Solve 20+0.1xleq0.2x

$$20 + 0.1 x \leq 0.2 x$$

Solve for x

$$x\geq 200$$

Solution Steps

Subtract 0.2x from both sides.

$$20+0.1x-0.2x\leq 0$$

Combine 0.1x and -0.2x to get -0.1x.

$$20-0.1x\leq 0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-0.1x\leq -20$$

Divide both sides by -0.1. Since -0.1 is negative, the inequality direction is changed.

$$x\geq \frac{-20}{-0.1}$$

Expand \frac{-20}{-0.1} by multiplying both numerator and the denominator by 10.

$$x\geq \frac{-200}{-1}$$

Fraction \frac{-200}{-1} can be simplified to 200 by removing the negative sign from both the numerator and the denominator.

$$x\geq 200$$