Solve 0.2m-0.1/0.3-0.9m+2/0.6leq0.2

$$\frac { 0.2 m - 0.1 } { 0.3 } - \frac { 0.9 m + 2 } { 0.6 } \leq 0.2$$

Solve for m

$$m\geq -\frac{116}{25}$$

Solution Steps

Divide each term of 0.2m-0.1 by 0.3 to get \frac{0.2m}{0.3}+\frac{-0.1}{0.3}.

$$\frac{0.2m}{0.3}+\frac{-0.1}{0.3}-\frac{0.9m+2}{0.6}\leq 0.2$$

Divide 0.2m by 0.3 to get \frac{2}{3}m.

$$\frac{2}{3}m+\frac{-0.1}{0.3}-\frac{0.9m+2}{0.6}\leq 0.2$$

Expand \frac{-0.1}{0.3} by multiplying both numerator and the denominator by 10.

$$\frac{2}{3}m+\frac{-1}{3}-\frac{0.9m+2}{0.6}\leq 0.2$$

Fraction \frac{-1}{3} can be rewritten as -\frac{1}{3} by extracting the negative sign.

$$\frac{2}{3}m-\frac{1}{3}-\frac{0.9m+2}{0.6}\leq 0.2$$

Divide each term of 0.9m+2 by 0.6 to get \frac{0.9m}{0.6}+\frac{2}{0.6}.

$$\frac{2}{3}m-\frac{1}{3}-\left(\frac{0.9m}{0.6}+\frac{2}{0.6}\right)\leq 0.2$$

Divide 0.9m by 0.6 to get 1.5m.

$$\frac{2}{3}m-\frac{1}{3}-\left(1.5m+\frac{2}{0.6}\right)\leq 0.2$$

Expand \frac{2}{0.6} by multiplying both numerator and the denominator by 10.

$$\frac{2}{3}m-\frac{1}{3}-\left(1.5m+\frac{20}{6}\right)\leq 0.2$$

Reduce the fraction \frac{20}{6} to lowest terms by extracting and canceling out 2.

$$\frac{2}{3}m-\frac{1}{3}-\left(1.5m+\frac{10}{3}\right)\leq 0.2$$

To find the opposite of 1.5m+\frac{10}{3}, find the opposite of each term.

$$\frac{2}{3}m-\frac{1}{3}-1.5m-\frac{10}{3}\leq 0.2$$

Combine \frac{2}{3}m and -1.5m to get -\frac{5}{6}m.

$$-\frac{5}{6}m-\frac{1}{3}-\frac{10}{3}\leq 0.2$$

Since -\frac{1}{3} and \frac{10}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{5}{6}m+\frac{-1-10}{3}\leq 0.2$$

Subtract 10 from -1 to get -11.

$$-\frac{5}{6}m-\frac{11}{3}\leq 0.2$$

Add \frac{11}{3} to both sides.

$$-\frac{5}{6}m\leq 0.2+\frac{11}{3}$$

Convert decimal number 0.2 to fraction \frac{2}{10}. Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.

$$-\frac{5}{6}m\leq \frac{1}{5}+\frac{11}{3}$$

Least common multiple of 5 and 3 is 15. Convert \frac{1}{5} and \frac{11}{3} to fractions with denominator 15.

$$-\frac{5}{6}m\leq \frac{3}{15}+\frac{55}{15}$$

Since \frac{3}{15} and \frac{55}{15} have the same denominator, add them by adding their numerators.

$$-\frac{5}{6}m\leq \frac{3+55}{15}$$

Add 3 and 55 to get 58.

$$-\frac{5}{6}m\leq \frac{58}{15}$$

Divide both sides by -\frac{5}{6}. Since -\frac{5}{6} is negative, the inequality direction is changed.

$$m\geq \frac{\frac{58}{15}}{-\frac{5}{6}}$$

Express \frac{\frac{58}{15}}{-\frac{5}{6}} as a single fraction.

$$m\geq \frac{58}{15\left(-\frac{5}{6}\right)}$$

Multiply 15 and -\frac{5}{6} to get -12.5.

$$m\geq \frac{58}{-12.5}$$

Expand \frac{58}{-12.5} by multiplying both numerator and the denominator by 10.

$$m\geq \frac{580}{-125}$$

Reduce the fraction \frac{580}{-125} to lowest terms by extracting and canceling out 5.

$$m\geq -\frac{116}{25}$$