Find the range of values of $x$ for which $|x - 4| \geq 2$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 2 - 2016

We can solve these equations by substitution or elimination. We'll use substitution:

1. From the second equation, express $x$:
   $x = -\frac{3y + 1}{2}$

2. Substitute this into the first equation:
   $\left(-\frac{3y + 1}{2}\right)^2 + \left(-\frac{3y + 1}{2}\right)y + 2y^2 = 4$

   Expanding this gives:

   $\frac{(3y + 1)^2}{4} - \frac{(3y + 1)y}{2} + 2y^2 = 4$

   In clearing the fractions by multiplying through by 4, we get:
   $\frac{(3y + 1)^2}{4} - 2(3y + 1)y + 8y^2 = 16$

   Continuing simplifies this down to a quadratic equation in $y$, which can be solved to find the values of $y$. From this, we can find corresponding $x$ values.