Solve 05leq5n-3/9n-3

$$05 \leq \frac { 5 n - 3 } { 9 n - 3 }$$

Solve for n

$$n\in (-\infty,\frac{1}{3})\cup [\frac{3}{5},\infty)$$

Solution Steps

Multiply 0 and 5 to get 0.

$$0\leq \frac{5n-3}{9n-3}$$

Swap sides so that all variable terms are on the left hand side. This changes the sign direction.

$$\frac{5n-3}{9n-3}\geq 0$$

For the quotient to be ≥0, 5n-3 and 9n-3 have to be both ≤0 or both ≥0, and 9n-3 cannot be zero. Consider the case when 5n-3\leq 0 and 9n-3 is negative.

$$5n-3\leq 0 9n-3<0$$

The solution satisfying both inequalities is n<\frac{1}{3}.

$$n<\frac{1}{3}$$

Consider the case when 5n-3\geq 0 and 9n-3 is positive.

$$5n-3\geq 0 9n-3>0$$

The solution satisfying both inequalities is n\geq \frac{3}{5}.

$$n\geq \frac{3}{5}$$

The final solution is the union of the obtained solutions.

$$n<\frac{1}{3}\text{; }n\geq \frac{3}{5}$$