{ ((x^2 + (x + 3y) = 11)),((y^2 + (y + 3x) = 29)) :} ⇒ x + y = ?


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Question Number 214499 by hardmath last updated on 10/Dec/24
${ ((x^2 + (x + 3y) = 11)),((y^2 + (y + 3x) = 29)) :} ⇒ x + y = ?$
${\begin{cases} x^{2} + (x + 3 y) = 11 \\ y^{2} + (y + 3 x) = 29 \end{cases} \Rightarrow x + y = ?$
Commented by Frix last updated on 10/Dec/24

$4 solutions$ $x + y = t$ $t^{4} + 4 t^{3} - 108 t^{2} + 352 t + 4 = 0$ $No nice solution .$
	Answered by A5T last updated on 10/Dec/24

	$(i) + (i i) \Rightarrow x^{2} + y^{2} + 4 (x + y) - 40 = 0$ $\Rightarrow \frac{{(x - y)}^{2} + {(x + y)}^{2}}{2} + 4 (x + y) - 40 = 0 . . . (i)$ $(i) - (i i) \Rightarrow x^{2} - y^{2} - 2 (x - y) + 18 = 0 . . . (i i)$ $x + y = p; x - y = q$ $\Rightarrow (i) : \frac{p^{2} + q^{2}}{2} + 4 p - 40 = 0 \Rightarrow p^{2} + 8 p + q^{2} - 80 = 0$ $(i i) : p q - 2 q + 18 = 0 \Rightarrow q = \frac{18}{2 - p}$ $q i n (i) \Rightarrow p^{2} + 8 p + \frac{324}{{(2 - p)}^{2}} - \frac{80 {(2 - p)}^{2}}{{(2 - p)}^{2}} = 0$ $\Rightarrow p^{4} + 4 p^{3} - 108 p^{2} + 352 p + 4 = 0$
	Commented by Rasheed.Sindhi last updated on 10/Dec/24

	$x^{2} + y^{2} = {(x + y)}^{2} - 2 x y$ $p e r h a p s y o u h a v e m i s s e d - 2 x y s i r!$
	Commented by A5T last updated on 10/Dec/24

	$T o o k a n o t h e r p a t h, t h a n k s .$
	Answered by Rasheed.Sindhi last updated on 11/Dec/24
	${ ((x^2 + (x + 3y) = 11)),((y^2 + (y + 3x) = 29)) :} ⇒ x + y = ? Adding: x^2 +y^2 +(x+y)+3(x+y)=40 (x+y)^2 −2xy+4(x+y)=40 but xy=(1/4){(x+y)^2 −(x−y)^2 } (x+y)^2 −2[(1/4){(x+y)^2 −(x−y)^2 }]+4(x+y)=40 let x+y=a,x−y=b a^2 −(1/2){a^2 −b^2 }+4a=40 a^2 +b^2 +8a=80......(i) Subtracting: x^2 −y^2 +(x−y)−3(x−y)=−18 (x−y)(x+y)−2(x−y)=−18 (x−y)(x+y−2)=−18 b(a−2)=−18.........(ii) b=((18)/(2−a)) putting this in (i) a^2 +(((18)/(2−a)))^2 +8a=80 a^2 (2−a)^2 +324+8a(2−a)^2 =80(2−a)^2 a^2 (a^2 −4a+4)+324+8a(a^2 −4a+4)=80a^2 −320a+320 a^2 (a^2 −4a+4)+324+8a(a^2 −4a+4)=80a^2 −320a+320 a^4 −4a^3 +4a^2 +8a^3 −32a^2 +32a+324−80a^2 +320a−320=0 a^4 +4a^3 −108a^2 +352a+4 ....$
	${\begin{cases} x^{2} + (x + 3 y) = 11 \\ y^{2} + (y + 3 x) = 29 \end{cases} \Rightarrow x + y = ?$ $Adding :$ $x^{2} + y^{2} + (x + y) + 3 (x + y) = 40$ ${(x + y)}^{2} - 2 xy + 4 (x + y) = 40$ $but xy = \frac{1}{4} {{(x + y)}^{2} - {(x - y)}^{2}}$ ${(x + y)}^{2} - 2 [\frac{1}{4} {{(x + y)}^{2} - {(x - y)}^{2}}] + 4 (x + y) = 40$ $let x + y = a, x - y = b$ $a^{2} - \frac{1}{2} {a^{2} - b^{2}} + 4 a = 40$ $a^{2} + b^{2} + 8 a = 80 . . . . . . (i)$ $Subtracting :$ $x^{2} - y^{2} + (x - y) - 3 (x - y) = - 18$ $(x - y) (x + y) - 2 (x - y) = - 18$ $(x - y) (x + y - 2) = - 18$ $b (a - 2) = - 18 . . . . . . . . . (ii)$ $b = \frac{18}{2 - a}$ $putting this in (i)$ $a^{2} + {(\frac{18}{2 - a})}^{2} + 8 a = 80$ $a^{2} {(2 - a)}^{2} + 324 + 8 a {(2 - a)}^{2} = 80 {(2 - a)}^{2}$ $a^{2} (a^{2} - 4 a + 4) + 324 + 8 a (a^{2} - 4 a + 4) = {80 a}^{2} - 320 a + 320$ $a^{2} (a^{2} - 4 a + 4) + 324 + 8 a (a^{2} - 4 a + 4) = {80 a}^{2} - 320 a + 320$ $a^{4} - {4 a}^{3} + {4 a}^{2} + {8 a}^{3} - {32 a}^{2} + 32 a + 324 - {80 a}^{2} + 320 a - 320 = 0$ $a^{4} + {4 a}^{3} - {108 a}^{2} + 352 a + 4$ $. . . .$
	Commented by A5T last updated on 10/Dec/24

	$a^{4} + 4 a^{3} - 108 a^{2} + 352 a + 4 = 0 s h o u l d b e t h e$ $c o r r e c t e q u a t i o n .$
	Answered by ajfour last updated on 11/Dec/24

	$x + 3 y = p$ $y + 3 x = q$ $4 (x + y) = p + q = 4 t$ $x = \frac{3 q - p}{8}$ $y = \frac{3 p - q}{8}$ $y - x = \frac{p - q}{2}$ ${(3 q - p)}^{2} + {(3 p - q)}^{2} + 256 t = 2560$ $10 (p^{2} + q^{2}) + 256 t = 2560 . . . (i)$ $o r 5 {(p + q)}^{2} + 5 {(p - q)}^{2} + 256 (t - 10) = 0$ $y^{2} - x^{2} + 2 (x - y) = 18$ $(y - x) (t - 2) = 18$ $(p - q) (t - 2) = 36 . . . . (i i)$ $80 t^{2} + \frac{5 \times 36 \times 36}{{(t - 2)}^{2}} + 256 (t - 10) = 0$ $80 {(t - 2)}^{2} + \frac{180 \times 36}{{(t - 2)}^{2}} + 24 \times 24 (t - 2)$ $= 2560 + 320 - 1152$ $\Rightarrow \frac{{(t - 2)}^{2}}{9} + \frac{9}{{(t - 2)}^{2}} + \frac{4}{5} (t - 2)$ $= \frac{1}{9} (32 + 4 - \frac{72}{5}) = \frac{12}{5}$ $l e t \frac{t - 2}{3} = z$ $z^{2} + \frac{1}{z^{2}} + \frac{12}{5} (z - 1) = 0$
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