Solve for x and y ax^2 + bxy + cy^2 = bx^2 + cxy + ay^2 = d.


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Question Number 216010 by MATHEMATICSAM last updated on 25/Jan/25

$Solve for x and y$ ${a x}^{2} + b x y + {c y}^{2} = {b x}^{2} + c x y + {a y}^{2} = d .$
	Answered by mr W last updated on 26/Jan/25
	$ax^2 +bxy+cy^2 =d ...(i) bx^2 +cxy+ay^2 =d ...(ii) (i)−(ii): (a−b)x^2 +(b−c)xy+(c−a)y^2 =0 (a−b)+(b−c)((y/x))+(c−a)((y/x))^2 =0 ⇒(y/x)=k=((c−b±(√((c−b)^2 −4(c−a)(a−b))))/(2(c−a))) =((c−b±(b+c−2a))/(2(c−a)))= { (1),(((a−b)/(c−a))) :} from (i): ax^2 +bkx^2 +ck^2 x^2 =d (a+bk+ck^2 )x^2 =d ⇒x=±(√(d/(a+bk+ck^2 ))) ⇒y=±k(√(d/(a+bk+ck^2 )))$
	${a x}^{2} + b x y + {c y}^{2} = d . . . (i)$ ${b x}^{2} + c x y + {a y}^{2} = d . . . (i i)$ $(i) - (i i) :$ $(a - b) x^{2} + (b - c) x y + (c - a) y^{2} = 0$ $(a - b) + (b - c) (\frac{y}{x}) + (c - a) {(\frac{y}{x})}^{2} = 0$ $\Rightarrow \frac{y}{x} = k = \frac{c - b \pm \sqrt{{(c - b)}^{2} - 4 (c - a) (a - b)}}{2 (c - a)}$ $= \frac{c - b \pm (b + c - 2 a)}{2 (c - a)} = {\begin{cases} 1 \\ \frac{a - b}{c - a} \end{cases}$ $f r o m (i) :$ ${a x}^{2} + {b k x}^{2} + {c k}^{2} x^{2} = d$ $(a + b k + {c k}^{2}) x^{2} = d$ $\Rightarrow x = \pm \sqrt{\frac{d}{a + b k + {c k}^{2}}}$ $\Rightarrow y = \pm k \sqrt{\frac{d}{a + b k + {c k}^{2}}}$
	Answered by Rasheed.Sindhi last updated on 26/Jan/25
	$ax^2 +bxy+cy^2 =d...(i) bx^2 +cxy+ay^2 =d...(ii) (i)−(ii): (a−b)x^2 +(b−c)xy+(c−a)y^2 =0 factors: (a−b)x^2 −(a−b)xy−(c−a)xy+(c−a)y^2 =0 (a−b)x(x−y)−(c−a)y(x−y)=0 (x−y)((a−b)x−(c−a)y)=0 x−y=0 ∣ (a−b)x−(c−a)y=0 determinant (((x=y ∣ x=(((c−a)y)/(a−b))))) x=y (i)⇒ax^2 +bx^2 +cx^2 =d ⇒x=y=±(√(d/(a+b+c))) ✓ x=(((c−a)y)/(a−b)) : a((((c−a)y)/(a−b)))^2 +b((((c−a)y)/(a−b)))y+cy^2 =d y^2 (((a(c−a)^2 +b(a−b)(c−a)+c(a−b)^2 )/((a−b)^2 )))=d y^2 =((d(a−b)^2 )/(a(c−a)^2 +b(a−b)(c−a)+c(a−b)^2 )) y=±(a−b)(√(d/(a(c−a)^2 +b(a−b)(c−a)+c(a−b)^2 ))) y=±(a−b)(√(d/(a^3 +ab(b−a)+ac(c−a)−abc))) x=±(((c−a))/(a−b)){(a−b)(√(d/(a^3 +ab(b−a)+ac(c−a)−abc))) } { ((x=±(c−a)(√(d/(a^3 +ab(b−a)+ac(c−a)−abc))))),((y=±(a−b)(√(d/(a^3 +ab(b−a)+ac(c−a)−abc))) )) :} ✓$
	${a x}^{2} + b x y + {c y}^{2} = d . . . (i)$ ${b x}^{2} + c x y + {a y}^{2} = d . . . (i i)$ $(i) - (i i) :$ $(a - b) x^{2} + (b - c) x y + (c - a) y^{2} = 0$ $f a c t o r s :$ $(a - b) x^{2} - (a - b) x y - (c - a) x y + (c - a) y^{2} = 0$ $(a - b) x (x - y) - (c - a) y (x - y) = 0$ $(x - y) ((a - b) x - (c - a) y) = 0$ $x - y = 0 ∣ (a - b) x - (c - a) y = 0$ $\begin{matrix} x = y ∣ x = \frac{(c - a) y}{a - b} \end{matrix}$ $x = y$ $(i) \Rightarrow {a x}^{2} + {b x}^{2} + {c x}^{2} = d$ $\Rightarrow x = y = \pm \sqrt{\frac{d}{a + b + c}} ✓$ $x = \frac{(c - a) y}{a - b} :$ $a {(\frac{(c - a) y}{a - b})}^{2} + b (\frac{(c - a) y}{a - b}) y + {c y}^{2} = d$ $y^{2} (\frac{a {(c - a)}^{2} + b (a - b) (c - a) + c {(a - b)}^{2}}{{(a - b)}^{2}}) = d$ $y^{2} = \frac{d {(a - b)}^{2}}{a {(c - a)}^{2} + b (a - b) (c - a) + c {(a - b)}^{2}}$ $y = \pm (a - b) \sqrt{\frac{d}{a {(c - a)}^{2} + b (a - b) (c - a) + c {(a - b)}^{2}}}$ $y = \pm (a - b) \sqrt{\frac{d}{a^{3} + a b (b - a) + a c (c - a) - a b c}}$ $x = \pm \frac{(c - a)}{a - b} {(a - b) \sqrt{\frac{d}{a^{3} + a b (b - a) + a c (c - a) - a b c}}}$ ${\begin{cases} x = \pm (c - a) \sqrt{\frac{d}{a^{3} + a b (b - a) + a c (c - a) - a b c}} \\ y = \pm (a - b) \sqrt{\frac{d}{a^{3} + a b (b - a) + a c (c - a) - a b c}} \end{cases} ✓$
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