# How do you solve 10 >29 - 3b?

Jun 22, 2016

$\frac{19}{3} < b$ or $6 \frac{1}{3} < b$

#### Explanation:

To solve $10 > 29 - 3 b$

Begin by isolating the term with the variable by additive inverse

$10 - 29 > \cancel{29} \cancel{- 29} - 3 b$

$- 19 > \left(- 3 b\right)$

Isolate the variable by multiplicative inverse

$- \frac{19}{-} 3 > \frac{\cancel{- 3} b}{\cancel{- 3}}$

When you divide by the negative you must flip the inequality sign

$\frac{19}{3} < b$ or $6 \frac{1}{3} < b$