Solve the simultaneous equation:
$\displaystyle x-2y+3=0$
$\displaystyle x^2+xy-27=0$
From the first equation, we can see that $\displaystyle x = 2y - 3$.
Substituting into the second equation:
$\displaystyle (2y - 3)^2 + (2y - 3)y - 27 = 0$
$\displaystyle 4y^2 - 12y + 9 + 2y^2 - 3y - 27 = 0$
$\displaystyle 6y^2 - 15y - 18 = 0$
$\displaystyle 2y^2 - 5y - 6 = 0$
$\displaystyle y^2 - \frac{5}{2}y - 3 = 0$
$\displaystyle y^2 - \frac{5}{2}y + \left(-\frac{5}{4}\right)^2 - \left(-\frac{5}{4}\right)^2 - 3 = 0$
$\displaystyle \left(y - \frac{5}{4}\right)^2 - \frac{25}{16} - \frac{48}{16} = 0$
$\displaystyle \left(y - \frac{5}{4}\right)^2 - \frac{73}{16} = 0$
$\displaystyle \left(y - \frac{5}{4}\right)^2 = \frac{73}{16}$
$\displaystyle y - \frac{5}{4} = \pm \frac{\sqrt{73}}{4}$
$\displaystyle y = \frac{5 \pm \sqrt{73}}{4}$.
So substitute $\displaystyle y = \frac{5 - \sqrt{73}}{4}$ and $\displaystyle y = \frac{5 + \sqrt{73}}{4}$
into $\displaystyle x = 2y - 3$
to find the two values of $\displaystyle x$.