Let the length of the airplane ticket be \( l \) cm and the width be \( w \) cm. We have two pieces of information:
- The area of the ticket: \[ lw = 286 \]
- The perimeter of the ticket: \[ 2l + 2w = 70 \]
We can simplify the perimeter equation: \[ l + w = 35 \]
Now we have the following system of equations:
- \( lw = 286 \)
- \( l + w = 35 \)
From the second equation, we can express \( w \) in terms of \( l \): \[ w = 35 - l \]
Now we can substitute this expression for \( w \) into the first equation: \[ l(35 - l) = 286 \] Expanding this gives: \[ 35l - l^2 = 286 \]
Rearranging this into standard quadratic form, we get: \[ l^2 - 35l + 286 = 0 \]
Now we can use the quadratic formula to solve for \( l \): \[ l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -35 \), and \( c = 286 \).
Calculating the discriminant: \[ b^2 - 4ac = (-35)^2 - 4 \cdot 1 \cdot 286 = 1225 - 1144 = 81 \]
Now substituting these values into the quadratic formula: \[ l = \frac{35 \pm \sqrt{81}}{2 \cdot 1} = \frac{35 \pm 9}{2} \]
Calculating the two possible values for \( l \):
- \( l = \frac{35 + 9}{2} = \frac{44}{2} = 22 \)
- \( l = \frac{35 - 9}{2} = \frac{26}{2} = 13 \)
Thus, we have two possible dimensions:
- If \( l = 22 \), then \( w = 35 - 22 = 13 \)
- If \( l = 13 \), then \( w = 35 - 13 = 22 \)
Therefore, the dimensions of the ticket are \( 22 \) centimeters by \( 13 \) centimeters.
Final answer: \[ 22 \text{ centimeters by } 13 \text{ centimeters} \]