Solve 900*6+99x<540*3x+80*3x

$$900 \times 6+99x < 540 \times 3x+80 \times 3x$$

Solve for x

$$x>\frac{1800}{587}$$

Solution Steps

Multiply 900 and 6 to get 5400.

$$5400+99x<540\times 3x+80\times 3x$$

Multiply 540 and 3 to get 1620. Multiply 80 and 3 to get 240.

$$5400+99x<1620x+240x$$

Combine 1620x and 240x to get 1860x.

$$5400+99x<1860x$$

Subtract 1860x from both sides.

$$5400+99x-1860x<0$$

Combine 99x and -1860x to get -1761x.

$$5400-1761x<0$$

Subtract 5400 from both sides. Anything subtracted from zero gives its negation.

$$-1761x<-5400$$

Divide both sides by -1761. Since -1761 is negative, the inequality direction is changed.

$$x>\frac{-5400}{-1761}$$

Reduce the fraction \frac{-5400}{-1761} to lowest terms by extracting and canceling out -3.

$$x>\frac{1800}{587}$$