$PQRS$ is a cyclic quadrilateral. We are given the measures of its angles in terms of $x$ and $y$. We need to find the value of $x$. The angles are: $\angle P = x$, $\angle Q = 2y - 30$, $\angle R = x + y$, and $\angle S = x$.

1. Problem Description

$PQRS$ is a cyclic quadrilateral. We are given the measures of its angles in terms of $x$ and $y$. We need to find the value of $x$. The angles are: $\angle P = x$, $\angle Q = 2y - 30$, $\angle R = x + y$, and $\angle S = x$.

2. Solution Steps

In a cyclic quadrilateral, the sum of opposite angles is $180^{\circ}$.

Therefore, $\angle P + \angle R = 180^{\circ}$ and $\angle Q + \angle S = 180^{\circ}$.

From $\angle P + \angle R = 180^{\circ}$, we have

$x + (x + y) = 180$

$2x + y = 180$ (1)

From $\angle Q + \angle S = 180^{\circ}$, we have

$(2y - 30) + x = 180$

$x + 2y = 210$ (2)

We now have a system of two linear equations in two variables, $x$ and $y$:

$2x + y = 180$ (1)

$x + 2y = 210$ (2)

Multiply equation (1) by 2 to eliminate $y$:

$4x + 2y = 360$ (3)

Subtract equation (2) from equation (3):

$(4x + 2y) - (x + 2y) = 360 - 210$

$3x = 150$

$x = \frac{150}{3}$

$x = 50$

Substitute $x = 50$ into equation (1):

$2(50) + y = 180$

$100 + y = 180$

$y = 180 - 100$

$y = 80$

Now, substitute $x = 50$ and $y = 80$ into the given angles to check if they are consistent.

$\angle P = x = 50^{\circ}$

$\angle Q = 2y - 30 = 2(80) - 30 = 160 - 30 = 130^{\circ}$

$\angle R = x + y = 50 + 80 = 130^{\circ}$

$\angle S = x = 50^{\circ}$

$\angle P + \angle R = 50 + 130 = 180^{\circ}$

$\angle Q + \angle S = 130 + 50 = 180^{\circ}$

The angles are consistent.

3. Final Answer

The value of $x$ is $50^{\circ}$.

The answer is A.