Solve 1/2y-1/8>3/4+6/5y

$$\frac { 1 } { 2 } y - \frac { 1 } { 8 } > \frac { 3 } { 4 } + \frac { 6 } { 5 } y$$

Solve for y

$$y<-\frac{5}{4}$$

Solution Steps

Subtract \frac{6}{5}y from both sides.

$$\frac{1}{2}y-\frac{1}{8}-\frac{6}{5}y>\frac{3}{4}$$

Combine \frac{1}{2}y and -\frac{6}{5}y to get -\frac{7}{10}y.

$$-\frac{7}{10}y-\frac{1}{8}>\frac{3}{4}$$

Add \frac{1}{8} to both sides.

$$-\frac{7}{10}y>\frac{3}{4}+\frac{1}{8}$$

Least common multiple of 4 and 8 is 8. Convert \frac{3}{4} and \frac{1}{8} to fractions with denominator 8.

$$-\frac{7}{10}y>\frac{6}{8}+\frac{1}{8}$$

Since \frac{6}{8} and \frac{1}{8} have the same denominator, add them by adding their numerators.

$$-\frac{7}{10}y>\frac{6+1}{8}$$

Add 6 and 1 to get 7.

$$-\frac{7}{10}y>\frac{7}{8}$$

Multiply both sides by -\frac{10}{7}, the reciprocal of -\frac{7}{10}. Since -\frac{7}{10} is negative, the inequality direction is changed.

$$y<\frac{7}{8}\left(-\frac{10}{7}\right)$$

Multiply \frac{7}{8} times -\frac{10}{7} by multiplying numerator times numerator and denominator times denominator.

$$y<\frac{7\left(-10\right)}{8\times 7}$$

Cancel out 7 in both numerator and denominator.

$$y<\frac{-10}{8}$$

Reduce the fraction \frac{-10}{8} to lowest terms by extracting and canceling out 2.

$$y<-\frac{5}{4}$$