Solve 3/2y-1geq8y-7/6

$$\frac { 3 } { 2 } y - 1 \geq 8 y - \frac { 7 } { 6 }$$

Solve for y

$$y\leq \frac{1}{39}$$

Solution Steps

Subtract 8y from both sides.

$$\frac{3}{2}y-1-8y\geq -\frac{7}{6}$$

Combine \frac{3}{2}y and -8y to get -\frac{13}{2}y.

$$-\frac{13}{2}y-1\geq -\frac{7}{6}$$

Add 1 to both sides.

$$-\frac{13}{2}y\geq -\frac{7}{6}+1$$

Convert 1 to fraction \frac{6}{6}.

$$-\frac{13}{2}y\geq -\frac{7}{6}+\frac{6}{6}$$

Since -\frac{7}{6} and \frac{6}{6} have the same denominator, add them by adding their numerators.

$$-\frac{13}{2}y\geq \frac{-7+6}{6}$$

Add -7 and 6 to get -1.

$$-\frac{13}{2}y\geq -\frac{1}{6}$$

Multiply both sides by -\frac{2}{13}, the reciprocal of -\frac{13}{2}. Since -\frac{13}{2} is negative, the inequality direction is changed.

$$y\leq -\frac{1}{6}\left(-\frac{2}{13}\right)$$

Multiply -\frac{1}{6} times -\frac{2}{13} by multiplying numerator times numerator and denominator times denominator.

$$y\leq \frac{-\left(-2\right)}{6\times 13}$$

Do the multiplications in the fraction \frac{-\left(-2\right)}{6\times 13}.

$$y\leq \frac{2}{78}$$

Reduce the fraction \frac{2}{78} to lowest terms by extracting and canceling out 2.

$$y\leq \frac{1}{39}$$