Solve 8+1/3-y-1/6leq1

$$\frac{ 8+1 }{ 3 } - \frac{ y-1 }{ 6 } \leq 1$$

Solve for y

$$y\geq 13$$

Solution Steps

Multiply both sides of the equation by 6, the least common multiple of 3,6. Since 6 is positive, the inequality direction remains the same.

$$2\left(8+1\right)-\left(y-1\right)\leq 6$$

Add 8 and 1 to get 9.

$$2\times 9-\left(y-1\right)\leq 6$$

Multiply 2 and 9 to get 18.

$$18-\left(y-1\right)\leq 6$$

To find the opposite of y-1, find the opposite of each term.

$$18-y-\left(-1\right)\leq 6$$

The opposite of -1 is 1.

$$18-y+1\leq 6$$

Add 18 and 1 to get 19.

$$19-y\leq 6$$

Subtract 19 from both sides.

$$-y\leq 6-19$$

Subtract 19 from 6 to get -13.

$$-y\leq -13$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$y\geq \frac{-13}{-1}$$

Fraction \frac{-13}{-1} can be simplified to 13 by removing the negative sign from both the numerator and the denominator.

$$y\geq 13$$