Graph 5(6+3r)+7>=127

$5 (6 + 3 r) + 7 \geq 127$

Simplify $5 (6 + 3 r) + 7$ .

Simplify each term.

Apply the distributive property.

$5 \cdot 6 + 5 (3 r) + 7 \geq 127$

Multiply $5$ by $6$ .

$30 + 5 (3 r) + 7 \geq 127$

Multiply $3$ by $5$ .

$30 + 15 r + 7 \geq 127$

Add $30$ and $7$ .

$15 r + 37 \geq 127$

Move all terms not containing $r$ to the right side of the inequality.

Subtract $37$ from both sides of the inequality.

$15 r \geq 127 - 37$

Subtract $37$ from $127$ .

$15 r \geq 90$

Divide each term in $15 r \geq 90$ by $15$ and simplify.

Divide each term in $15 r \geq 90$ by $15$ .

$\frac{15 r}{15} \geq \frac{90}{15}$

Simplify the left side.

Cancel the common factor of $15$ .

Cancel the common factor.

$\frac{15 r}{15} \geq \frac{90}{15}$

Divide $r$ by $1$ .

$r \geq \frac{90}{15}$

Simplify the right side.

Divide $90$ by $15$ .

$r \geq 6$