# How do you solve -3(r-11) + 15 ≥ 9?

Apr 29, 2015

First work out the brackets:

$- 3 \cdot r - 3 \cdot \left(- 11\right) + 15 \ge 9 \to$

$- 3 r + 33 + 15 \ge 9 \to$

Remember, adding and subtracting may allways be done, so we add $3 r$ to both sides, and we subtract $9$:

$33 + 15 - 9 \ge 3 r \to 39 \ge 3 r \to$

Divide by $3$:

$13 \ge r \to r \le 13$