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

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Answer 1 · 2016-12-04T14:34:56

$r > -5/8$

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:

$1/7r + 53/56 > (8/8)*(6/7)$

$1/7r + 53/56 > 48/56$

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

$1/7r + 53/56 - 53/56 > 48/56 - 53/56$

$1/7r + 0 > -5/56$

$1/7r > -5/56$

$7 * 1/7r > 7 * -5/56$

$r > 7 * -5/(8*7)$

$r > cancel(7) * -5/(8*cancel(7))$

$r > -5/8$

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