 0$ $This case has no roots$ $ii) (m, n) = (6, 5) \Leftrightarrow (u, v) \in {(5, 1), (1, 5)}$ $+ (u, v) = (5, 1) \Leftrightarrow {\begin{cases} x + y = 5 \\ xy = 1 \end{cases} \Leftrightarrow {\begin{cases} x + y = 5 \\ {(x - y)}^{2} = {(x + y)}^{2} - 4 xy = 21 \end{cases}$ $\Leftrightarrow {\begin{cases} x + y = 5 \\ x - y = \pm \sqrt{21} \end{cases} \Leftrightarrow (x, y) \in {(\frac{5 + \sqrt{21}}{2}, \frac{5 - \sqrt{21}}{2}), (\frac{5 - \sqrt{21}}{2}, \frac{5 + \sqrt{21}}{2})}$ $+ (u, v) = (1, 5) \Leftrightarrow {\begin{cases} x + y = 1 \\ xy = 5 \end{cases} \Rightarrow {(x - y)}^{2} = {(x + y)}^{2} - 4 xy = - 19 < 0$ $so this case has no roots$ $Thus, the given system has four roots :$ $(x, y) \in {(1, 2), (2, 1), (\frac{5 + \sqrt{21}}{2}, \frac{5 - \sqrt{21}}{2}), (\frac{5 - \sqrt{21}}{2}, \frac{5 + \sqrt{21}}{2})$
	Commented by 1549442205PVT last updated on 01/Aug/20

	$Ok, excuse me . Thank you sir .$
	Commented by PRITHWISH SEN 2 last updated on 31/Jul/20

	$\frac{5 + \sqrt{21}}{2} = \frac{10 + 2 \sqrt{21}}{4} = \frac{{(\sqrt{7})}^{2} + {(\sqrt{3})}^{2} + 2 . \sqrt{7} . \sqrt{3}}{4}$ $= {(\frac{\sqrt{7} + \sqrt{3}}{2})}^{2}$
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