Solve 8+6/5yleq8/5y+12

$$8 + \frac { 6 } { 5 } y \leq \frac { 8 } { 5 } y + 12$$

Solve for y

$$y\geq -10$$

Solution Steps

Subtract \frac{8}{5}y from both sides.

$$8+\frac{6}{5}y-\frac{8}{5}y\leq 12$$

Combine \frac{6}{5}y and -\frac{8}{5}y to get -\frac{2}{5}y.

$$8-\frac{2}{5}y\leq 12$$

Subtract 8 from both sides.

$$-\frac{2}{5}y\leq 12-8$$

Subtract 8 from 12 to get 4.

$$-\frac{2}{5}y\leq 4$$

Multiply both sides by -\frac{5}{2}, the reciprocal of -\frac{2}{5}. Since -\frac{2}{5} is negative, the inequality direction is changed.

$$y\geq 4\left(-\frac{5}{2}\right)$$

Express 4\left(-\frac{5}{2}\right) as a single fraction.

$$y\geq \frac{4\left(-5\right)}{2}$$

Multiply 4 and -5 to get -20.

$$y\geq \frac{-20}{2}$$

Divide -20 by 2 to get -10.

$$y\geq -10$$