Solve 45x-3/-45leq0

$$\frac { 45 x - 3 } { - 45 } \leq 0$$

Solve for x

$$x\geq \frac{1}{15}$$

Solution Steps

Multiply both sides by -45. Since -45 is negative, the inequality direction is changed. Anything times zero gives zero.

$$45x-3\geq 0$$

Add 3 to both sides. Anything plus zero gives itself.

$$45x\geq 3$$

Divide both sides by 45. Since 45 is positive, the inequality direction remains the same.

$$x\geq \frac{3}{45}$$

Reduce the fraction \frac{3}{45} to lowest terms by extracting and canceling out 3.

$$x\geq \frac{1}{15}$$