Solve (0.1x+0.7)div0.3>(0.03x+0.02)div0.04

$$(0.1x+0.7) \div 0.3 > (0.03x+0.02) \div 0.04$$

Solve for x

$$x<\frac{22}{5}$$

Solution Steps

Divide each term of 0.1x+0.7 by 0.3 to get \frac{0.1x}{0.3}+\frac{0.7}{0.3}.

$$\frac{0.1x}{0.3}+\frac{0.7}{0.3}>\frac{0.03x+0.02}{0.04}$$

Divide 0.1x by 0.3 to get \frac{1}{3}x.

$$\frac{1}{3}x+\frac{0.7}{0.3}>\frac{0.03x+0.02}{0.04}$$

Expand \frac{0.7}{0.3} by multiplying both numerator and the denominator by 10.

$$\frac{1}{3}x+\frac{7}{3}>\frac{0.03x+0.02}{0.04}$$

Divide each term of 0.03x+0.02 by 0.04 to get \frac{0.03x}{0.04}+\frac{0.02}{0.04}.

$$\frac{1}{3}x+\frac{7}{3}>\frac{0.03x}{0.04}+\frac{0.02}{0.04}$$

Divide 0.03x by 0.04 to get 0.75x.

$$\frac{1}{3}x+\frac{7}{3}>0.75x+\frac{0.02}{0.04}$$

Expand \frac{0.02}{0.04} by multiplying both numerator and the denominator by 100.

$$\frac{1}{3}x+\frac{7}{3}>0.75x+\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{1}{3}x+\frac{7}{3}>0.75x+\frac{1}{2}$$

Subtract 0.75x from both sides.

$$\frac{1}{3}x+\frac{7}{3}-0.75x>\frac{1}{2}$$

Combine \frac{1}{3}x and -0.75x to get -\frac{5}{12}x.

$$-\frac{5}{12}x+\frac{7}{3}>\frac{1}{2}$$

Subtract \frac{7}{3} from both sides.

$$-\frac{5}{12}x>\frac{1}{2}-\frac{7}{3}$$

Least common multiple of 2 and 3 is 6. Convert \frac{1}{2} and \frac{7}{3} to fractions with denominator 6.

$$-\frac{5}{12}x>\frac{3}{6}-\frac{14}{6}$$

Since \frac{3}{6} and \frac{14}{6} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{5}{12}x>\frac{3-14}{6}$$

Subtract 14 from 3 to get -11.

$$-\frac{5}{12}x>-\frac{11}{6}$$

Divide both sides by -\frac{5}{12}. Since -\frac{5}{12} is negative, the inequality direction is changed.

$$x<\frac{-\frac{11}{6}}{-\frac{5}{12}}$$

Express \frac{-\frac{11}{6}}{-\frac{5}{12}} as a single fraction.

$$x<\frac{-11}{6\left(-\frac{5}{12}\right)}$$

Multiply 6 and -\frac{5}{12} to get -2.5.

$$x<\frac{-11}{-2.5}$$

Expand \frac{-11}{-2.5} by multiplying both numerator and the denominator by 10.

$$x<\frac{-110}{-25}$$

Reduce the fraction \frac{-110}{-25} to lowest terms by extracting and canceling out -5.

$$x<\frac{22}{5}$$