1. **State the problem:** Solve the system of linear equations: $$\begin{cases} N_{N} + N_{D} = 50 \\ 5N_{N} + 10 N_{D} = 450 \end{cases}$$ 2. **Use substitution or elimination method:** Here, we use elimination. 3. Multiply the first equation by 5 to align coefficients of $N_N$: $$5(N_{N} + N_{D}) = 5 \times 50 \Rightarrow 5N_{N} + 5N_{D} = 250$$ 4. Subtract this from the second equation: $$\cancel{5N_{N}} + 10N_{D} - (\cancel{5N_{N}} + 5N_{D}) = 450 - 250$$ $$10N_{D} - 5N_{D} = 200$$ $$5N_{D} = 200$$ 5. Solve for $N_D$: $$N_{D} = \frac{200}{5} = 40$$ 6. Substitute $N_D = 40$ back into the first equation: $$N_{N} + 40 = 50$$ $$N_{N} = 50 - 40 = 10$$ 7. **Final answer:** $$N_{N} = 10, \quad N_{D} = 40$$ This means there are 10 units of $N_N$ and 40 units of $N_D$ that satisfy both equations.