Solve 18x+42geq12x-3

$$18x+42 \geq 12x-3$$

Solve for x

$$x\geq -\frac{15}{2}$$

Solution Steps

Subtract 12x from both sides.

$$18x+42-12x\geq -3$$

Combine 18x and -12x to get 6x.

$$6x+42\geq -3$$

Subtract 42 from both sides.

$$6x\geq -3-42$$

Subtract 42 from -3 to get -45.

$$6x\geq -45$$

Divide both sides by 6. Since 6 is positive, the inequality direction remains the same.

$$x\geq \frac{-45}{6}$$

Reduce the fraction \frac{-45}{6} to lowest terms by extracting and canceling out 3.

$$x\geq -\frac{15}{2}$$