Solve 6+0.1xleq0.12x

$$6 + 0.1 x \leq 0.12 x$$

Solve for x

$$x\geq 300$$

Solution Steps

Subtract 0.12x from both sides.

$$6+0.1x-0.12x\leq 0$$

Combine 0.1x and -0.12x to get -0.02x.

$$6-0.02x\leq 0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-0.02x\leq -6$$

Divide both sides by -0.02. Since -0.02 is negative, the inequality direction is changed.

$$x\geq \frac{-6}{-0.02}$$

Expand \frac{-6}{-0.02} by multiplying both numerator and the denominator by 100.

$$x\geq \frac{-600}{-2}$$

Divide -600 by -2 to get 300.

$$x\geq 300$$