Solve ngeqn*(350/350+n-1)

$$n \geq n * ( 350 / 350 + n - 1 )$$

Solve for n

$$n\in \begin{bmatrix}0,1\end{bmatrix}$$

Solution Steps

Divide 350 by 350 to get 1.

$$n\geq n\left(1+n-1\right)$$

Subtract 1 from 1 to get 0.

$$n\geq nn$$

Multiply n and n to get n^{2}.

$$n\geq n^{2}$$

Subtract n^{2} from both sides.

$$n-n^{2}\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in n-n^{2} positive. Since -1 is negative, the inequality direction is changed.

$$-n+n^{2}\leq 0$$

Factor out n.

$$n\left(n-1\right)\leq 0$$

For the product to be ≤0, one of the values n and n-1 has to be ≥0 and the other has to be ≤0. Consider the case when n\geq 0 and n-1\leq 0.

$$n\geq 0 n-1\leq 0$$

The solution satisfying both inequalities is n\in \left[0,1\right].

$$n\in \begin{bmatrix}0,1\end{bmatrix}$$

Consider the case when n\leq 0 and n-1\geq 0.

$$n-1\geq 0 n\leq 0$$

This is false for any n.

$$n\in \emptyset$$

The final solution is the union of the obtained solutions.

$$n\in \begin{bmatrix}0,1\end{bmatrix}$$