# How do you solve 3(2-b)<10-3(b-6)?

Sep 2, 2017

The inequality is satisfied for all real b.
$\left\{b \in \mathbb{R}\right\}$

#### Explanation:

After rearranging:

$3 \left(2 - b\right) < 10 - 3 \left(b - 6\right) \implies - 22 - 3 b < - 3 b$

Adding $3 b$ to both sides :

$- 22 < - 3 b + 3 b$

for any real $b$

$- 3 b + 3 b = 0$

$- 22 < 0$ is always true.

Hope this helps.