Solve 1/3(x+1)-1>1/5(x+4)

$$\frac { 1 } { 3 } ( x + 1 ) - 1 > \frac { 1 } { 5 } ( x + 4 )$$

Solve for x

$$x>11$$

Solution Steps

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

$$\frac{1}{3}x+\frac{1}{3}-1>\frac{1}{5}\left(x+4\right)$$

Convert 1 to fraction \frac{3}{3}.

$$\frac{1}{3}x+\frac{1}{3}-\frac{3}{3}>\frac{1}{5}\left(x+4\right)$$

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

$$\frac{1}{3}x+\frac{1-3}{3}>\frac{1}{5}\left(x+4\right)$$

Subtract 3 from 1 to get -2.

$$\frac{1}{3}x-\frac{2}{3}>\frac{1}{5}\left(x+4\right)$$

Use the distributive property to multiply \frac{1}{5} by x+4.

$$\frac{1}{3}x-\frac{2}{3}>\frac{1}{5}x+\frac{1}{5}\times 4$$

Multiply \frac{1}{5} and 4 to get \frac{4}{5}.

$$\frac{1}{3}x-\frac{2}{3}>\frac{1}{5}x+\frac{4}{5}$$

Subtract \frac{1}{5}x from both sides.

$$\frac{1}{3}x-\frac{2}{3}-\frac{1}{5}x>\frac{4}{5}$$

Combine \frac{1}{3}x and -\frac{1}{5}x to get \frac{2}{15}x.

$$\frac{2}{15}x-\frac{2}{3}>\frac{4}{5}$$

Add \frac{2}{3} to both sides.

$$\frac{2}{15}x>\frac{4}{5}+\frac{2}{3}$$

Least common multiple of 5 and 3 is 15. Convert \frac{4}{5} and \frac{2}{3} to fractions with denominator 15.

$$\frac{2}{15}x>\frac{12}{15}+\frac{10}{15}$$

Since \frac{12}{15} and \frac{10}{15} have the same denominator, add them by adding their numerators.

$$\frac{2}{15}x>\frac{12+10}{15}$$

Add 12 and 10 to get 22.

$$\frac{2}{15}x>\frac{22}{15}$$

Multiply both sides by \frac{15}{2}, the reciprocal of \frac{2}{15}. Since \frac{2}{15} is positive, the inequality direction remains the same.

$$x>\frac{22}{15}\times \frac{15}{2}$$

Multiply \frac{22}{15} times \frac{15}{2} by multiplying numerator times numerator and denominator times denominator.

$$x>\frac{22\times 15}{15\times 2}$$

Cancel out 15 in both numerator and denominator.

$$x>\frac{22}{2}$$

Divide 22 by 2 to get 11.

$$x>11$$