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Let the given equation be
\n" ); document.write( "$$x^2 + 3xy + 2y^2 = 1 - 12x + 5y$$
\n" ); document.write( "We can factor the left side as
\n" ); document.write( "$$(x+y)(x+2y) = 1 - 12x + 5y$$
\n" ); document.write( "Let $u = x+y$ and $v = x+2y$. Then $u+y=x+y$ so $y = v-u$ and $x = u-y = u-(v-u) = 2u-v$.
\n" ); document.write( "Substituting into the given equation, we have
\n" ); document.write( "$$uv = 1 - 12(2u-v) + 5(v-u)$$
\n" ); document.write( "$$uv = 1 - 24u + 12v + 5v - 5u$$
\n" ); document.write( "$$uv = 1 - 29u + 17v$$
\n" ); document.write( "$$uv - 17v = 1 - 29u$$
\n" ); document.write( "$$v(u-17) = 1 - 29u$$
\n" ); document.write( "If $u \neq 17$, we can solve for $v$:
\n" ); document.write( "$$v = \frac{1-29u}{u-17}$$
\n" ); document.write( "Then $y = v-u = \frac{1-29u}{u-17} - u = \frac{1-29u-u(u-17)}{u-17} = \frac{1-29u-u^2+17u}{u-17} = \frac{-u^2-12u+1}{u-17}$
\n" ); document.write( "And $x = 2u-v = 2u - \frac{1-29u}{u-17} = \frac{2u(u-17)-(1-29u)}{u-17} = \frac{2u^2-34u-1+29u}{u-17} = \frac{2u^2-5u-1}{u-17}$\r
\n" ); document.write( "\n" ); document.write( "Since $x$ and $y$ are real numbers, we must have $u \neq 17$.
\n" ); document.write( "We are interested in the possible values of $u = x+y$.
\n" ); document.write( "The equation can be rewritten as
\n" ); document.write( "$$x^2 + 3xy + 2y^2 + 12x - 5y - 1 = 0$$
\n" ); document.write( "Consider this as a quadratic in $x$:
\n" ); document.write( "$$x^2 + (3y+12)x + (2y^2-5y-1) = 0$$
\n" ); document.write( "For $x$ to be real, the discriminant must be non-negative:
\n" ); document.write( "$$(3y+12)^2 - 4(2y^2-5y-1) \ge 0$$
\n" ); document.write( "$$9y^2 + 72y + 144 - 8y^2 + 20y + 4 \ge 0$$
\n" ); document.write( "$$y^2 + 92y + 148 \ge 0$$
\n" ); document.write( "The roots of $y^2 + 92y + 148 = 0$ are
\n" ); document.write( "$$y = \frac{-92 \pm \sqrt{92^2 - 4(148)}}{2} = \frac{-92 \pm \sqrt{8464 - 592}}{2} = \frac{-92 \pm \sqrt{7872}}{2} = -46 \pm \sqrt{1968}$$
\n" ); document.write( "$$y = -46 \pm 4\sqrt{123}$$
\n" ); document.write( "Thus $y \le -46 - 4\sqrt{123}$ or $y \ge -46 + 4\sqrt{123}$.
\n" ); document.write( "Consider the equation as a quadratic in $y$:
\n" ); document.write( "$$2y^2 + (3x-5)y + (x^2+12x-1) = 0$$
\n" ); document.write( "The discriminant must be non-negative:
\n" ); document.write( "$$(3x-5)^2 - 4(2)(x^2+12x-1) \ge 0$$
\n" ); document.write( "$$9x^2 - 30x + 25 - 8x^2 - 96x + 8 \ge 0$$
\n" ); document.write( "$$x^2 - 126x + 33 \ge 0$$
\n" ); document.write( "The roots of $x^2 - 126x + 33 = 0$ are
\n" ); document.write( "$$x = \frac{126 \pm \sqrt{126^2 - 4(33)}}{2} = \frac{126 \pm \sqrt{15876 - 132}}{2} = \frac{126 \pm \sqrt{15744}}{2} = 63 \pm \sqrt{3936}$$
\n" ); document.write( "Thus $x \le 63 - \sqrt{3936}$ or $x \ge 63 + \sqrt{3936}$.\r
\n" ); document.write( "\n" ); document.write( "If $u=17$, then $0=1-29(17)=-492$, which is impossible. So $u\ne 17$ for all real solutions.
\n" ); document.write( "We need to find the range of $u$.
\n" ); document.write( "Let $y=ax+b$.
\n" ); document.write( "Then $x^2+3x(ax+b)+2(ax+b)^2=1-12x+5(ax+b)$.
\n" ); document.write( "$x^2+3ax^2+3bx+2a^2x^2+4abx+2b^2=1-12x+5ax+5b$.
\n" ); document.write( "$(1+3a+2a^2)x^2+(3b+4ab+12-5a)x+2b^2-5b-1=0$.
\n" ); document.write( "For real solutions, the discriminant must be non-negative.
\n" ); document.write( "$(3b+4ab+12-5a)^2-4(1+3a+2a^2)(2b^2-5b-1)\ge 0$.
\n" ); document.write( "$u=x+y=x+ax+b=(1+a)x+b$.
\n" ); document.write( "We want to find $u$.
\n" ); document.write( "If $x+y=17$, then $v=\frac{1-29(17)}{17-17}$. This is impossible.\r
\n" ); document.write( "\n" ); document.write( "We have $u=x+y$. Let $x+y=c$ and $x=c-y$.
\n" ); document.write( "$(c-y)^2+3(c-y)y+2y^2=1-12(c-y)+5y$.
\n" ); document.write( "$c^2-2cy+y^2+3cy-3y^2+2y^2=1-12c+12y+5y$.
\n" ); document.write( "$c^2+cy=1-12c+17y$.
\n" ); document.write( "$y(c-17)=c^2+12c-1$.
\n" ); document.write( "$y=\frac{c^2+12c-1}{c-17}$.
\n" ); document.write( "$x=c-y = c-\frac{c^2+12c-1}{c-17} = \frac{c(c-17)-c^2-12c+1}{c-17} = \frac{-29c+1}{c-17}$.
\n" ); document.write( "For real $x$ and $y$, $c\ne 17$.
\n" ); document.write( "Therefore $x+y$ can be any real number except 17.\r
\n" ); document.write( "\n" ); document.write( "Final Answer: The final answer is $\boxed{(-\infty,17)\cup(17,\infty)}$
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