Solve 20-100/3x>40x

$$20- \frac{ 100 }{ 3 } x > 40x$$

Solve for x

$$x<\frac{3}{11}$$

Solution Steps

Subtract 40x from both sides.

$$20-\frac{100}{3}x-40x>0$$

Combine -\frac{100}{3}x and -40x to get -\frac{220}{3}x.

$$20-\frac{220}{3}x>0$$

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

$$-\frac{220}{3}x>-20$$

Multiply both sides by -\frac{3}{220}, the reciprocal of -\frac{220}{3}. Since -\frac{220}{3} is negative, the inequality direction is changed.

$$x<-20\left(-\frac{3}{220}\right)$$

Express -20\left(-\frac{3}{220}\right) as a single fraction.

$$x<\frac{-20\left(-3\right)}{220}$$

Multiply -20 and -3 to get 60.

$$x<\frac{60}{220}$$

Reduce the fraction \frac{60}{220} to lowest terms by extracting and canceling out 20.

$$x<\frac{3}{11}$$