We are given the system of equations for $x$ and $y$:

$18x^2 + 2y = 1 \quad \text{and} \quad x + y = 1$

We need to find the minimum value of $x + y$. Let’s start by solving the system step by step.

Step 1: Solve for $y$ from the second equation

The second equation is:

$x + y = 1$

Solving for $y$, we get:

$y = 1 - x$

Step 2: Substitute $y = 1 - x$ into the first equation

Now, substitute $y = 1 - x$ into the first equation:

$18x^2 + 2(1 - x) = 1$

Simplifying this:

$18x^2 + 2 - 2x = 1$

$18x^2 - 2x + 2 = 1$

$18x^2 - 2x + 1 = 0$

Step 3: Solve the quadratic equation

We need to solve the quadratic equation:

$18x^2 - 2x + 1 = 0$

We can use the quadratic formula to find the values of $x$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $18x^2 - 2x + 1 = 0$, we have $a = 18$, $b = -2$, and $c = 1$. Plugging these values into the quadratic formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(18)(1)}}{2(18)}$ $x = \frac{2 \pm \sqrt{4 - 72}}{36}$ $x = \frac{2 \pm \sqrt{-68}}{36}$

Since the discriminant is negative ($\sqrt{-68}$), there are no real solutions for $x$. Therefore, this system does not have a real solution for $x$ and $y$. Hence, the minimum value of $x + y$ is not attainable in real numbers.

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