Solve -9/8y-1leq5/6y+3/8

$$- \frac { 9 } { 8 } y - 1 \leq \frac { 5 } { 6 } y + \frac { 3 } { 8 }$$

Solve for y

$$y\geq -\frac{33}{47}$$

Solution Steps

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

$$-\frac{9}{8}y-1-\frac{5}{6}y\leq \frac{3}{8}$$

Combine -\frac{9}{8}y and -\frac{5}{6}y to get -\frac{47}{24}y.

$$-\frac{47}{24}y-1\leq \frac{3}{8}$$

Add 1 to both sides.

$$-\frac{47}{24}y\leq \frac{3}{8}+1$$

Convert 1 to fraction \frac{8}{8}.

$$-\frac{47}{24}y\leq \frac{3}{8}+\frac{8}{8}$$

Since \frac{3}{8} and \frac{8}{8} have the same denominator, add them by adding their numerators.

$$-\frac{47}{24}y\leq \frac{3+8}{8}$$

Add 3 and 8 to get 11.

$$-\frac{47}{24}y\leq \frac{11}{8}$$

Multiply both sides by -\frac{24}{47}, the reciprocal of -\frac{47}{24}. Since -\frac{47}{24} is negative, the inequality direction is changed.

$$y\geq \frac{11}{8}\left(-\frac{24}{47}\right)$$

Multiply \frac{11}{8} times -\frac{24}{47} by multiplying numerator times numerator and denominator times denominator.

$$y\geq \frac{11\left(-24\right)}{8\times 47}$$

Do the multiplications in the fraction \frac{11\left(-24\right)}{8\times 47}.

$$y\geq \frac{-264}{376}$$

Reduce the fraction \frac{-264}{376} to lowest terms by extracting and canceling out 8.

$$y\geq -\frac{33}{47}$$