Solve 1/3(x+3)<1/5(3x+2)

$$\frac { 1 } { 3 } ( x + 3 ) < \frac { 1 } { 5 } ( 3 x + 2 )$$

Solve for x

$$x>\frac{9}{4}$$

Solution Steps

Use the distributive property to multiply \frac{1}{3} by x+3.

$$\frac{1}{3}x+\frac{1}{3}\times 3<\frac{1}{5}\left(3x+2\right)$$

Cancel out 3 and 3.

$$\frac{1}{3}x+1<\frac{1}{5}\left(3x+2\right)$$

Use the distributive property to multiply \frac{1}{5} by 3x+2.

$$\frac{1}{3}x+1<\frac{1}{5}\times 3x+\frac{1}{5}\times 2$$

Multiply \frac{1}{5} and 3 to get \frac{3}{5}.

$$\frac{1}{3}x+1<\frac{3}{5}x+\frac{1}{5}\times 2$$

Multiply \frac{1}{5} and 2 to get \frac{2}{5}.

$$\frac{1}{3}x+1<\frac{3}{5}x+\frac{2}{5}$$

Subtract \frac{3}{5}x from both sides.

$$\frac{1}{3}x+1-\frac{3}{5}x<\frac{2}{5}$$

Combine \frac{1}{3}x and -\frac{3}{5}x to get -\frac{4}{15}x.

$$-\frac{4}{15}x+1<\frac{2}{5}$$

Subtract 1 from both sides.

$$-\frac{4}{15}x<\frac{2}{5}-1$$

Convert 1 to fraction \frac{5}{5}.

$$-\frac{4}{15}x<\frac{2}{5}-\frac{5}{5}$$

Since \frac{2}{5} and \frac{5}{5} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{4}{15}x<\frac{2-5}{5}$$

Subtract 5 from 2 to get -3.

$$-\frac{4}{15}x<-\frac{3}{5}$$

Multiply both sides by -\frac{15}{4}, the reciprocal of -\frac{4}{15}. Since -\frac{4}{15} is negative, the inequality direction is changed.

$$x>-\frac{3}{5}\left(-\frac{15}{4}\right)$$

Multiply -\frac{3}{5} times -\frac{15}{4} by multiplying numerator times numerator and denominator times denominator.

$$x>\frac{-3\left(-15\right)}{5\times 4}$$

Do the multiplications in the fraction \frac{-3\left(-15\right)}{5\times 4}.

$$x>\frac{45}{20}$$

Reduce the fraction \frac{45}{20} to lowest terms by extracting and canceling out 5.

$$x>\frac{9}{4}$$