Step 1: Start from the general distortion energy formula.
For any three principal stresses $\sigma_1, \sigma_2, \sigma_3$, the von Mises equivalent stress is $\sigma_{eq}^2 = \frac{1}{2}\left[(\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2\right]$.
For this plate the stress is biaxial, so $\sigma_3 = 0$, $\sigma_1 = \sigma$, $\sigma_2 = -2\sigma$.
Step 2: Plug the three principal stresses into the formula.
$(\sigma_1-\sigma_2)^2 = (\sigma-(-2\sigma))^2 = (3\sigma)^2 = 9\sigma^2$.
$(\sigma_2-\sigma_3)^2 = (-2\sigma-0)^2 = 4\sigma^2$.
$(\sigma_3-\sigma_1)^2 = (0-\sigma)^2 = \sigma^2$.
Add these: $9\sigma^2+4\sigma^2+\sigma^2 = 14\sigma^2$.
Step 3: Get the equivalent stress.
$\sigma_{eq}^2 = \dfrac{14\sigma^2}{2} = 7\sigma^2$, so $\sigma_{eq} = \sigma\sqrt{7}$, the same value the direct biaxial formula gives.
Step 4: Bring in the allowable stress and factor of safety.
The plate must not yield, so the design limit is $\sigma_{eq} \le \dfrac{S_{yt}}{n}$ with $S_{yt} = 550$ MPa and $n = 2$, giving $\sigma_{eq} \le 275$ MPa.
Setting $\sigma\sqrt{7} = 275$ gives $\sigma = 275/\sqrt{7} = 275 \times 0.37796 = 103.94$ MPa.
Final Answer:
Both routes land on the same safe stress limit for the plate.
\[ \boxed{\sigma \approx 103.9 \text{ MPa}} \]