# How do you solve the inequality (x +4) + (x+6)> 3x - 1?

Feb 4, 2016

$x < 11$

#### Explanation:

1. Add like terms.

$\left(x + 4\right) + \left(x + 6\right) > 3 x - 1$

$2 x + 10 > 3 x - 1$

2. Subtract 10 from both sides.

$2 x + 10$ $\textcolor{red}{- 10} > 3 x - 1$ $\textcolor{red}{- 10}$

$2 x > 3 x - 11$

3. Subtract 3x from both sides.

$2 x$ $\textcolor{red}{- 3 x} > 3 x$ $\textcolor{red}{- 3 x} - 11$

$- x$$>$$- 11$

4. Divide both sides by -1.

$- x \textcolor{red}{\div - 1}$$>$$- 11 \textcolor{red}{\div - 1}$

$x > 11$

5. Flip the inequality sign.
Whenever you divide by a negative number, you must flip the inequality sign, even if you divide a negative number by another negative number.

$\textcolor{g r e e n}{x < 11}$