Solve 4x+40leq5.5x

$$4x+40 \leq 5.5x$$

Solve for x

$$x\geq \frac{80}{3}$$

Solution Steps

Subtract 5.5x from both sides.

$$4x+40-5.5x\leq 0$$

Combine 4x and -5.5x to get -1.5x.

$$-1.5x+40\leq 0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$-1.5x\leq -40$$

Divide both sides by -1.5. Since -1.5 is negative, the inequality direction is changed.

$$x\geq \frac{-40}{-1.5}$$

Expand \frac{-40}{-1.5} by multiplying both numerator and the denominator by 10.

$$x\geq \frac{-400}{-15}$$

Reduce the fraction \frac{-400}{-15} to lowest terms by extracting and canceling out -5.

$$x\geq \frac{80}{3}$$