In a parallelogram, opposite sides are equal in length. Given the sides you provided, we can set up equations based on the lengths of sides \( PQ \), \( QR \), \( RS \), and \( SP \).
Let’s denote:
- \( PQ = PTe = y \)
- \( QR = TR = 33x + 1 \)
- \( RS = QTe = 33y \)
- \( SP = TS = 77x + 99 \)
From the properties of a parallelogram, we know that:
- Opposite sides are equal:
- \( y = 33y \) (from \( PQ \) and \( RS \))
- \( 33x + 1 = 77x + 99 \) (from \( QR \) and \( SP \))
Solving for \( y \):
From the equation \( y = 33y \): \[ y - 33y = 0 \ -32y = 0 \ y = 0 \]
Solving for \( x \):
Now substitute \( y \) back to find \( x \) using the equation \( 33x + 1 = 77x + 99 \): \[ 33x + 1 = 77x + 99 \ 1 - 99 = 77x - 33x \ -98 = 44x \ x = -\frac{98}{44} \ x = -\frac{49}{22} \]
Thus, the values we found are: \[ x = -\frac{49}{22}, \quad y = 0 \]
Therefore, the final answer is: \[ x = -\frac{49}{22}, \quad y = 0 \]