# How do you solve \frac { 1} { 7} r + \frac { 53} { 56} > \frac { 6} { 7}?

Dec 4, 2016

$r > - \frac{5}{8}$

#### Explanation:

First, change each constant to have the same common denominator, in this case 56, by multiplying it by the correct for of $1$ if necessary:

$\frac{1}{7} r + \frac{53}{56} > \left(\frac{8}{8}\right) \cdot \left(\frac{6}{7}\right)$

$\frac{1}{7} r + \frac{53}{56} > \frac{48}{56}$

Now we can solve for $r$ using the necessary mathematics while keeping the inequality balanced:

$\frac{1}{7} r + \frac{53}{56} - \frac{53}{56} > \frac{48}{56} - \frac{53}{56}$

$\frac{1}{7} r + 0 > - \frac{5}{56}$

$\frac{1}{7} r > - \frac{5}{56}$

$7 \cdot \frac{1}{7} r > 7 \cdot - \frac{5}{56}$

$r > 7 \cdot - \frac{5}{8 \cdot 7}$

$r > \cancel{7} \cdot - \frac{5}{8 \cdot \cancel{7}}$

$r > - \frac{5}{8}$